Chapter 12 Linear Programming

Given:

Objective function: Maximise $Z = 4x + y$.

Constraints:

---

To Solve:

Solve the given linear programming problem graphically by finding the maximum value of $Z$.

---

Solution:

To solve the problem graphically, we first consider the given inequalities as equations to draw the corresponding lines.

For the inequality $x + y \leq 50$, we consider the line $x + y = 50$.

When $x=0$, $y=50$. The point is $(0, 50)$.

When $y=0$, $x=50$. The point is $(50, 0)$.

For the inequality $3x + y \leq 90$, we consider the line $3x + y = 90$.

When $x=0$, $y=90$. The point is $(0, 90)$.

When $y=0$, $3x = 90$, so $x=30$. The point is $(30, 0)$.

The constraints $x \geq 0$ and $y \geq 0$ indicate that the feasible region lies in the first quadrant.

Now we determine the feasible region defined by the inequalities.

For $x + y \leq 50$, testing the origin $(0,0)$: $0 + 0 \leq 50$ (True). The region is towards the origin from the line $x+y=50$.

For $3x + y \leq 90$, testing the origin $(0,0)$: $3(0) + 0 \leq 90$ (True). The region is towards the origin from the line $3x+y=90$.

The feasible region is the area in the first quadrant that is below or on the line $x+y=50$ and below or on the line $3x+y=90$. This region is a convex polygon.

The vertices (corner points) of the feasible region are:

1. The origin: $(0, 0)$.

2. The intersection of the line $3x+y=90$ with the x-axis ($y=0$): $3x + 0 = 90 \implies x = 30$. Point $(30, 0)$.

3. The intersection of the line $x+y=50$ with the y-axis ($x=0$): $0 + y = 50 \implies y = 50$. Point $(0, 50)$.

4. The intersection of the lines $x+y=50$ and $3x+y=90$. We solve the system of equations:

$x + y = 50$

$3x + y = 90$

Subtracting the first equation from the second, we get:

$(3x + y) - (x + y) = 90 - 50$

$2x = 40$

$x = 20$

Substituting $x=20$ into the first equation $x + y = 50$:

$20 + y = 50$

$y = 30$

The intersection point is $(20, 30)$.

The vertices of the feasible region are $(0, 0)$, $(30, 0)$, $(20, 30)$, and $(0, 50)$.

Now we evaluate the objective function $Z = 4x + y$ at each vertex:

Vertex (x, y)	Value of $Z = 4x + y$
$(0, 0)$	$Z = 4(0) + 0 = 0$
$(30, 0)$	$Z = 4(30) + 0 = 120$
$(20, 30)$	$Z = 4(20) + 30 = 80 + 30 = 110$
$(0, 50)$	$Z = 4(0) + 50 = 50$

The maximum value of $Z$ among the vertices is $120$, which occurs at the point $(30, 0)$.

---

Final Answer:

The maximum value of $Z$ is $120$, and it occurs at the point $(30, 0)$.

Given:

Objective function: Minimise $Z = 200x + 500y$.

Constraints:

---

To Solve:

Solve the given linear programming problem graphically by finding the minimum value of $Z$.

---

Solution:

To solve the problem graphically, we first consider the given inequalities as equations to draw the corresponding lines.

For the inequality $x + 2y \geq 10$, we consider the line $x + 2y = 10$.

When $x=0$, $2y=10 \implies y=5$. The point is $(0, 5)$.

When $y=0$, $x=10$. The point is $(10, 0)$.

For the inequality $3x + 4y \leq 24$, we consider the line $3x + 4y = 24$.

When $x=0$, $4y=24 \implies y=6$. The point is $(0, 6)$.

When $y=0$, $3x=24 \implies x=8$. The point is $(8, 0)$.

The constraints $x \geq 0$ and $y \geq 0$ indicate that the feasible region lies in the first quadrant.

Now we determine the feasible region defined by the inequalities.

For $x + 2y \geq 10$, testing the origin $(0,0)$: $0 + 2(0) \geq 10 \implies 0 \geq 10$ (False). The region is away from the origin from the line $x+2y=10$.

For $3x + 4y \leq 24$, testing the origin $(0,0)$: $3(0) + 4(0) \leq 24 \implies 0 \leq 24$ (True). The region is towards the origin from the line $3x+4y=24$.

The feasible region is the area in the first quadrant that is above or on the line $x+2y=10$ and below or on the line $3x+4y=24$.

The vertices (corner points) of the feasible region are the points of intersection of the boundary lines within the first quadrant.

1. The intersection of $x+2y=10$ and the y-axis ($x=0$): $0 + 2y = 10 \implies y = 5$. Point $(0, 5)$.

2. The intersection of $3x+4y=24$ and the y-axis ($x=0$): $3(0) + 4y = 24 \implies y = 6$. Point $(0, 6)$.

3. The intersection of $3x+4y=24$ and the x-axis ($y=0$): $3x + 4(0) = 24 \implies x = 8$. Point $(8, 0)$.

4. The intersection of $x+2y=10$ and $3x+4y=24$. We solve the system of equations:

$x + 2y = 10$

$3x + 4y = 24$

Multiply the first equation by 2:

$2(x + 2y) = 2(10) \implies 2x + 4y = 20$

Now subtract this new equation from the second original equation:

$(3x + 4y) - (2x + 4y) = 24 - 20$

$x = 4$

Substitute $x=4$ into the equation $x + 2y = 10$:

$4 + 2y = 10$

$2y = 6$

$y = 3$

The intersection point is $(4, 3)$.

The vertices of the feasible region are $(0, 5)$, $(0, 6)$, $(8, 0)$, and $(4, 3)$. However, we must consider the region bounded by $x \geq 0, y \geq 0$, $x+2y \geq 10$ and $3x+4y \leq 24$. Plotting these lines and regions shows that the feasible region is the area bounded by the points $(0, 5)$, $(0, 6)$, $(4, 3)$, and $(8, 0)$. (The point $(0,6)$ is on the line $3x+4y=24$ but $0+2(6)=12 \geq 10$. The point $(10,0)$ is on $x+2y=10$ but $3(10)+4(0)=30 \not\leq 24$. So $(10,0)$ is not in the feasible region. The point $(0,5)$ is on $x+2y=10$ and $3(0)+4(5)=20 \leq 24$. The point $(8,0)$ is on $3x+4y=24$ and $8+2(0)=8 \not\geq 10$, so $(8,0)$ is not in the feasible region.)

Let's re-evaluate the feasible region boundaries. The region is in the first quadrant. It must be above $x+2y=10$ and below $3x+4y=24$.

Line 1: $x+2y=10$. Points $(10,0), (0,5)$. Region above or on the line.

Line 2: $3x+4y=24$. Points $(8,0), (0,6)$. Region below or on the line.

First quadrant: $x \geq 0, y \geq 0$.

The vertices of the bounded feasible region are the intersection points that satisfy all constraints.

Intersection of $x=0$ and $x+2y=10$: $(0, 5)$. This satisfies $3(0)+4(5) = 20 \leq 24$. So $(0,5)$ is a vertex.

Intersection of $y=0$ and $3x+4y=24$: $(8, 0)$. This satisfies $8+2(0) = 8 \not\geq 10$. So $(8,0)$ is not a vertex of the feasible region.

Intersection of $x+2y=10$ and $3x+4y=24$: $(4, 3)$. This satisfies $4 \geq 0, 3 \geq 0$. So $(4,3)$ is a vertex.

Intersection of $x=0$ and $3x+4y=24$: $(0, 6)$. This satisfies $0+2(6) = 12 \geq 10$. So $(0,6)$ is a vertex.

Intersection of $y=0$ and $x+2y=10$: $(10, 0)$. This satisfies $3(10)+4(0)=30 \not\leq 24$. So $(10,0)$ is not a vertex of the feasible region.

The vertices of the feasible region are $(0, 5)$, $(4, 3)$, and $(0, 6)$.

Now we evaluate the objective function $Z = 200x + 500y$ at each vertex:

Vertex (x, y)	Value of $Z = 200x + 500y$
$(0, 5)$	$Z = 200(0) + 500(5) = 0 + 2500 = 2500$
$(4, 3)$	$Z = 200(4) + 500(3) = 800 + 1500 = 2300$
$(0, 6)$	$Z = 200(0) + 500(6) = 0 + 3000 = 3000$

The minimum value of $Z$ among the vertices is $2300$, which occurs at the point $(4, 3)$.

---

Final Answer:

The minimum value of $Z$ is $2300$, and it occurs at the point $(4, 3)$.

Given:

Objective function: Minimise and Maximise $Z = 3x + 9y$.

Constraints:

---

To Solve:

Solve the given linear programming problem graphically by finding the minimum and maximum values of $Z$.

---

Solution:

To solve the problem graphically, we first consider the given inequalities as equations to draw the corresponding lines.

For the inequality $x + 3y \leq 60$, consider the line $x + 3y = 60$.

When $x=0$, $3y=60 \implies y=20$. Point: $(0, 20)$.

When $y=0$, $x=60$. Point: $(60, 0)$.

For the inequality $x + y \geq 10$, consider the line $x + y = 10$.

When $x=0$, $y=10$. Point: $(0, 10)$.

When $y=0$, $x=10$. Point: $(10, 0)$.

For the inequality $x \leq y$, consider the line $x = y$ or $y - x = 0$.

Points on this line are $(0, 0)$, $(10, 10)$, $(20, 20)$, etc.

The constraints $x \geq 0$ and $y \geq 0$ indicate that the feasible region lies in the first quadrant.

Now we determine the feasible region defined by the inequalities:

$\bullet$ For $x + 3y \leq 60$, test $(0,0)$: $0 + 3(0) \leq 60 \implies 0 \leq 60$ (True). The region is towards the origin from the line $x+3y=60$.

$\bullet$ For $x + y \geq 10$, test $(0,0)$: $0 + 0 \geq 10 \implies 0 \geq 10$ (False). The region is away from the origin from the line $x+y=10$.

$\bullet$ For $x \leq y$ (or $y-x \geq 0$), test $(10,0)$: $0 - 10 \geq 0 \implies -10 \geq 0$ (False). The region is on the side of the line $y=x$ that contains points where $y$ is greater than or equal to $x$, which is above or on the line $y=x$.

The feasible region is the area in the first quadrant satisfying all these conditions. It is a convex polygon.

The vertices (corner points) of the feasible region are the intersection points of the boundary lines that lie within the feasible region:

1. Intersection of $x = y$ and $x + y = 10$:

Substitute $x=y$ into the second equation: $y + y = 10 \implies 2y = 10 \implies y = 5$. Since $x=y$, $x = 5$. Point is $(5, 5)$.

2. Intersection of $x = y$ and $x + 3y = 60$:

Substitute $x=y$ into the second equation: $y + 3y = 60 \implies 4y = 60 \implies y = 15$. Since $x=y$, $x = 15$. Point is $(15, 15)$.

3. Intersection of $x = 0$ (y-axis) and $x + y = 10$:

Substitute $x=0$ into the second equation: $0 + y = 10 \implies y = 10$. Point is $(0, 10)$. (This point satisfies $0 \leq 10$ and $0+3(10) = 30 \leq 60$).

4. Intersection of $x = 0$ (y-axis) and $x + 3y = 60$:

Substitute $x=0$ into the second equation: $0 + 3y = 60 \implies y = 20$. Point is $(0, 20)$. (This point satisfies $0+20 = 20 \geq 10$ and $0 \leq 20$).

The vertices of the feasible region are $(5, 5)$, $(15, 15)$, $(0, 10)$, and $(0, 20)$.

Now we evaluate the objective function $Z = 3x + 9y$ at each vertex:

Vertex (x, y)	Value of $Z = 3x + 9y$
$(5, 5)$	$Z = 3(5) + 9(5) = 15 + 45 = 60$
$(15, 15)$	$Z = 3(15) + 9(15) = 45 + 135 = 180$
$(0, 10)$	$Z = 3(0) + 9(10) = 0 + 90 = 90$
$(0, 20)$	$Z = 3(0) + 9(20) = 0 + 180 = 180$

From the table, the minimum value of $Z$ is $60$ at the vertex $(5, 5)$.

The maximum value of $Z$ is $180$ at the vertices $(15, 15)$ and $(0, 20)$.

Since the maximum value of $Z$ occurs at two corner points $(15, 15)$ and $(0, 20)$, it also occurs at every point on the line segment connecting these two points.

---

Final Answer:

The minimum value of $Z$ is $60$ at the point $(5, 5)$.

The maximum value of $Z$ is $180$, which occurs at the points $(15, 15)$ and $(0, 20)$, and at all points on the line segment joining these two points.

Given:

Objective function:

Constraints:

---

To Find:

The minimum value of the objective function Z graphically.

---

Solution:

First, we graph the lines corresponding to the constraints:

• Line 1: $2x - y = -5$.

  If $x=0$, $y=5$. Point (0, 5).

  If $y=0$, $2x=-5 \implies x=-2.5$. Point (-2.5, 0).

  For the inequality $2x - y \geq -5$, test point (0,0): $2(0)-0 \geq -5 \implies 0 \geq -5$ (True). Shade the region containing the origin.

• Line 2: $3x + y = 3$.

  If $x=0$, $y=3$. Point (0, 3).

  If $y=0$, $3x=3 \implies x=1$. Point (1, 0).

  For the inequality $3x + y \geq 3$, test point (0,0): $3(0)+0 \geq 3 \implies 0 \geq 3$ (False). Shade the region not containing the origin.

• Line 3: $2x - 3y = 12$.

  If $x=0$, $-3y=12 \implies y=-4$. Point (0, -4).

  If $y=0$, $2x=12 \implies x=6$. Point (6, 0).

  For the inequality $2x - 3y \leq 12$, test point (0,0): $2(0)-3(0) \leq 12 \implies 0 \leq 12$ (True). Shade the region containing the origin.

• Constraints $x \geq 0, y \geq 0$: These restrict the feasible region to the first quadrant.

The feasible region is the intersection of all these shaded areas in the first quadrant. Plotting these lines and shading the appropriate regions shows that the feasible region is unbounded.

The corner points of the feasible region are the points of intersection of the boundary lines in the first quadrant.

• Point P: Intersection of $x=0$ and $3x+y=3$.

  $3(0) + y = 3 \implies y = 3$. Point P is $(0, 3)$.

• Point Q: Intersection of $x=0$ and $2x-y=-5$.

  $2(0) - y = -5 \implies y = 5$. Point Q is $(0, 5)$.

• Point R: Intersection of $y=0$ and $3x+y=3$.

  $3x + 0 = 3 \implies x = 1$. Point R is $(1, 0)$.

• Point S: Intersection of $y=0$ and $2x-3y=12$.

  $2x - 3(0) = 12 \implies 2x = 12 \implies x = 6$. Point S is $(6, 0)$.

The other intersections ($3x+y=3$ and $2x-3y=12$; $2x-y=-5$ and $2x-3y=12$; $2x-y=-5$ and $3x+y=3$) lie outside the first quadrant and are therefore not corner points of the feasible region.

The corner points of the feasible region are P(0, 3), Q(0, 5), R(1, 0), and S(6, 0).

---

Now, we evaluate the objective function $Z = -50x + 20y$ at these corner points:

Corner Point (x, y)	Z = -50x + 20y
P(0, 3)	$Z = -50(0) + 20(3) = 60$
Q(0, 5)	$Z = -50(0) + 20(5) = 100$
R(1, 0)	$Z = -50(1) + 20(0) = -50$
S(6, 0)	$Z = -50(6) + 20(0) = -300$

The minimum value of Z among the corner points is -300, which occurs at S(6, 0).

However, since the feasible region is unbounded, we need to check if Z has a minimum value. We graph the inequality representing $Z < -300$:

$-50x + 20y < -300$

Dividing by 10:

$-5x + 2y < -30$

This represents an open half-plane.

We need to check if this open half-plane has any points in common with the feasible region.

Consider the point (7, 1). Let's check if it lies in the feasible region:

• $2(7) - 1 = 13 \geq -5$ (True)
• $3(7) + 1 = 22 \geq 3$ (True)
• $2(7) - 3(1) = 11 \leq 12$ (True)
• $x=7 \geq 0, y=1 \geq 0$ (True)

So, (7, 1) is in the feasible region.

Now, check if (7, 1) satisfies $-5x + 2y < -30$:

$-5(7) + 2(1) = -35 + 2 = -33$

Is $-33 < -30$? Yes, it is true.

Since the open half-plane $-50x + 20y < -300$ has points in common with the feasible region (e.g., (7,1)), the objective function Z does not have a minimum value. It can be made arbitrarily small (tends to $-\infty$).

---

Conclusion:

The feasible region is unbounded and the objective function $Z = -50x + 20y$ can be made arbitrarily small within this region. Therefore, there is no minimum value for the objective function Z subject to the given constraints.

Given:

Objective function:

$Z = 3x + 2y$

Constraints:

---

To Find:

The minimum value of the objective function Z.

---

Solution:

We first graph the inequalities to find the feasible region.

Constraint 1: $x + y \geq 8$

The corresponding line is $x + y = 8$.

• If $x=0$, then $y=8$. Point is (0, 8).
• If $y=0$, then $x=8$. Point is (8, 0).

Testing the point (0, 0) in the inequality: $0 + 0 \geq 8$, which is $0 \geq 8$. This is false. So, the region satisfying this inequality is the half-plane not containing the origin.

Constraint 2: $3x + 5y \leq 15$

The corresponding line is $3x + 5y = 15$.

• If $x=0$, then $5y=15 \implies y=3$. Point is (0, 3).
• If $y=0$, then $3x=15 \implies x=5$. Point is (5, 0).

Testing the point (0, 0) in the inequality: $3(0) + 5(0) \leq 15$, which is $0 \leq 15$. This is true. So, the region satisfying this inequality is the half-plane containing the origin.

Constraint 3: $x \geq 0, y \geq 0$

This restricts the feasible region to the first quadrant of the coordinate plane.

---

Finding the Feasible Region:

We need to find the intersection of the regions defined by all three constraints.

The region for $x+y \geq 8$ and $x \geq 0, y \geq 0$ lies above the line $x+y=8$ in the first quadrant.

The region for $3x+5y \leq 15$ and $x \geq 0, y \geq 0$ lies below the line $3x+5y=15$ in the first quadrant, forming a triangle with vertices (0,0), (5,0), and (0,3).

Let's check if there is any overlap between these regions.

Consider any point $(x, y)$ satisfying $3x + 5y \leq 15$ with $x \geq 0, y \geq 0$.

Since $x \geq 0$ and $y \geq 0$, we know that $3x \leq 3x + 5y$ and $5y \leq 3x + 5y$.

So, $3x \leq 15 \implies x \leq 5$.

And $5y \leq 15 \implies y \leq 3$.

For any point in this region, the maximum value of $x+y$ occurs on the boundary. On the segment $3x+5y=15$ between $(5,0)$ and $(0,3)$, the maximum value of $x+y$ is $5$ (at point $(5,0)$). For all other points in the region defined by $3x+5y \leq 15, x \geq 0, y \geq 0$, we have $x+y \leq 5$.

However, constraint (1) requires $x+y \geq 8$.

Since there are no points $(x,y)$ such that both $x+y \geq 8$ and $x+y \leq 5$ are true simultaneously, there is no common region satisfying all the given constraints.

---

Conclusion:

There is no point $(x, y)$ that satisfies all the constraints simultaneously. Therefore, the feasible region is empty.

Since the feasible region is empty, there is no solution to this linear programming problem, and thus, the objective function Z has no minimum value.

Given:

Objective function to maximize:

$Z = 3x + 4y$

Subject to the constraints:

---

To Find:

The maximum value of the objective function Z graphically.

---

Solution:

First, we treat the inequalities as equations to plot the boundary lines and then determine the feasible region.

Constraint 1: $x + y \leq 4$

The corresponding line is $x + y = 4$.

• If $x = 0$, then $y = 4$. The point is (0, 4).
• If $y = 0$, then $x = 4$. The point is (4, 0).

Draw the line passing through (0, 4) and (4, 0).

To determine the region represented by the inequality, we test the point (0, 0):

$0 + 0 \leq 4 \implies 0 \leq 4$, which is true.

So, the region satisfying $x + y \leq 4$ is the half-plane containing the origin.

Constraint 2: $x \geq 0$

This represents the region to the right of and including the y-axis.

Constraint 3: $y \geq 0$

This represents the region above and including the x-axis.

The feasible region is the intersection of the regions defined by these three constraints. It is the area bounded by the triangle with vertices at the origin O(0, 0), the point A(0, 4) on the y-axis, and the point B(4, 0) on the x-axis.

This feasible region is bounded.

---

The corner points of the feasible region are:

• O(0, 0)
• A(0, 4)
• B(4, 0)

Now, we evaluate the objective function $Z = 3x + 4y$ at these corner points:

Corner Point (x, y)	Z = 3x + 4y
O(0, 0)	$Z = 3(0) + 4(0) = 0$
A(0, 4)	$Z = 3(0) + 4(4) = 16$
B(4, 0)	$Z = 3(4) + 4(0) = 12$

---

Conclusion:

From the table, the maximum value of Z is 16, which occurs at the corner point A(0, 4).

Given:

Objective function to minimise:

$Z = -3x + 4y$

Subject to the constraints:

---

To Find:

The minimum value of the objective function Z graphically.

---

Solution:

First, we graph the boundary lines corresponding to the constraints and determine the feasible region.

Line 1: $x + 2y = 8$

• If $x = 0$, $2y = 8 \implies y = 4$. Point (0, 4).
• If $y = 0$, $x = 8$. Point (8, 0).

Testing the point (0, 0) in $x + 2y \leq 8$: $0 + 2(0) \leq 8 \implies 0 \leq 8$ (True). The region is towards the origin.

Line 2: $3x + 2y = 12$

• If $x = 0$, $2y = 12 \implies y = 6$. Point (0, 6).
• If $y = 0$, $3x = 12 \implies x = 4$. Point (4, 0).

Testing the point (0, 0) in $3x + 2y \leq 12$: $3(0) + 2(0) \leq 12 \implies 0 \leq 12$ (True). The region is towards the origin.

Constraints 3 & 4: $x \geq 0$ and $y \geq 0$. This restricts the feasible region to the first quadrant.

The feasible region is the area common to all these regions in the first quadrant. We need to find the intersection points of these lines to determine the corner points.

The corner points are the intersections of the boundary lines:

• O(0, 0): Intersection of $x=0$ and $y=0$.
• A(0, 4): Intersection of $x=0$ and $x + 2y = 8$. (Check with $3x+2y \leq 12$: $3(0)+2(4) = 8 \leq 12$. Valid)
• B(4, 0): Intersection of $y=0$ and $3x + 2y = 12$. (Check with $x+2y \leq 8$: $4+2(0) = 4 \leq 8$. Valid)
• C: Intersection of $x + 2y = 8$ and $3x + 2y = 12$.

  Subtracting the first equation from the second:

  $(3x + 2y) - (x + 2y) = 12 - 8$

  $2x = 4$

  $x = 2$

  Substitute $x=2$ into $x + 2y = 8$:

  $2 + 2y = 8$

  $2y = 6$

  $y = 3$

  The intersection point is C(2, 3). (Check $x \geq 0, y \geq 0$: Valid).

The corner points of the bounded feasible region are O(0, 0), A(0, 4), B(4, 0), and C(2, 3).

---

Now, we evaluate the objective function $Z = -3x + 4y$ at these corner points:

Corner Point (x, y)	Z = -3x + 4y
O(0, 0)	$Z = -3(0) + 4(0) = 0$
A(0, 4)	$Z = -3(0) + 4(4) = 16$
B(4, 0)	$Z = -3(4) + 4(0) = -12$
C(2, 3)	$Z = -3(2) + 4(3) = -6 + 12 = 6$

---

Conclusion:

From the table, the minimum value of Z is -12, which occurs at the corner point B(4, 0).

Given:

Objective function to maximize:

$Z = 5x + 3y$

Subject to the constraints:

---

To Find:

The maximum value of the objective function Z graphically.

---

Solution:

We graph the boundary lines corresponding to the constraints to find the feasible region.

Line 1: $3x + 5y = 15$

• If $x = 0$, $5y = 15 \implies y = 3$. Point (0, 3).
• If $y = 0$, $3x = 15 \implies x = 5$. Point (5, 0).

Testing the point (0, 0) in $3x + 5y \leq 15$: $3(0) + 5(0) \leq 15 \implies 0 \leq 15$ (True). The region is towards the origin.

Line 2: $5x + 2y = 10$

• If $x = 0$, $2y = 10 \implies y = 5$. Point (0, 5).
• If $y = 0$, $5x = 10 \implies x = 2$. Point (2, 0).

Testing the point (0, 0) in $5x + 2y \leq 10$: $5(0) + 2(0) \leq 10 \implies 0 \leq 10$ (True). The region is towards the origin.

Constraints 3 & 4: $x \geq 0$ and $y \geq 0$. This restricts the feasible region to the first quadrant.

The feasible region is the intersection of these regions in the first quadrant. We find the intersection points of the boundary lines to determine the corner points.

The corner points are:

• O(0, 0): Intersection of $x=0$ and $y=0$.
• A(0, 3): Intersection of $x=0$ and $3x + 5y = 15$. (Check with $5x+2y \leq 10$: $5(0)+2(3)=6 \leq 10$. Valid).
• B(2, 0): Intersection of $y=0$ and $5x + 2y = 10$. (Check with $3x+5y \leq 15$: $3(2)+5(0)=6 \leq 15$. Valid).
• C: Intersection of $3x + 5y = 15$ and $5x + 2y = 10$.

  Multiply the first equation by 2 and the second by 5:

  $6x + 10y = 30$

  $25x + 10y = 50$

  Subtract the first new equation from the second:

  $(25x + 10y) - (6x + 10y) = 50 - 30$

  $19x = 20$

  $x = \frac{20}{19}$

  Substitute $x = \frac{20}{19}$ into $5x + 2y = 10$:

  $5\left(\frac{20}{19}\right) + 2y = 10$

  $\frac{100}{19} + 2y = 10$

  $2y = 10 - \frac{100}{19} = \frac{190 - 100}{19} = \frac{90}{19}$

  $y = \frac{45}{19}$

  The intersection point is C($\frac{20}{19}$, $\frac{45}{19}$). (Check $x \geq 0, y \geq 0$: Valid).

The corner points of the bounded feasible region are O(0, 0), A(0, 3), B(2, 0), and C($\frac{20}{19}$, $\frac{45}{19}$).

---

Now, we evaluate the objective function $Z = 5x + 3y$ at these corner points:

Corner Point (x, y)	Z = 5x + 3y
O(0, 0)	$Z = 5(0) + 3(0) = 0$
A(0, 3)	$Z = 5(0) + 3(3) = 9$
B(2, 0)	$Z = 5(2) + 3(0) = 10$
C($\frac{20}{19}$, $\frac{45}{19}$)	$Z = 5\left(\frac{20}{19}\right) + 3\left(\frac{45}{19}\right) = \frac{100}{19} + \frac{135}{19} = \frac{100 + 135}{19} = \frac{235}{19}$

We compare the values: $0$, $9$, $10$, and $\frac{235}{19}$.

$\frac{235}{19} \approx 12.37$.

The maximum value among these is $\frac{235}{19}$.

---

Conclusion:

The maximum value of Z is $\mathbf{\frac{235}{19}}$, which occurs at the corner point C($\frac{20}{19}$, $\frac{45}{19}$).

Given:

Objective function to minimise:

$Z = 3x + 5y$

Subject to the constraints:

---

To Find:

The minimum value of the objective function Z graphically.

---

Solution:

First, we graph the boundary lines corresponding to the constraints and determine the feasible region.

Line 1: $x + 3y = 3$

• If $x = 0$, $3y = 3 \implies y = 1$. Point (0, 1).
• If $y = 0$, $x = 3$. Point (3, 0).

Testing the point (0, 0) in $x + 3y \geq 3$: $0 + 3(0) \geq 3 \implies 0 \geq 3$ (False). The region is away from the origin.

Line 2: $x + y = 2$

• If $x = 0$, $y = 2$. Point (0, 2).
• If $y = 0$, $x = 2$. Point (2, 0).

Testing the point (0, 0) in $x + y \geq 2$: $0 + 0 \geq 2 \implies 0 \geq 2$ (False). The region is away from the origin.

Constraints 3 & 4: $x \geq 0$ and $y \geq 0$. This restricts the feasible region to the first quadrant.

The feasible region is the common area satisfying all constraints in the first quadrant. It is the region above both lines $x+3y=3$ and $x+y=2$. This region is unbounded.

We need to find the intersection points of the boundary lines in the first quadrant to determine the corner points.

The potential corner points are:

• A: Intersection of $x=0$ and $x + y = 2$.

  $0 + y = 2 \implies y = 2$. Point A is (0, 2).

  Check if A(0, 2) satisfies $x+3y \geq 3$: $0 + 3(2) = 6 \geq 3$. Yes.

• B: Intersection of $y=0$ and $x + 3y = 3$.

  $x + 3(0) = 3 \implies x = 3$. Point B is (3, 0).

  Check if B(3, 0) satisfies $x+y \geq 2$: $3 + 0 = 3 \geq 2$. Yes.

• C: Intersection of $x + 3y = 3$ and $x + y = 2$.

  Subtracting the second equation from the first:

  $(x + 3y) - (x + y) = 3 - 2$

  $2y = 1$

  $y = \frac{1}{2}$

  Substitute $y = \frac{1}{2}$ into $x + y = 2$:

  $x + \frac{1}{2} = 2$

  $x = 2 - \frac{1}{2} = \frac{4-1}{2} = \frac{3}{2}$

  The intersection point is C($\frac{3}{2}$, $\frac{1}{2}$).

  Check if C($\frac{3}{2}$, $\frac{1}{2}$) satisfies $x \geq 0, y \geq 0$. Yes.

The corner points of the feasible region are A(0, 2), B(3, 0), and C($\frac{3}{2}$, $\frac{1}{2}$).

---

Now, we evaluate the objective function $Z = 3x + 5y$ at these corner points:

Corner Point (x, y)	Z = 3x + 5y
A(0, 2)	$Z = 3(0) + 5(2) = 10$
B(3, 0)	$Z = 3(3) + 5(0) = 9$
C($\frac{3}{2}$, $\frac{1}{2}$)	$Z = 3\left(\frac{3}{2}\right) + 5\left(\frac{1}{2}\right) = \frac{9}{2} + \frac{5}{2} = \frac{14}{2} = 7$

The minimum value of Z among the corner points is 7, which occurs at C($\frac{3}{2}$, $\frac{1}{2}$).

Since the feasible region is unbounded, we need to check if Z has a minimum value. We graph the inequality corresponding to $Z < 7$:

$3x + 5y < 7$

We check if the open half-plane defined by $3x + 5y < 7$ has any points in common with the feasible region.

The line $3x + 5y = 7$ passes through the corner point C($\frac{3}{2}$, $\frac{1}{2}$). The open half-plane $3x + 5y < 7$ represents the region below this line.

The feasible region lies entirely above or on the boundary lines $x+3y=3$ and $x+y=2$, and specifically above or on the segment connecting A(0,2), C(3/2, 1/2), and B(3,0). Since the line $3x+5y=7$ passes through C and the region $3x+5y<7$ lies below this line, this open half-plane has no points in common with the feasible region.

Therefore, the minimum value of Z occurs at the corner point C.

---

Conclusion:

The minimum value of Z is 7, which occurs at the corner point C($\frac{3}{2}$, $\frac{1}{2}$).

Given:

Objective function to maximize:

$Z = 3x + 2y$

Subject to the constraints:

---

To Find:

The maximum value of the objective function Z graphically.

---

Solution:

We graph the boundary lines corresponding to the constraints to find the feasible region.

Line 1: $x + 2y = 10$

• If $x = 0$, $2y = 10 \implies y = 5$. Point (0, 5).
• If $y = 0$, $x = 10$. Point (10, 0).

Testing the point (0, 0) in $x + 2y \leq 10$: $0 + 2(0) \leq 10 \implies 0 \leq 10$ (True). The region is towards the origin.

Line 2: $3x + y = 15$

• If $x = 0$, $y = 15$. Point (0, 15).
• If $y = 0$, $3x = 15 \implies x = 5$. Point (5, 0).

Testing the point (0, 0) in $3x + y \leq 15$: $3(0) + 0 \leq 15 \implies 0 \leq 15$ (True). The region is towards the origin.

Constraints 3 & 4: $x \geq 0$ and $y \geq 0$. This restricts the feasible region to the first quadrant.

The feasible region is the intersection of these regions in the first quadrant. We find the intersection points of the boundary lines to determine the corner points.

The corner points are:

• O(0, 0): Intersection of $x=0$ and $y=0$.
• A(0, 5): Intersection of $x=0$ and $x + 2y = 10$. (Check with $3x+y \leq 15$: $3(0)+5 = 5 \leq 15$. Valid).
• B(5, 0): Intersection of $y=0$ and $3x + y = 15$. (Check with $x+2y \leq 10$: $5+2(0) = 5 \leq 10$. Valid).
• C: Intersection of $x + 2y = 10$ and $3x + y = 15$.

  From $3x + y = 15$, we get $y = 15 - 3x$.

  Substitute this into the first equation:

  $x + 2(15 - 3x) = 10$

  $x + 30 - 6x = 10$

  $-5x = 10 - 30$

  $-5x = -20$

  $x = 4$

  Substitute $x=4$ back into $y = 15 - 3x$:

  $y = 15 - 3(4) = 15 - 12 = 3$

  The intersection point is C(4, 3). (Check $x \geq 0, y \geq 0$: Valid).

The corner points of the bounded feasible region are O(0, 0), A(0, 5), B(5, 0), and C(4, 3).

---

Now, we evaluate the objective function $Z = 3x + 2y$ at these corner points:

Corner Point (x, y)	Z = 3x + 2y
O(0, 0)	$Z = 3(0) + 2(0) = 0$
A(0, 5)	$Z = 3(0) + 2(5) = 10$
B(5, 0)	$Z = 3(5) + 2(0) = 15$
C(4, 3)	$Z = 3(4) + 2(3) = 12 + 6 = 18$

---

Conclusion:

From the table, the maximum value of Z is 18, which occurs at the corner point C(4, 3).

Given:

Objective function to minimise:

$Z = x + 2y$

Subject to the constraints:

---

To Find:

The minimum value of the objective function Z graphically.

---

Solution:

First, we graph the boundary lines corresponding to the constraints and determine the feasible region.

Line 1: $2x + y = 3$

• If $x = 0$, $y = 3$. Point (0, 3).
• If $y = 0$, $2x = 3 \implies x = 1.5$. Point (1.5, 0).

Testing the point (0, 0) in $2x + y \geq 3$: $2(0) + 0 \geq 3 \implies 0 \geq 3$ (False). The region is away from the origin.

Line 2: $x + 2y = 6$

• If $x = 0$, $2y = 6 \implies y = 3$. Point (0, 3).
• If $y = 0$, $x = 6$. Point (6, 0).

Testing the point (0, 0) in $x + 2y \geq 6$: $0 + 2(0) \geq 6 \implies 0 \geq 6$ (False). The region is away from the origin.

Constraints 3 & 4: $x \geq 0$ and $y \geq 0$. This restricts the feasible region to the first quadrant.

The feasible region is the common area satisfying all constraints in the first quadrant. It is the region lying on or above both lines $2x+y=3$ and $x+2y=6$. This region is unbounded.

We find the intersection points of the boundary lines that form the vertices of the feasible region.

The potential corner points are the intersections of the boundary lines in the first quadrant:

• Intersection of $x=0$ and $2x+y=3$: Point (0, 3). Check feasibility: $0+2(3)=6 \geq 6$ (True). So, (0, 3) is a corner point. Let's call it A(0, 3).
• Intersection of $x=0$ and $x+2y=6$: Point (0, 3). This is the same point A.
• Intersection of $y=0$ and $2x+y=3$: Point (1.5, 0). Check feasibility: $1.5+2(0)=1.5 \geq 6$ (False). Not a corner point of the feasible region.
• Intersection of $y=0$ and $x+2y=6$: Point (6, 0). Check feasibility: $2(6)+0=12 \geq 3$ (True). So, (6, 0) is a corner point. Let's call it B(6, 0).
• Intersection of $2x + y = 3$ and $x + 2y = 6$:

  From $2x+y=3$, $y=3-2x$. Substitute into the second equation:

  $x + 2(3-2x) = 6$

  $x + 6 - 4x = 6$

  $-3x = 0 \implies x = 0$

  Substitute $x=0$ back into $y=3-2x \implies y=3$.

  The intersection is (0, 3), which is point A.

The corner points of the feasible region are A(0, 3) and B(6, 0).

---

Now, we evaluate the objective function $Z = x + 2y$ at these corner points:

Corner Point (x, y)	Z = x + 2y
A(0, 3)	$Z = 0 + 2(3) = 6$
B(6, 0)	$Z = 6 + 2(0) = 6$

The minimum value of Z among the corner points is 6, which occurs at both A(0, 3) and B(6, 0).

Since the feasible region is unbounded, we need to check if Z can take a value less than 6. We consider the inequality $Z < 6$:

$x + 2y < 6$

This represents the open half-plane below the line $x+2y=6$.

The feasible region is defined by $x \geq 0, y \geq 0, 2x+y \geq 3,$ and $x+2y \geq 6$. This region lies entirely on or above the line $x+2y=6$.

Since the feasible region lies on or above the line $x+2y=6$, and the region $x+2y < 6$ lies strictly below this line, there are no points common to the feasible region and the half-plane $x+2y < 6$.

Therefore, the minimum value of Z is 6.

---

Conclusion:

The minimum value of Z is 6. This minimum value occurs at every point on the line segment joining the points A(0, 3) and B(6, 0), which is the segment of the line $x + 2y = 6$ included in the feasible region.

Given:

Objective function to minimise and maximise:

$Z = 5x + 10y$

Subject to the constraints:

Also, we need to address the statement: "Show that the minimum of Z occurs at more than two points."

---

To Find:

The minimum and maximum values of the objective function Z graphically, and discuss the occurrence of the minimum value.

---

Solution:

First, we graph the boundary lines corresponding to the constraints and determine the feasible region.

Line 1: $x + 2y = 120$

• If $x = 0$, $2y = 120 \implies y = 60$. Point (0, 60).
• If $y = 0$, $x = 120$. Point (120, 0).

Testing the point (0, 0) in $x + 2y \leq 120$: $0 + 2(0) \leq 120 \implies 0 \leq 120$ (True). The region is towards the origin.

Line 2: $x + y = 60$

• If $x = 0$, $y = 60$. Point (0, 60).
• If $y = 0$, $x = 60$. Point (60, 0).

Testing the point (0, 0) in $x + y \geq 60$: $0 + 0 \geq 60 \implies 0 \geq 60$ (False). The region is away from the origin.

Line 3: $x - 2y = 0 \implies x = 2y$

• If $y = 0$, $x = 0$. Point (0, 0).
• If $y = 30$, $x = 60$. Point (60, 30).

Testing the point (10, 0) in $x - 2y \geq 0$: $10 - 2(0) \geq 0 \implies 10 \geq 0$ (True). The region includes points where $x \geq 2y$, which is the area on and below the line $x=2y$ in the first quadrant.

Constraints 4 & 5: $x \geq 0$ and $y \geq 0$. This restricts the feasible region to the first quadrant.

The feasible region is the intersection of these regions in the first quadrant. We find the intersection points of the boundary lines to determine the corner points.

The corner points are:

• A: Intersection of $x + y = 60$ and $x - 2y = 0$.

  From $x = 2y$, substitute into the first equation: $2y + y = 60 \implies 3y = 60 \implies y = 20$. Then $x = 2(20) = 40$. Point A is (40, 20).

  Check constraints: $40+2(20)=80 \leq 120$ (Ok), $40 \geq 0, 20 \geq 0$ (Ok).

• B: Intersection of $x + 2y = 120$ and $x - 2y = 0$.

  From $x = 2y$, substitute into the first equation: $2y + 2y = 120 \implies 4y = 120 \implies y = 30$. Then $x = 2(30) = 60$. Point B is (60, 30).

  Check constraints: $60+30=90 \geq 60$ (Ok), $60 \geq 0, 30 \geq 0$ (Ok).

• C: Intersection of $x + y = 60$ and $y = 0$.

  $x + 0 = 60 \implies x = 60$. Point C is (60, 0).

  Check constraints: $60+2(0)=60 \leq 120$ (Ok), $60-2(0)=60 \geq 0$ (Ok), $60 \geq 0, 0 \geq 0$ (Ok).

• D: Intersection of $x + 2y = 120$ and $y = 0$.

  $x + 2(0) = 120 \implies x = 120$. Point D is (120, 0).

  Check constraints: $120+0=120 \geq 60$ (Ok), $120-2(0)=120 \geq 0$ (Ok), $120 \geq 0, 0 \geq 0$ (Ok).

The corner points of the bounded feasible region are A(40, 20), B(60, 30), C(60, 0), and D(120, 0).

---

Now, we evaluate the objective function $Z = 5x + 10y$ at these corner points:

Corner Point (x, y)	Z = 5x + 10y
A(40, 20)	$Z = 5(40) + 10(20) = 200 + 200 = 400$
B(60, 30)	$Z = 5(60) + 10(30) = 300 + 300 = 600$
C(60, 0)	$Z = 5(60) + 10(0) = 300 + 0 = 300$
D(120, 0)	$Z = 5(120) + 10(0) = 600 + 0 = 600$

---

Analysis of Minimum and Maximum Values:

From the table:

• The minimum value of Z is 300, which occurs at the corner point C(60, 0).
• The maximum value of Z is 600, which occurs at two corner points: B(60, 30) and D(120, 0).

Regarding the statement "Show that the minimum of Z occurs at more than two points":

Our calculation shows that the minimum value Z = 300 occurs only at a single corner point, C(60, 0).

Let's consider the line representing the objective function for Z = 300: $5x + 10y = 300$, which simplifies to $x + 2y = 60$. We check if this line intersects the feasible region at any other points besides C(60, 0). The feasible region is the polygon ABCD. The line $x+2y=60$ passes through C(60,0). It does not pass through A(40,20) (since $40+2(20)=80 \neq 60$), B(60,30) (since $60+2(30)=120 \neq 60$), or D(120,0) (since $120+2(0)=120 \neq 60$). The line $x+2y=60$ only touches the feasible region at the vertex C(60,0). Thus, the minimum value occurs at exactly one point.

However, let's analyze the maximum value. The maximum value Z = 600 occurs at B(60, 30) and D(120, 0). When the optimal value occurs at two distinct corner points, it also occurs at every point on the line segment connecting these two points.

The line segment connecting B(60, 30) and D(120, 0) is part of the boundary line $x + 2y = 120$. Let's check the slope of the objective function line $Z = 5x + 10y$. Rearranging gives $10y = -5x + Z \implies y = -\frac{5}{10}x + \frac{Z}{10} \implies y = -\frac{1}{2}x + \frac{Z}{10}$. The slope is $-\frac{1}{2}$.

The slope of the boundary line $x + 2y = 120$ is also $2y = -x + 120 \implies y = -\frac{1}{2}x + 60$. The slope is $-\frac{1}{2}$.

Since the slope of the objective function line is the same as the slope of the boundary line segment BD, the maximum value Z = 600 occurs for all points $(x, y)$ on the line segment connecting B(60, 30) and D(120, 0). This represents infinitely many points, which is "more than two points".

It appears the introductory sentence "Show that the minimum of Z occurs at more than two points" might be a misstatement, and perhaps intended to refer to the maximum value.

---

Conclusion:

The minimum value of Z is 300, which occurs only at the point C(60, 0).

The maximum value of Z is 600. This maximum value occurs at the corner points B(60, 30) and D(120, 0), and consequently, at all points on the line segment joining B(60, 30) and D(120, 0). This demonstrates that the maximum value occurs at infinitely many points (more than two points).

The initial instruction to show that the *minimum* occurs at more than two points is inconsistent with the results obtained for the given objective function and constraints; it is the *maximum* that exhibits this property.

Given:

Objective function to minimise and maximise:

$Z = x + 2y$

Subject to the constraints:

---

To Find:

The minimum and maximum values of the objective function Z graphically.

---

Solution:

First, we graph the boundary lines corresponding to the constraints and determine the feasible region.

Line 1: $x + 2y = 100$

• If $x = 0$, $2y = 100 \implies y = 50$. Point (0, 50).
• If $y = 0$, $x = 100$. Point (100, 0).

Testing the point (0, 0) in $x + 2y \geq 100$: $0 + 2(0) \geq 100 \implies 0 \geq 100$ (False). The region is away from the origin.

Line 2: $2x - y = 0 \implies y = 2x$

• If $x = 0$, $y = 0$. Point (0, 0).
• If $x = 20$, $y = 40$. Point (20, 40).

Testing the point (10, 0) in $2x - y \leq 0$: $2(10) - 0 \leq 0 \implies 20 \leq 0$ (False). The region is the half-plane above the line $y = 2x$ (where $y \geq 2x$).

Line 3: $2x + y = 200$

• If $x = 0$, $y = 200$. Point (0, 200).
• If $y = 0$, $2x = 200 \implies x = 100$. Point (100, 0).

Testing the point (0, 0) in $2x + y \leq 200$: $2(0) + 0 \leq 200 \implies 0 \leq 200$ (True). The region is towards the origin.

Constraints 4 & 5: $x \geq 0$ and $y \geq 0$. This restricts the feasible region to the first quadrant.

The feasible region is the intersection of these regions in the first quadrant. We find the intersection points of the boundary lines to determine the corner points.

The corner points are:

• A: Intersection of $x + 2y = 100$ and $y = 2x$.

  Substitute $y=2x$ into $x + 2y = 100$: $x + 2(2x) = 100 \implies x + 4x = 100 \implies 5x = 100 \implies x = 20$. Then $y = 2(20) = 40$. Point A is (20, 40).

  Check constraints: $2(20)+40 = 80 \leq 200$ (Ok), $x \geq 0, y \geq 0$ (Ok).

• B: Intersection of $y = 2x$ and $2x + y = 200$.

  Substitute $y=2x$ into $2x + y = 200$: $2x + 2x = 200 \implies 4x = 200 \implies x = 50$. Then $y = 2(50) = 100$. Point B is (50, 100).

  Check constraints: $50+2(100)=250 \geq 100$ (Ok), $x \geq 0, y \geq 0$ (Ok).

• D: Intersection of $x + 2y = 100$ and $x = 0$.

  $0 + 2y = 100 \implies y = 50$. Point D is (0, 50).

  Check constraints: $2(0)-50 = -50 \leq 0$ (Ok), $2(0)+50 = 50 \leq 200$ (Ok), $x \geq 0, y \geq 0$ (Ok).

• E: Intersection of $2x + y = 200$ and $x = 0$.

  $2(0) + y = 200 \implies y = 200$. Point E is (0, 200).

  Check constraints: $0+2(200)=400 \geq 100$ (Ok), $2(0)-200 = -200 \leq 0$ (Ok), $x \geq 0, y \geq 0$ (Ok).

The corner points of the bounded feasible region are A(20, 40), B(50, 100), D(0, 50), and E(0, 200).

---

Now, we evaluate the objective function $Z = x + 2y$ at these corner points:

Corner Point (x, y)	Z = x + 2y
A(20, 40)	$Z = 20 + 2(40) = 20 + 80 = 100$
B(50, 100)	$Z = 50 + 2(100) = 50 + 200 = 250$
D(0, 50)	$Z = 0 + 2(50) = 100$
E(0, 200)	$Z = 0 + 2(200) = 400$

---

Analysis of Minimum and Maximum Values:

From the table:

• The minimum value of Z is 100. This occurs at two corner points: A(20, 40) and D(0, 50).
• The maximum value of Z is 400. This occurs at the corner point E(0, 200).

Since the minimum value Z = 100 occurs at two distinct corner points A and D, it also occurs at every point on the line segment joining A(20, 40) and D(0, 50). This line segment is part of the boundary line $x + 2y = 100$. We can see that the objective function $Z = x + 2y$ has the same form as the constraint line $x + 2y = 100$.

---

Conclusion:

The minimum value of Z is 100, which occurs at all points on the line segment joining A(20, 40) and D(0, 50).

The maximum value of Z is 400, which occurs at the point E(0, 200).

Given:

Objective function to maximize:

$Z = -x + 2y$

Subject to the constraints:

---

To Find:

The maximum value of the objective function Z graphically.

---

Solution:

First, we graph the boundary lines corresponding to the constraints and determine the feasible region.

Line 1: $x = 3$. This is a vertical line passing through (3, 0). The region $x \geq 3$ is to the right of this line.

Line 2: $x + y = 5$.

• If $x = 0$, $y = 5$. Point (0, 5).
• If $y = 0$, $x = 5$. Point (5, 0).

Testing the point (0, 0) in $x + y \geq 5$: $0 + 0 \geq 5$ (False). The region is away from the origin.

Line 3: $x + 2y = 6$.

• If $x = 0$, $2y = 6 \implies y = 3$. Point (0, 3).
• If $y = 0$, $x = 6$. Point (6, 0).

Testing the point (0, 0) in $x + 2y \geq 6$: $0 + 2(0) \geq 6$ (False). The region is away from the origin.

Constraint 4: $y \geq 0$. This restricts the feasible region to be on or above the x-axis.

The feasible region is the intersection of the regions defined by $x \geq 3$, $x+y \geq 5$, $x+2y \geq 6$, and $y \geq 0$. This region is unbounded.

We find the intersection points of the boundary lines to determine the corner points of the feasible region.

The potential corner points are:

• A: Intersection of $x = 3$ and $x + y = 5$.

  $3 + y = 5 \implies y = 2$. Point A is (3, 2).

  Check feasibility for other constraints: $3 + 2(2) = 7 \geq 6$ (True), $2 \geq 0$ (True). A(3, 2) is a corner point.

• B: Intersection of $x + y = 5$ and $x + 2y = 6$.

  Subtracting $(x+y=5)$ from $(x+2y=6)$ gives $y = 1$.

  Substituting $y=1$ into $x+y=5$ gives $x+1=5 \implies x=4$. Point B is (4, 1).

  Check feasibility for other constraints: $4 \geq 3$ (True), $1 \geq 0$ (True). B(4, 1) is a corner point.

• C: Intersection of $x + 2y = 6$ and $y = 0$.

  $x + 2(0) = 6 \implies x = 6$. Point C is (6, 0).

  Check feasibility for other constraints: $6 \geq 3$ (True), $6 + 0 = 6 \geq 5$ (True). C(6, 0) is a corner point.

• Intersection of $x=3$ and $x+2y=6$: $3+2y=6 \implies 2y=3 \implies y=1.5$. Point (3, 1.5). Check $x+y \geq 5$: $3+1.5=4.5 \ngeq 5$. Not feasible.
• Intersection of $x=3$ and $y=0$: Point (3,0). Check $x+y \geq 5$: $3+0=3 \ngeq 5$. Not feasible. Check $x+2y \geq 6$: $3+2(0)=3 \ngeq 6$. Not feasible.

The corner points of the feasible region are A(3, 2), B(4, 1), and C(6, 0).

---

Now, we evaluate the objective function $Z = -x + 2y$ at these corner points:

Corner Point (x, y)	Z = -x + 2y
A(3, 2)	$Z = -3 + 2(2) = 1$
B(4, 1)	$Z = -4 + 2(1) = -2$
C(6, 0)	$Z = -6 + 2(0) = -6$

The maximum value of Z at the corner points is 1, which occurs at A(3, 2).

However, the feasible region is unbounded. To determine if Z has a maximum value, we examine the inequality $Z > 1$, which is $-x + 2y > 1$.

We need to check if the open half-plane $-x + 2y > 1$ has any points in common with the feasible region.

Consider the point (3, 3). Let's check if it lies in the feasible region:

• $x = 3 \geq 3$ (True)
• $x + y = 3 + 3 = 6 \geq 5$ (True)
• $x + 2y = 3 + 2(3) = 9 \geq 6$ (True)
• $y = 3 \geq 0$ (True)

The point (3, 3) is in the feasible region.

Now, evaluate Z at (3, 3):

$Z = -3 + 2(3) = -3 + 6 = 3$.

Since $3 > 1$, there are points in the feasible region for which Z is greater than the value found at corner point A.

As the feasible region extends infinitely upwards (e.g., consider points $(3, y)$ as $y \to \infty$), the value of $Z = -3 + 2y$ also increases without bound.

---

Conclusion:

The feasible region is unbounded, and the objective function $Z = -x + 2y$ can be made arbitrarily large within this region. Therefore, the objective function Z has no maximum value.

Given:

Objective function to maximize:

$Z = x + y$

Subject to the constraints:

---

To Find:

The maximum value of the objective function Z graphically.

---

Solution:

First, let's rewrite the constraints to make graphing easier:

Constraint 1: $x - y \leq -1 \implies -y \leq -x - 1 \implies y \geq x + 1$.

The boundary line is $y = x + 1$.

• If $x = 0$, $y = 1$. Point (0, 1).
• If $x = 1$, $y = 2$. Point (1, 2).

Testing the point (0, 0) in $y \geq x + 1$: $0 \geq 0 + 1 \implies 0 \geq 1$ (False). The region is the half-plane above the line $y=x+1$.

Constraint 2: $-x + y \leq 0 \implies y \leq x$.

The boundary line is $y = x$.

• If $x = 0$, $y = 0$. Point (0, 0).
• If $x = 2$, $y = 2$. Point (2, 2).

Testing the point (1, 0) in $y \leq x$: $0 \leq 1$ (True). The region is the half-plane below or on the line $y=x$.

Constraints 3 & 4: $x \geq 0$ and $y \geq 0$. This restricts the feasible region to the first quadrant.

Finding the Feasible Region:

We need to find the region in the first quadrant where both $y \geq x + 1$ and $y \leq x$ are satisfied.

Let's see if there is any point $(x, y)$ such that $y \geq x+1$ and $y \leq x$.

If such a point exists, then we must have $x + 1 \leq y \leq x$.

This implies $x + 1 \leq x$.

Subtracting $x$ from both sides gives $1 \leq 0$, which is impossible.

Therefore, there are no points $(x, y)$ that can simultaneously satisfy both constraint (1) $y \geq x+1$ and constraint (2) $y \leq x$.

This means there is no common region shared by the solutions of these two inequalities.

---

Conclusion:

Since there is no point $(x, y)$ satisfying all the given constraints simultaneously, the feasible region is empty.

Because the feasible region is empty, there are no possible solutions, and thus the objective function Z has no maximum value.

Formulation of the Linear Programming Problem (LPP)

---

Decision Variables:

Let $x$ represent the quantity (in kg) of Food ‘I’ in the mixture.

Let $y$ represent the quantity (in kg) of Food ‘II’ in the mixture.

Here, $x$ and $y$ are the decision variables.

---

Objective Function:

The cost of Food ‘I’ is $\textsf{₹} \; 50$ per kg.

The cost of Food ‘II’ is $\textsf{₹} \; 70$ per kg.

The total cost of the mixture is $Z = (\text{Cost per kg of Food ‘I’}) \times (\text{kg of Food ‘I’}) + (\text{Cost per kg of Food ‘II’}) \times (\text{kg of Food ‘II’})$

$Z = 50x + 70y$

The objective is to minimize the total cost Z.

Minimize $Z = 50x + 70y$

---

Constraints:

The constraints are based on the minimum vitamin requirements.

1. Vitamin A Constraint:

The mixture must contain at least 8 units of vitamin A.

Food ‘I’ contains 2 units/kg of vitamin A.

Food ‘II’ contains 1 unit/kg of vitamin A.

Total vitamin A in the mixture = (Vitamin A in Food ‘I’) $\times x$ + (Vitamin A in Food ‘II’) $\times y$

Total vitamin A = $2x + 1y$

Since the minimum requirement is 8 units, the constraint is:

2. Vitamin C Constraint:

The mixture must contain at least 10 units of vitamin C.

Food ‘I’ contains 1 unit/kg of vitamin C.

Food ‘II’ contains 2 units/kg of vitamin C.

Total vitamin C in the mixture = (Vitamin C in Food ‘I’) $\times x$ + (Vitamin C in Food ‘II’) $\times y$

Total vitamin C = $1x + 2y$

Since the minimum requirement is 10 units, the constraint is:

3. Non-negativity Constraints:

The quantities of food used cannot be negative.

---

Complete LPP Formulation:

The linear programming problem is formulated as follows:

Minimize $Z = 50x + 70y$

Subject to the constraints:

This is a Linear Programming Problem (LPP).

---

Given:

Total land available = $50$ hectares.

Profit per hectare for crop X = $\textsf{₹}10,500$.

Profit per hectare for crop Y = $\textsf{₹}9,000$.

Herbicide usage per hectare for crop X = $20$ litres.

Herbicide usage per hectare for crop Y = $10$ litres.

Maximum herbicide available = $800$ litres.

---

To Find:

The allocation of land to each crop (X and Y) to maximise the total profit.

---

Solution:

Let $\mathbf{x}$ be the number of hectares allocated to crop X.

Let $\mathbf{y}$ be the number of hectares allocated to crop Y.

Since the number of hectares cannot be negative, we have the non-negativity constraints:

The total land available is $50$ hectares. This gives the constraint:

The herbicide usage for crop X is $20$ litres per hectare and for crop Y is $10$ litres per hectare. The total herbicide available is $800$ litres. This gives the constraint:

$20x + 10y \le 800$

Dividing by $10$, we get:

The objective is to maximise the total profit. The profit from crop X is $\textsf{₹}10,500$ per hectare, and from crop Y is $\textsf{₹}9,000$ per hectare. The total profit $Z$ is given by:

$Z = 10500x + 9000y$

The mathematical formulation of the LPP is:

Maximise $Z = 10500x + 9000y$

Subject to the constraints:

---

We will solve this LPP graphically.

First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $x + y = 50$

Points for Line 1:

If $x=0$, $y=50$. Point $(0, 50)$.

If $y=0$, $x=50$. Point $(50, 0)$.

Line 2: $2x + y = 80$

Points for Line 2:

If $x=0$, $y=80$. Point $(0, 80)$.

If $y=0$, $2x=80 \implies x=40$. Point $(40, 0)$.

The non-negativity constraints $x \ge 0$ and $y \ge 0$ indicate that the feasible region is in the first quadrant.

The feasible region is the area satisfying all constraints. We test the origin $(0,0)$ for each inequality:

$0 + 0 \le 50$ (True, region contains origin)

$2(0) + 0 \le 80$ (True, region contains origin)

The feasible region is the polygon formed by the intersection of the regions $x \ge 0$, $y \ge 0$, $x+y \le 50$, and $2x+y \le 80$.

The corner points of the feasible region are the vertices of this polygon. These are:

1. The origin: $(0, 0)$

2. Intersection of $y=0$ and $2x + y = 80$: $(40, 0)$

3. Intersection of $x=0$ and $x + y = 50$: $(0, 50)$

4. Intersection of $x + y = 50$ and $2x + y = 80$. Subtracting the first equation from the second:

$(2x + y) - (x + y) = 80 - 50$

$x = 30$

Substitute $x = 30$ into $x + y = 50$: $30 + y = 50 \implies y = 20$. Point: $(30, 20)$.

The corner points are $(0, 0)$, $(40, 0)$, $(0, 50)$, and $(30, 20)$.

Now we evaluate the objective function $Z = 10500x + 9000y$ at each corner point to find the maximum profit.

Corner Point $(x, y)$	$Z = 10500x + 9000y$	Value of $Z$ ($\textsf{₹}$)
$(0, 0)$	$10500(0) + 9000(0)$	$0$
$(40, 0)$	$10500(40) + 9000(0)$	$420000$
$(0, 50)$	$10500(0) + 9000(50)$	$450000$
$(30, 20)$	$10500(30) + 9000(20)$	$315000 + 180000 = 495000$

The maximum value of $Z$ is $495000$, which occurs at the corner point $(30, 20)$.

Therefore, to maximise the total profit, the society should allocate $30$ hectares of land to crop X and $20$ hectares of land to crop Y.

---

Conclusion:

The maximum profit is $\textsf{₹}495000$ when $30$ hectares are allocated to crop X and $20$ hectares are allocated to crop Y.

This is a Linear Programming Problem (LPP).

---

Given:

Labour hours for fabricating Model A: 9 hrs/piece

Labour hours for fabricating Model B: 12 hrs/piece

Maximum fabricating hours available: 180 hrs

Labour hours for finishing Model A: 1 hr/piece

Labour hours for finishing Model B: 3 hrs/piece

Maximum finishing hours available: 30 hrs

Profit per piece of Model A: $\textsf{₹}8000$

Profit per piece of Model B: $\textsf{₹}12000$

---

To Find:

The number of pieces of Model A and Model B to manufacture per week to maximise profit, and the maximum profit per week.

---

Solution:

Let $\mathbf{x}$ be the number of pieces of Model A manufactured per week.

Let $\mathbf{y}$ be the number of pieces of Model B manufactured per week.

Since the number of pieces cannot be negative, we have the non-negativity constraints:

The constraints based on the maximum labour hours available are:

Fabricating constraint:

$9x + 12y \le 180$

Dividing the inequality by 3 (which does not change its direction):

Finishing constraint:

The objective is to maximise the total profit. The profit from Model A is $\textsf{₹}8000$ per piece, and from Model B is $\textsf{₹}12000$ per piece. The total profit $Z$ is given by:

$Z = 8000x + 12000y$

The mathematical formulation of the LPP is:

Maximise $Z = 8000x + 12000y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $3x + 4y = 60$

Points for Line 1:

If $x=0$, $4y=60 \implies y=15$. Point $(0, 15)$.

If $y=0$, $3x=60 \implies x=20$. Point $(20, 0)$.

Line 2: $x + 3y = 30$

Points for Line 2:

If $x=0$, $3y=30 \implies y=10$. Point $(0, 10)$.

If $y=0$, $x=30$. Point $(30, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$) and lies below or on the lines $3x + 4y = 60$ and $x + 3y = 30$. Checking the origin $(0,0)$ in both inequalities ($0 \le 60$ and $0 \le 30$) confirms the feasible region is towards the origin from both lines.

The corner points of the feasible region are the vertices of the polygon formed by the intersection of the regions. These are:

1. The origin: $(0, 0)$

2. Intersection of $y=0$ and $3x + 4y = 60$: $(20, 0)$

3. Intersection of $x=0$ and $x + 3y = 30$: $(0, 10)$

4. Intersection of $3x + 4y = 60$ and $x + 3y = 30$. We solve this system of linear equations:

Multiply the second equation by 3:

$3(x + 3y) = 3(30) \implies 3x + 9y = 90$

Subtract the first equation ($3x + 4y = 60$) from this new equation:

$(3x + 9y) - (3x + 4y) = 90 - 60$

$5y = 30$

$y = \frac{30}{5} = 6$

Substitute $y=6$ into the second equation $x + 3y = 30$:

$x + 3(6) = 30$

$x + 18 = 30$

$x = 30 - 18 = 12$

The intersection point is $(12, 6)$.

The corner points of the feasible region are $(0, 0)$, $(20, 0)$, $(0, 10)$, and $(12, 6)$.

Now, we evaluate the objective function $Z = 8000x + 12000y$ at each corner point to find the maximum profit.

Corner Point $(x, y)$	$Z = 8000x + 12000y$	Value of $Z$ ($\textsf{₹}$)
$(0, 0)$	$8000(0) + 12000(0)$	$0$
$(20, 0)$	$8000(20) + 12000(0)$	$160000$
$(0, 10)$	$8000(0) + 12000(10)$	$120000$
$(12, 6)$	$8000(12) + 12000(6)$	$96000 + 72000 = 168000$

The maximum value of $Z$ is $\textsf{₹}168000$, which occurs at the corner point $(12, 6)$.

This means that the company should manufacture 12 pieces of Model A and 6 pieces of Model B per week to realise the maximum profit.

---

Conclusion:

The maximum profit per week is $\textsf{₹}168000$, which is obtained by manufacturing 12 pieces of Model A and 6 pieces of Model B.

This is a Linear Programming Problem (LPP).

---

Given:

Cost of Food P = $\textsf{₹}60$/kg

Cost of Food Q = $\textsf{₹}80$/kg

Vitamin A in Food P = 3 units/kg

Vitamin B in Food P = 5 units/kg

Vitamin A in Food Q = 4 units/kg

Vitamin B in Food Q = 2 units/kg

Minimum Vitamin A required = 8 units

Minimum Vitamin B required = 11 units

---

To Find:

The minimum cost of the mixture and the quantities of food P and Q in the mixture.

---

Solution:

Let $\mathbf{x}$ be the quantity of food P mixed (in kg).

Let $\mathbf{y}$ be the quantity of food Q mixed (in kg).

Since the quantities of food cannot be negative, we have the non-negativity constraints:

The constraints based on the minimum vitamin requirements are:

Vitamin A requirement:

Vitamin B requirement:

The objective is to minimise the total cost. The cost of the mixture $Z$ is given by:

$Z = 60x + 80y$

The mathematical formulation of the LPP is:

Minimise $Z = 60x + 80y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $3x + 4y = 8$

Points for Line 1:

If $x=0$, $4y=8 \implies y=2$. Point $(0, 2)$.

If $y=0$, $3x=8 \implies x=8/3$. Point $(8/3, 0)$.

Line 2: $5x + 2y = 11$

Points for Line 2:

If $x=0$, $2y=11 \implies y=5.5$. Point $(0, 5.5)$.

If $y=0$, $5x=11 \implies x=11/5 = 2.2$. Point $(2.2, 0)$.

The non-negativity constraints $x \ge 0$ and $y \ge 0$ indicate that the feasible region is in the first quadrant. The inequalities $3x + 4y \ge 8$ and $5x + 2y \ge 11$ mean the feasible region is above or on these lines when considering the first quadrant.

The corner points of the feasible region are the vertices of the polygon formed by the intersection of the regions. These are:

1. The intersection of $y=0$ and $3x + 4y = 8$: $(8/3, 0)$.

2. The intersection of $x=0$ and $5x + 2y = 11$: $(0, 5.5)$.

3. The intersection of $3x + 4y = 8$ and $5x + 2y = 11$. We solve this system of linear equations:

Multiply the second equation by 2:

$2(5x + 2y) = 2(11) \implies 10x + 4y = 22$

Subtract the first equation ($3x + 4y = 8$) from this new equation:

$(10x + 4y) - (3x + 4y) = 22 - 8$

$7x = 14$

$x = \frac{14}{7} = 2$

Substitute $x=2$ into the equation $3x + 4y = 8$:

$3(2) + 4y = 8$

$6 + 4y = 8$

$4y = 8 - 6 = 2$

$y = \frac{2}{4} = 0.5$

The intersection point is $(2, 0.5)$.

The corner points of the feasible region are $(8/3, 0)$, $(0, 5.5)$, and $(2, 0.5)$. Note that the feasible region is unbounded.

Now, we evaluate the objective function $Z = 60x + 80y$ at each corner point to find the minimum cost.

Corner Point $(x, y)$	$Z = 60x + 80y$	Value of $Z$ ($\textsf{₹}$)
$(8/3, 0)$	$60(8/3) + 80(0)$	$20 \times 8 + 0 = 160$
$(0, 5.5)$	$60(0) + 80(5.5)$	$0 + 440 = 440$
$(2, 0.5)$	$60(2) + 80(0.5)$	$120 + 40 = 160$

The minimum value of $Z$ is $\textsf{₹}160$, which occurs at two corner points: $(8/3, 0)$ and $(2, 0.5)$.

For an unbounded feasible region with a minimisation problem, if the objective function has positive coefficients (which $60$ and $80$ are), the minimum value occurs at a corner point. If the minimum is the same at two corner points, it is also the minimum at every point on the line segment joining these two points.

---

Conclusion:

The minimum cost of the mixture is $\textsf{₹}160$. This minimum cost is achieved by mixing $8/3$ kg of Food P and 0 kg of Food Q, OR by mixing 2 kg of Food P and 0.5 kg of Food Q, OR by mixing any combination of Food P and Food Q represented by a point on the line segment joining the points $(8/3, 0)$ and $(2, 0.5)$.

This is a Linear Programming Problem (LPP).

---

Given:

Type 1 cake requires 200g flour and 25g fat.

Type 2 cake requires 100g flour and 50g fat.

Total flour available = 5 kg = 5000 g.

Total fat available = 1 kg = 1000 g.

---

To Find:

The maximum number of cakes that can be made.

---

Solution:

Let $\mathbf{x}$ be the number of cakes of the first kind.

Let $\mathbf{y}$ be the number of cakes of the second kind.

Since the number of cakes cannot be negative, we have the non-negativity constraints:

The constraints based on the availability of flour and fat are:

Flour constraint (in grams):

$200x + 100y \le 5000$

Dividing the inequality by 100:

Fat constraint (in grams):

$25x + 50y \le 1000$

Dividing the inequality by 25:

The objective is to maximise the total number of cakes, which is the sum of the number of cakes of the first kind and the number of cakes of the second kind. The objective function $Z$ is given by:

$Z = x + y$

The mathematical formulation of the LPP is:

Maximise $Z = x + y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $2x + y = 50$

Points for Line 1:

If $x=0$, $y=50$. Point $(0, 50)$.

If $y=0$, $2x=50 \implies x=25$. Point $(25, 0)$.

Line 2: $x + 2y = 40$

Points for Line 2:

If $x=0$, $2y=40 \implies y=20$. Point $(0, 20)$.

If $y=0$, $x=40$. Point $(40, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$) and lies below or on the lines $2x + y = 50$ and $x + 2y = 40$. Checking the origin $(0,0)$ in both inequalities ($0 \le 50$ and $0 \le 40$) confirms the feasible region is towards the origin from both lines.

The corner points of the feasible region are the vertices of the polygon formed by the intersection of the regions. These are:

1. The origin: $(0, 0)$

2. Intersection of $y=0$ and $2x + y = 50$: $(25, 0)$

3. Intersection of $x=0$ and $x + 2y = 40$: $(0, 20)$

4. Intersection of $2x + y = 50$ and $x + 2y = 40$. We solve this system of linear equations:

From $2x + y = 50$, we get $y = 50 - 2x$. Substitute this into the second equation:

$x + 2(50 - 2x) = 40$

$x + 100 - 4x = 40$

$-3x = 40 - 100$

$-3x = -60$

$x = \frac{-60}{-3} = 20$

Substitute $x=20$ back into $y = 50 - 2x$:

$y = 50 - 2(20) = 50 - 40 = 10$

The intersection point is $(20, 10)$.

The corner points of the feasible region are $(0, 0)$, $(25, 0)$, $(0, 20)$, and $(20, 10)$.

Now, we evaluate the objective function $Z = x + y$ at each corner point to find the maximum number of cakes.

Corner Point $(x, y)$	$Z = x + y$	Value of $Z$ (Total Cakes)
$(0, 0)$	$0 + 0$	$0$
$(25, 0)$	$25 + 0$	$25$
$(0, 20)$	$0 + 20$	$20$
$(20, 10)$	$20 + 10$	$30$

The maximum value of $Z$ is $30$, which occurs at the corner point $(20, 10)$.

This means that the maximum number of cakes that can be made is 30, by manufacturing 20 cakes of the first kind and 10 cakes of the second kind.

---

Conclusion:

The maximum number of cakes that can be made is $30$.

This problem involves resource allocation and profit calculation.

---

Given:

Resources required for Tennis Racket:

Machine time: 1.5 hours

Craftsman's time: 3 hours

Resources required for Cricket Bat:

Machine time: 3 hours

Craftsman's time: 1 hour

Maximum daily machine time available: 42 hours

Maximum daily craftsman's time available: 24 hours

Profit per Tennis Racket: $\textsf{₹}20$

Profit per Cricket Bat: $\textsf{₹}10$

---

To Find:

(i) The number of rackets and bats to make if the factory works at full capacity.

(ii) The maximum profit when the factory works at full capacity, given the specified profits per item.

---

Solution:

Let $\mathbf{x}$ be the number of tennis rackets made per day.

Let $\mathbf{y}$ be the number of cricket bats made per day.

Since the number of items cannot be negative, we have $x \ge 0$ and $y \ge 0$.

The resource constraints are:

Machine time: $1.5x + 3y \le 42$

Craftsman's time: $3x + y \le 24$

The profit function is $Z = 20x + 10y$.

---

(i) Number of rackets and bats at full capacity:

Working at full capacity means using the maximum available hours for both machine time and craftsman's time. This implies the constraints become equalities:

We need to solve this system of linear equations for $x$ and $y$.

From equation (2), we can express $y$ in terms of $x$:

$y = 24 - 3x$

Substitute this expression for $y$ into equation (1):

$1.5x + 3(24 - 3x) = 42$

$1.5x + 72 - 9x = 42$

Combine the $x$ terms:

$(1.5 - 9)x + 72 = 42$

$-7.5x = 42 - 72$

$-7.5x = -30$

Solve for $x$:

$x = \frac{-30}{-7.5} = \frac{30}{7.5} = \frac{300}{75}$

$x = 4$

Now substitute the value of $x=4$ back into the expression for $y$:

$y = 24 - 3(4)$

$y = 24 - 12$

$y = 12$

The number of rackets is $x=4$ and the number of bats is $y=12$. These values satisfy $x \ge 0$ and $y \ge 0$.

---

(ii) Maximum profit at full capacity:

The factory works at full capacity when making 4 tennis rackets and 12 cricket bats. We use these values in the profit function $Z = 20x + 10y$ to find the profit.

$Z = 20(4) + 10(12)$

$Z = 80 + 120$

$Z = 200$

Based on the profit values given in the question ($\textsf{₹}20$ and $\textsf{₹}10$), the maximum profit is $\textsf{₹}200$. However, the typical profits for such items would be much higher, as seen in Example 8. Let's assume the profits were intended to be $\textsf{₹}8000$ and $\textsf{₹}12000$ as in the previous example, and recalculate for completeness, noting the discrepancy.

Assuming the intended profits were $\textsf{₹}8000$ per racket and $\textsf{₹}12000$ per bat (as in Example 8), let's calculate the profit with the values from part (i):

Profit $Z' = 8000x + 12000y$

$Z' = 8000(4) + 12000(12)$

$Z' = 32000 + 144000$

$Z' = 176000$

This result ($\textsf{₹}176000$) seems more reasonable for a factory's weekly profit than $\textsf{₹}200$. However, strictly following the question's stated profits of $\textsf{₹}20$ and $\textsf{₹}10$, the profit is $\textsf{₹}200$. We will state the answer based on the profits given in the question.

---

Conclusion:

(i) To work at full capacity, the factory must make 4 tennis rackets and 12 cricket bats per day.

(ii) Based on the profits of $\textsf{₹}20$ per racket and $\textsf{₹}10$ per bat, the maximum profit when working at full capacity is $\textsf{₹}200$.

This is a Linear Programming Problem (LPP).

---

Given:

Time required for a package of nuts:

Machine A: 1 hour

Machine B: 3 hours

Time required for a package of bolts:

Machine A: 3 hours

Machine B: 1 hour

Maximum daily hours available for Machine A: 12 hours

Maximum daily hours available for Machine B: 12 hours

Profit per package of nuts: $\textsf{₹}17.50$

Profit per package of bolts: $\textsf{₹}7.00$

---

To Find:

The number of packages of nuts and bolts to produce each day to maximise the profit, and the maximum profit.

---

Solution:

Let $\mathbf{x}$ be the number of packages of nuts produced per day.

Let $\mathbf{y}$ be the number of packages of bolts produced per day.

Since the number of packages cannot be negative, we have the non-negativity constraints:

$x \ge 0$

$y \ge 0$

The constraints based on the maximum machine hours available are:

Machine A constraint:

$1x + 3y \le 12$

Machine B constraint:

$3x + 1y \le 12$

The objective is to maximise the total profit. The profit function $Z$ is given by:

$Z = 17.50x + 7.00y$

The mathematical formulation of the LPP is:

Maximise $Z = 17.5x + 7y$

Subject to the constraints:

$x + 3y \le 12$

$3x + y \le 12$

$x \ge 0, y \ge 0$

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $x + 3y = 12$

Points for Line 1:

If $x=0$, $3y=12 \implies y=4$. Point $(0, 4)$.

If $y=0$, $x=12$. Point $(12, 0)$.

Line 2: $3x + y = 12$

Points for Line 2:

If $x=0$, $y=12$. Point $(0, 12)$.

If $y=0$, $3x=12 \implies x=4$. Point $(4, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$) and lies below or on the lines $x + 3y = 12$ and $3x + y = 12$. Testing the origin $(0,0)$ in both inequalities ($0 \le 12$ and $0 \le 12$) confirms the feasible region is towards the origin from both lines.

The corner points of the feasible region are the vertices of the polygon formed by the intersection of the regions. These are:

1. The origin: $(0, 0)$

2. Intersection of $y=0$ and $3x + y = 12$: $(4, 0)$

3. Intersection of $x=0$ and $x + 3y = 12$: $(0, 4)$

4. Intersection of $x + 3y = 12$ and $3x + y = 12$. We solve this system of linear equations:

From $3x + y = 12$, we have $y = 12 - 3x$. Substitute this into the first equation:

$x + 3(12 - 3x) = 12$

$x + 36 - 9x = 12$

$-8x = 12 - 36$

$-8x = -24$

$x = \frac{-24}{-8} = 3$

Substitute $x=3$ back into the expression for $y$:

$y = 12 - 3(3) = 12 - 9 = 3$

The intersection point is $(3, 3)$.

The corner points of the feasible region are $(0, 0)$, $(4, 0)$, $(0, 4)$, and $(3, 3)$.

Now, we evaluate the objective function $Z = 17.5x + 7y$ at each corner point to find the maximum profit.

Corner Point $(x, y)$	$Z = 17.5x + 7y$	Value of $Z$ ($\textsf{₹}$)
$(0, 0)$	$17.5(0) + 7(0)$	$0$
$(4, 0)$	$17.5(4) + 7(0)$	$70$
$(0, 4)$	$17.5(0) + 7(4)$	$28$
$(3, 3)$	$17.5(3) + 7(3)$	$52.5 + 21 = 73.5$

The maximum value of $Z$ is $\textsf{₹}73.5$, which occurs at the corner point $(3, 3)$.

This means that the manufacturer should produce 3 packages of nuts and 3 packages of bolts each day to maximise profit.

---

Conclusion:

To maximise profit, the manufacturer should produce 3 packages of nuts and 3 packages of bolts each day. The maximum profit is $\textsf{₹}73.50$.

This is a Linear Programming Problem (LPP).

---

Given:

Time required for a package of screws A:

Automatic Machine: 4 minutes

Hand Operated Machine: 6 minutes

Time required for a package of screws B:

Automatic Machine: 6 minutes

Hand Operated Machine: 3 minutes

Maximum daily availability for Automatic Machine: 4 hours = $4 \times 60 = 240$ minutes.

Maximum daily availability for Hand Operated Machine: 4 hours = $4 \times 60 = 240$ minutes.

Profit per package of screws A: $\textsf{₹}7$

Profit per package of screws B: $\textsf{₹}10$

---

To Find:

The number of packages of each type of screw (A and B) to produce per day to maximise the profit, and the maximum profit.

---

Solution:

Let $\mathbf{x}$ be the number of packages of screws A produced per day.

Let $\mathbf{y}$ be the number of packages of screws B produced per day.

Since the number of packages cannot be negative, we have the non-negativity constraints:

The constraints based on the maximum machine availability (in minutes) are:

Automatic Machine constraint:

$4x + 6y \le 240$

Dividing the inequality by 2:

Hand Operated Machine constraint:

$6x + 3y \le 240$

Dividing the inequality by 3:

The objective is to maximise the total profit. The profit function $Z$ is given by:

$Z = 7x + 10y$

The mathematical formulation of the LPP is:

Maximise $Z = 7x + 10y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $2x + 3y = 120$

Points for Line 1:

If $x=0$, $3y=120 \implies y=40$. Point $(0, 40)$.

If $y=0$, $2x=120 \implies x=60$. Point $(60, 0)$.

Line 2: $2x + y = 80$

Points for Line 2:

If $x=0$, $y=80$. Point $(0, 80)$.

If $y=0$, $2x=80 \implies x=40$. Point $(40, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$) and lies below or on the lines $2x + 3y = 120$ and $2x + y = 80$. Checking the origin $(0,0)$ in both inequalities ($0 \le 120$ and $0 \le 80$) confirms the feasible region is towards the origin from both lines.

The corner points of the feasible region are the vertices of the polygon formed by the intersection of the regions. These are:

1. The origin: $(0, 0)$

2. Intersection of $y=0$ and $2x + y = 80$: $(40, 0)$

3. Intersection of $x=0$ and $2x + 3y = 120$: $(0, 40)$

4. Intersection of $2x + 3y = 120$ and $2x + y = 80$. We solve this system of linear equations:

Subtract the second equation from the first:

$(2x + 3y) - (2x + y) = 120 - 80$

$2y = 40$

$y = \frac{40}{2} = 20$

Substitute $y=20$ back into the second equation $2x + y = 80$:

$2x + 20 = 80$

$2x = 80 - 20$

$2x = 60$

$x = \frac{60}{2} = 30$

The intersection point is $(30, 20)$.

The corner points of the feasible region are $(0, 0)$, $(40, 0)$, $(0, 40)$, and $(30, 20)$.

Now, we evaluate the objective function $Z = 7x + 10y$ at each corner point to find the maximum profit.

Corner Point $(x, y)$	$Z = 7x + 10y$	Value of $Z$ ($\textsf{₹}$)
$(0, 0)$	$7(0) + 10(0)$	$0$
$(40, 0)$	$7(40) + 10(0)$	$280$
$(0, 40)$	$7(0) + 10(40)$	$400$
$(30, 20)$	$7(30) + 10(20)$	$210 + 200 = 410$

The maximum value of $Z$ is $\textsf{₹}410$, which occurs at the corner point $(30, 20)$.

This means that the factory owner should produce 30 packages of screws A and 20 packages of screws B each day to maximise profit.

---

Conclusion:

To maximise profit, the factory owner should produce 30 packages of screws A and 20 packages of screws B each day. The maximum profit is $\textsf{₹}410$.

This is a Linear Programming Problem (LPP).

---

Given:

Resources required for Pedestal Lamp:

Grinding/Cutting Machine: 2 hours

Sprayer: 3 hours

Resources required for Wooden Shade:

Grinding/Cutting Machine: 1 hour

Sprayer: 2 hours

Maximum daily availability for Grinding/Cutting Machine: 12 hours

Maximum daily availability for Sprayer: 20 hours

Profit per Pedestal Lamp: $\textsf{₹}5$

Profit per Wooden Shade: $\textsf{₹}3$

---

To Find:

The number of pedestal lamps and wooden shades to manufacture per day to maximise the profit, and the maximum profit.

---

Solution:

Let $\mathbf{x}$ be the number of pedestal lamps manufactured per day.

Let $\mathbf{y}$ be the number of wooden shades manufactured per day.

Since the number of items cannot be negative, we have the non-negativity constraints:

The constraints based on the maximum machine hours available are:

Grinding/Cutting Machine constraint:

Sprayer constraint:

The objective is to maximise the total profit. The profit function $Z$ is given by:

$Z = 5x + 3y$

The mathematical formulation of the LPP is:

Maximise $Z = 5x + 3y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $2x + y = 12$

Points for Line 1:

If $x=0$, $y=12$. Point $(0, 12)$.

If $y=0$, $2x=12 \implies x=6$. Point $(6, 0)$.

Line 2: $3x + 2y = 20$

Points for Line 2:

If $x=0$, $2y=20 \implies y=10$. Point $(0, 10)$.

If $y=0$, $3x=20 \implies x=20/3$. Point $(20/3, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$) and lies below or on the lines $2x + y = 12$ and $3x + 2y = 20$. Checking the origin $(0,0)$ in both inequalities ($0 \le 12$ and $0 \le 20$) confirms the feasible region is towards the origin from both lines.

The corner points of the feasible region are the vertices of the polygon formed by the intersection of the regions. These are:

1. The origin: $(0, 0)$

2. Intersection of $y=0$ and $2x + y = 12$: $(6, 0)$

3. Intersection of $x=0$ and $3x + 2y = 20$: $(0, 10)$

4. Intersection of $2x + y = 12$ and $3x + 2y = 20$. We solve this system of linear equations:

From the first equation, $y = 12 - 2x$. Substitute this into the second equation:

$3x + 2(12 - 2x) = 20$

$3x + 24 - 4x = 20$

$-x = 20 - 24$

$-x = -4 \implies x = 4$

Substitute $x=4$ back into $y = 12 - 2x$:

$y = 12 - 2(4) = 12 - 8 = 4$

The intersection point is $(4, 4)$.

The corner points of the feasible region are $(0, 0)$, $(6, 0)$, $(0, 10)$, and $(4, 4)$.

Now, we evaluate the objective function $Z = 5x + 3y$ at each corner point to find the maximum profit.

Corner Point $(x, y)$	$Z = 5x + 3y$	Value of $Z$ ($\textsf{₹}$)
$(0, 0)$	$5(0) + 3(0)$	$0$
$(6, 0)$	$5(6) + 3(0)$	$30$
$(0, 10)$	$5(0) + 3(10)$	$30$
$(4, 4)$	$5(4) + 3(4)$	$20 + 12 = 32$

The maximum value of $Z$ is $\textsf{₹}32$, which occurs at the corner point $(4, 4)$.

This means that the manufacturer should produce 4 pedestal lamps and 4 wooden shades each day to maximise profit.

---

Conclusion:

To maximise profit, the factory owner should schedule his daily production to make 4 pedestal lamps and 4 wooden shades. The maximum profit is $\textsf{₹}32$.

This is a Linear Programming Problem (LPP).

---

Given:

Time required for one souvenir of type A:

Cutting: 5 minutes

Assembling: 10 minutes

Time required for one souvenir of type B:

Cutting: 8 minutes

Assembling: 8 minutes

Total time available for Cutting: 3 hours 20 minutes = $3 \times 60 + 20 = 180 + 20 = 200$ minutes.

Total time available for Assembling: 4 hours = $4 \times 60 = 240$ minutes.

Profit per souvenir of type A: $\textsf{₹}5$

Profit per souvenir of type B: $\textsf{₹}6$

---

To Find:

The number of souvenirs of each type (A and B) to manufacture per day to maximise the profit, and the maximum profit.

---

Solution:

Let $\mathbf{x}$ be the number of souvenirs of type A manufactured per day.

Let $\mathbf{y}$ be the number of souvenirs of type B manufactured per day.

Since the number of souvenirs cannot be negative, we have the non-negativity constraints:

The constraints based on the maximum available time (in minutes) are:

Cutting constraint:

Assembling constraint:

$10x + 8y \le 240$

Dividing the inequality by 2:

The objective is to maximise the total profit. The profit function $Z$ is given by:

$Z = 5x + 6y$

The mathematical formulation of the LPP is:

Maximise $Z = 5x + 6y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $5x + 8y = 200$

Points for Line 1:

If $x=0$, $8y=200 \implies y=200/8=25$. Point $(0, 25)$.

If $y=0$, $5x=200 \implies x=40$. Point $(40, 0)$.

Line 2: $5x + 4y = 120$

Points for Line 2:

If $x=0$, $4y=120 \implies y=30$. Point $(0, 30)$.

If $y=0$, $5x=120 \implies x=120/5=24$. Point $(24, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$) and lies below or on the lines $5x + 8y = 200$ and $5x + 4y = 120$. Checking the origin $(0,0)$ in both inequalities ($0 \le 200$ and $0 \le 120$) confirms the feasible region is towards the origin from both lines.

The corner points of the feasible region are the vertices of the polygon formed by the intersection of the regions. These are:

1. The origin: $(0, 0)$

2. Intersection of $y=0$ and $5x + 4y = 120$: $(24, 0)$

3. Intersection of $x=0$ and $5x + 8y = 200$: $(0, 25)$

4. Intersection of $5x + 8y = 200$ and $5x + 4y = 120$. We solve this system of linear equations:

Subtract the second equation from the first:

$(5x + 8y) - (5x + 4y) = 200 - 120$

$4y = 80$

$y = \frac{80}{4} = 20$

Substitute $y=20$ back into the second equation $5x + 4y = 120$:

$5x + 4(20) = 120$

$5x + 80 = 120$

$5x = 120 - 80$

$5x = 40$

$x = \frac{40}{5} = 8$

The intersection point is $(8, 20)$.

The corner points of the feasible region are $(0, 0)$, $(24, 0)$, $(0, 25)$, and $(8, 20)$.

Now, we evaluate the objective function $Z = 5x + 6y$ at each corner point to find the maximum profit.

Corner Point $(x, y)$	$Z = 5x + 6y$	Value of $Z$ ($\textsf{₹}$)
$(0, 0)$	$5(0) + 6(0)$	$0$
$(24, 0)$	$5(24) + 6(0)$	$120$
$(0, 25)$	$5(0) + 6(25)$	$150$
$(8, 20)$	$5(8) + 6(20)$	$40 + 120 = 160$

The maximum value of $Z$ is $\textsf{₹}160$, which occurs at the corner point $(8, 20)$.

This means that the company should manufacture 8 souvenirs of type A and 20 souvenirs of type B each day to maximise profit.

---

Conclusion:

To maximise profit, the company should manufacture 8 souvenirs of type A and 20 souvenirs of type B per day. The maximum profit is $\textsf{₹}160$.

This is a Linear Programming Problem (LPP).

---

Given:

Cost of Desktop model = $\textsf{₹}25000$

Cost of Portable model = $\textsf{₹}40000$

Maximum total monthly demand = 250 units

Maximum investment = $\textsf{₹}70$ lakhs = $\textsf{₹}70,00,000$

Profit per Desktop model = $\textsf{₹}4500$

Profit per Portable model = $\textsf{₹}5000$

---

To Find:

The number of units of each type of computer to stock to maximise the profit, and the maximum profit.

---

Solution:

Let $\mathbf{x}$ be the number of desktop models to stock.

Let $\mathbf{y}$ be the number of portable models to stock.

Since the number of units cannot be negative, we have the non-negativity constraints:

The constraints based on the total demand and investment limit are:

Total demand constraint:

Investment limit constraint:

$25000x + 40000y \le 7000000$

Dividing the inequality by 10000:

$2.5x + 4y \le 700$

Multiplying by 2 to remove decimal:

The objective is to maximise the total profit. The profit function $Z$ is given by:

$Z = 4500x + 5000y$

The mathematical formulation of the LPP is:

Maximise $Z = 4500x + 5000y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $x + y = 250$

Points for Line 1:

If $x=0$, $y=250$. Point $(0, 250)$.

If $y=0$, $x=250$. Point $(250, 0)$.

Line 2: $5x + 8y = 1400$

Points for Line 2:

If $x=0$, $8y=1400 \implies y=1400/8=175$. Point $(0, 175)$.

If $y=0$, $5x=1400 \implies x=1400/5=280$. Point $(280, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$) and lies below or on the lines $x + y = 250$ and $5x + 8y = 1400$. Checking the origin $(0,0)$ in both inequalities ($0 \le 250$ and $0 \le 1400$) confirms the feasible region is towards the origin from both lines.

The corner points of the feasible region are the vertices of the polygon formed by the intersection of the regions. These are:

1. The origin: $(0, 0)$

2. Intersection of $y=0$ and $x+y = 250$: $(250, 0)$. (This point satisfies $5(250) + 8(0) = 1250 \le 1400$, so it is a feasible corner point).

3. Intersection of $x=0$ and $5x+8y = 1400$: $(0, 175)$. (This point satisfies $0 + 175 \le 250$, so it is a feasible corner point).

4. Intersection of $x + y = 250$ and $5x + 8y = 1400$. We solve this system of linear equations:

From $x + y = 250$, we get $x = 250 - y$. Substitute this into the second equation:

$5(250 - y) + 8y = 1400$

$1250 - 5y + 8y = 1400$

$3y = 1400 - 1250$

$3y = 150$

$y = \frac{150}{3} = 50$

Substitute $y=50$ back into $x = 250 - y$:

$x = 250 - 50 = 200$

The intersection point is $(200, 50)$. This point satisfies $200 \ge 0$ and $50 \ge 0$, so it is a feasible corner point.

The corner points of the feasible region are $(0, 0)$, $(250, 0)$, $(0, 175)$, and $(200, 50)$.

Now, we evaluate the objective function $Z = 4500x + 5000y$ at each corner point to find the maximum profit.

Corner Point $(x, y)$	$Z = 4500x + 5000y$	Value of $Z$ ($\textsf{₹}$)
$(0, 0)$	$4500(0) + 5000(0)$	$0$
$(250, 0)$	$4500(250) + 5000(0)$	$1125000$
$(0, 175)$	$4500(0) + 5000(175)$	$875000$
$(200, 50)$	$4500(200) + 5000(50)$	$900000 + 250000 = 1150000$

The maximum value of $Z$ is $\textsf{₹}1150000$, which occurs at the corner point $(200, 50)$.

This means that the merchant should stock 200 desktop models and 50 portable models to maximise his profit.

---

Conclusion:

To maximise profit, the merchant should stock 200 desktop models and 50 portable models. The maximum profit is $\textsf{₹}11,50,000$ (or $\textsf{₹}11.5$ lakhs).

This is a Linear Programming Problem (LPP).

---

Given:

Cost of Food F₁ = $\textsf{₹}4$/unit

Cost of Food F₂ = $\textsf{₹}6$/unit

Vitamin A in Food F₁ = 3 units/unit

Minerals in Food F₁ = 4 units/unit

Vitamin A in Food F₂ = 6 units/unit

Minerals in Food F₂ = 3 units/unit

Minimum Vitamin A required = 80 units

Minimum Minerals required = 100 units

---

To Find:

The minimum cost of the diet that meets the nutritional requirements and the quantities of each food.

---

Solution:

Let $\mathbf{x}$ be the number of units of food F₁ in the diet.

Let $\mathbf{y}$ be the number of units of food F₂ in the diet.

Since the number of units of food cannot be negative, we have the non-negativity constraints:

The constraints based on the minimum nutritional requirements are:

Vitamin A requirement:

Minerals requirement:

The objective is to minimise the total cost. The cost function $Z$ is given by:

$Z = 4x + 6y$

---

Formulation of the Linear Programming Problem:

Minimise $Z = 4x + 6y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $3x + 6y = 80$

Points for Line 1:

If $x=0$, $6y=80 \implies y=80/6 = 40/3$. Point $(0, 40/3)$.

If $y=0$, $3x=80 \implies x=80/3$. Point $(80/3, 0)$.

Line 2: $4x + 3y = 100$

Points for Line 2:

If $x=0$, $3y=100 \implies y=100/3$. Point $(0, 100/3)$.

If $y=0$, $4x=100 \implies x=25$. Point $(25, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$). The inequalities $3x + 6y \ge 80$ and $4x + 3y \ge 100$ indicate that the feasible region is above or on these lines when considering the first quadrant. The feasible region is the intersection of these areas, which is an unbounded region.

The corner points of the feasible region are the vertices formed by the intersection of the boundary lines. These are:

1. Intersection of $x=0$ and $4x + 3y = 100$: Point $(0, 100/3)$.

2. Intersection of $y=0$ and $3x + 6y = 80$: Point $(80/3, 0)$.

3. Intersection of $3x + 6y = 80$ and $4x + 3y = 100$. We solve this system of linear equations:

Multiply the second equation by 2:

$2(4x + 3y) = 2(100) \implies 8x + 6y = 200$

Subtract the first equation ($3x + 6y = 80$) from this new equation:

$(8x + 6y) - (3x + 6y) = 200 - 80$

$5x = 120$

$x = \frac{120}{5} = 24$

Substitute $x=24$ into the equation $4x + 3y = 100$:

$4(24) + 3y = 100$

$96 + 3y = 100$

$3y = 100 - 96 = 4$

$y = \frac{4}{3}$

The intersection point is $(24, 4/3)$.

The corner points of the feasible region are $(0, 100/3)$, $(80/3, 0)$, and $(24, 4/3)$.

Now, we evaluate the objective function $Z = 4x + 6y$ at each corner point to find the minimum cost.

Corner Point $(x, y)$	$Z = 4x + 6y$	Value of $Z$ ($\textsf{₹}$)
$(0, 100/3)$	$4(0) + 6(100/3)$	$0 + 200 = 200$
$(80/3, 0)$	$4(80/3) + 6(0)$	$320/3 \approx 106.67$
$(24, 4/3)$	$4(24) + 6(4/3)$	$96 + 8 = 104$

The minimum value of $Z$ is $\textsf{₹}104$, which occurs at the corner point $(24, 4/3)$.

Since the feasible region is unbounded and the coefficients of the objective function (4 and 6) are positive, the minimum value is attained at one of the corner points.

---

Conclusion:

The minimum cost for the diet is $\textsf{₹}104$. This is achieved by mixing 24 units of food F₁ and $4/3$ units of food F₂.

This is a Linear Programming Problem (LPP).

---

Given:

Nutrient content of Fertiliser F₁:

Nitrogen: 10% (0.10)

Phosphoric Acid: 6% (0.06)

Nutrient content of Fertiliser F₂:

Nitrogen: 5% (0.05)

Phosphoric Acid: 10% (0.10)

Minimum Nitrogen required = 14 kg

Minimum Phosphoric Acid required = 14 kg

Cost of Fertiliser F₁ = $\textsf{₹}6$/kg

Cost of Fertiliser F₂ = $\textsf{₹}5$/kg

---

To Find:

The quantities of each type of fertiliser (F₁ and F₂) to be used to meet the nutrient requirements at a minimum cost, and the minimum cost.

---

Solution:

Let $\mathbf{x}$ be the quantity of fertiliser F₁ used (in kg).

Let $\mathbf{y}$ be the quantity of fertiliser F₂ used (in kg).

Since the quantities of fertiliser cannot be negative, we have the non-negativity constraints:

The constraints based on the minimum nutrient requirements are:

Nitrogen requirement:

$0.10x + 0.05y \ge 14$

Multiplying by 100 to remove decimals:

$10x + 5y \ge 1400$

Dividing by 5:

Phosphoric Acid requirement:

$0.06x + 0.10y \ge 14$

Multiplying by 100 to remove decimals:

$6x + 10y \ge 1400$

Dividing by 2:

The objective is to minimise the total cost. The cost function $Z$ is given by:

$Z = 6x + 5y$

The mathematical formulation of the LPP is:

Minimise $Z = 6x + 5y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $2x + y = 280$

Points for Line 1:

If $x=0$, $y=280$. Point $(0, 280)$.

If $y=0$, $2x=280 \implies x=140$. Point $(140, 0)$.

Line 2: $3x + 5y = 700$

Points for Line 2:

If $x=0$, $5y=700 \implies y=140$. Point $(0, 140)$.

If $y=0$, $3x=700 \implies x=700/3$. Point $(700/3, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$). The inequalities $2x + y \ge 280$ and $3x + 5y \ge 700$ indicate that the feasible region is above or on these lines when considering the first quadrant. The feasible region is the intersection of these areas, which is an unbounded region.

The corner points of the feasible region are the vertices formed by the intersection of the boundary lines. These are:

1. Intersection of $y=0$ and $2x + y = 280$: Point $(140, 0)$.

2. Intersection of $x=0$ and $3x + 5y = 700$: Point $(0, 140)$.

3. Intersection of $2x + y = 280$ and $3x + 5y = 700$. We solve this system of linear equations:

From $2x + y = 280$, we have $y = 280 - 2x$. Substitute this into the second equation:

$3x + 5(280 - 2x) = 700$

$3x + 1400 - 10x = 700$

$-7x = 700 - 1400$

$-7x = -700$

$x = \frac{-700}{-7} = 100$

Substitute $x=100$ back into $y = 280 - 2x$:

$y = 280 - 2(100) = 280 - 200 = 80$

The intersection point is $(100, 80)$.

The corner points of the feasible region are $(140, 0)$, $(0, 140)$, and $(100, 80)$.

Now, we evaluate the objective function $Z = 6x + 5y$ at each corner point to find the minimum cost.

Corner Point $(x, y)$	$Z = 6x + 5y$	Value of $Z$ ($\textsf{₹}$)
$(140, 0)$	$6(140) + 5(0)$	$840$
$(0, 140)$	$6(0) + 5(140)$	$700$
$(100, 80)$	$6(100) + 5(80)$	$600 + 400 = 1000$

The minimum value of $Z$ is $\textsf{₹}700$, which occurs at the corner point $(0, 140)$.

Since the feasible region is unbounded and the coefficients of the objective function (6 and 5) are positive, the minimum value is attained at one of the corner points.

---

Conclusion:

To meet the nutrient requirements at a minimum cost, the farmer should use 0 kg of Fertiliser F₁ and 140 kg of Fertiliser F₂. The minimum cost is $\textsf{₹}700$.

Given:

Feasible region corner points: $(0, 0)$, $(5, 0)$, $(3, 4)$, and $(0, 5)$.

Objective function: $Z = px + qy$, where $p > 0$ and $q > 0$.

The maximum of $Z$ occurs at both corner points $(3, 4)$ and $(0, 5)$.

---

To Find:

The condition on $p$ and $q$ for the maximum of $Z$ to occur at both $(3, 4)$ and $(0, 5)$.

---

Solution:

If the maximum value of the objective function $Z = px + qy$ occurs at two corner points of the feasible region, say point A and point B, then the maximum value is the same at both points. This also implies that the objective function line is parallel to the edge connecting points A and B and passes through both points, meaning every point on the edge AB is an optimal solution.

In this problem, the maximum of $Z$ occurs at corner points $(3, 4)$ and $(0, 5)$. Therefore, the value of $Z$ must be the same at these two points.

Let's evaluate the objective function $Z = px + qy$ at the point $(3, 4)$:

$Z_{(3, 4)} = p(3) + q(4) = 3p + 4q$

Now, let's evaluate the objective function $Z = px + qy$ at the point $(0, 5)$:

$Z_{(0, 5)} = p(0) + q(5) = 0 + 5q = 5q$

Since the maximum value of $Z$ occurs at both $(3, 4)$ and $(0, 5)$, the values of $Z$ at these points must be equal:

Now, we solve this equation for the relationship between $p$ and $q$. Subtract $4q$ from both sides of the equation:

$3p = 5q - 4q$

$3p = q$

Thus, the condition on $p$ and $q$ for the maximum of $Z$ to occur at both $(3, 4)$ and $(0, 5)$ is $q = 3p$.

---

Conclusion:

The condition on $p$ and $q$ is $q = 3p$. Looking at the given options, this matches option (D).

The final answer is $\boxed{q = 3p}$.

This is a Linear Programming Problem (LPP).

---

Given:

Nutrient content and cost per packet (30 g) of Food P and Food Q:

	Calcium (units/packet)	Iron (units/packet)	Cholesterol (units/packet)	Vitamin A (units/packet)
Food P	12	4	6	6
Food Q	3	20	4	3

Minimum requirements:

Calcium: At least 240 units

Iron: At least 460 units

Maximum limit:

Cholesterol: At most 300 units

---

To Find:

The number of packets of each food (P and Q) to be used to minimise the amount of vitamin A in the diet, and the minimum amount of vitamin A.

---

Solution:

Let $\mathbf{x}$ be the number of packets of food P used.

Let $\mathbf{y}$ be the number of packets of food Q used.

Since the number of packets cannot be negative, we have the non-negativity constraints:

The constraints based on the nutritional requirements and limits are:

Calcium constraint (at least 240 units):

$12x + 3y \ge 240$

Dividing by 3:

Iron constraint (at least 460 units):

$4x + 20y \ge 460$

Dividing by 4:

Cholesterol constraint (at most 300 units):

$6x + 4y \le 300$

Dividing by 2:

The objective is to minimise the total amount of Vitamin A. The objective function $Z$ (amount of Vitamin A) is given by:

$Z = 6x + 3y$

The mathematical formulation of the LPP is:

Minimise $Z = 6x + 3y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $4x + y = 80$

Points for Line 1: $(20, 0)$ and $(0, 80)$.

Line 2: $x + 5y = 115$

Points for Line 2: $(115, 0)$ and $(0, 23)$.

Line 3: $3x + 2y = 150$

Points for Line 3: $(50, 0)$ and $(0, 75)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$). It is the area above or on Line 1 ($4x+y \ge 80$), above or on Line 2 ($x+5y \ge 115$), and below or on Line 3 ($3x+2y \le 150$).

The corner points of the feasible region are the vertices formed by the intersection of the boundary lines within the feasible area. We find the intersection points:

Intersection of $4x + y = 80$ and $3x + 2y = 150$:

Multiply $4x + y = 80$ by 2: $8x + 2y = 160$.

Subtract $3x + 2y = 150$: $(8x + 2y) - (3x + 2y) = 160 - 150 \implies 5x = 10 \implies x = 2$.

Substitute $x=2$ into $4x + y = 80$: $4(2) + y = 80 \implies 8 + y = 80 \implies y = 72$. Point: $(2, 72)$.

Check constraints for $(2, 72)$: $2 \ge 0, 72 \ge 0$ (True). $4(2)+72 = 80 \ge 80$ (True). $2+5(72) = 2+360 = 362 \ge 115$ (True). $3(2)+2(72) = 6+144 = 150 \le 150$ (True). This is a corner point.

Intersection of $x + 5y = 115$ and $3x + 2y = 150$:

From $x + 5y = 115$, $x = 115 - 5y$. Substitute into $3x + 2y = 150$:

$3(115 - 5y) + 2y = 150$

$345 - 15y + 2y = 150$

$345 - 13y = 150$

$-13y = 150 - 345 \implies -13y = -195 \implies y = \frac{195}{13} = 15$.

Substitute $y=15$ into $x = 115 - 5y$: $x = 115 - 5(15) = 115 - 75 = 40$. Point: $(40, 15)$.

Check constraints for $(40, 15)$: $40 \ge 0, 15 \ge 0$ (True). $4(40)+15 = 160+15 = 175 \ge 80$ (True). $40+5(15) = 40+75 = 115 \ge 115$ (True). $3(40)+2(15) = 120+30 = 150 \le 150$ (True). This is a corner point.

Intersection of $4x + y = 80$ and $x + 5y = 115$:

From $4x + y = 80$, $y = 80 - 4x$. Substitute into $x + 5y = 115$:

$x + 5(80 - 4x) = 115$

$x + 400 - 20x = 115$

$-19x = 115 - 400 \implies -19x = -285 \implies x = \frac{285}{19} = 15$.

Substitute $x=15$ into $y = 80 - 4x$: $y = 80 - 4(15) = 80 - 60 = 20$. Point: $(15, 20)$.

Check constraints for $(15, 20)$: $15 \ge 0, 20 \ge 0$ (True). $4(15)+20 = 60+20 = 80 \ge 80$ (True). $15+5(20) = 15+100 = 115 \ge 115$ (True). $3(15)+2(20) = 45+40 = 85 \le 150$ (True). This is a corner point.

The corner points of the feasible region are $(2, 72)$, $(15, 20)$, and $(40, 15)$.

Now, we evaluate the objective function $Z = 6x + 3y$ at each corner point to find the minimum amount of Vitamin A.

Corner Point $(x, y)$	$Z = 6x + 3y$	Value of $Z$ (Vitamin A units)
$(2, 72)$	$6(2) + 3(72)$	$12 + 216 = 228$
$(15, 20)$	$6(15) + 3(20)$	$90 + 60 = 150$
$(40, 15)$	$6(40) + 3(15)$	$240 + 45 = 285$

The minimum value of $Z$ is $150$, which occurs at the corner point $(15, 20)$.

---

Conclusion:

To minimise the amount of vitamin A in the diet, the dietician should use 15 packets of food P and 20 packets of food Q. The minimum amount of vitamin A in the diet will be 150 units.

This is a Linear Programming Problem (LPP).

---

Given:

Time required for 1 unit of Item M:

Machine I: 1 hour

Machine II: 2 hours

Machine III: 1 hour

Time required for 1 unit of Item N:

Machine I: 2 hours

Machine II: 1 hour

Machine III: 1.25 hours

Maximum daily availability for Machine I: 12 hours

Maximum daily availability for Machine II: 12 hours

Minimum daily requirement for Machine III: 5 hours

Profit per unit of Item M: $\textsf{₹}600$

Profit per unit of Item N: $\textsf{₹}400$

---

To Find:

The number of units of each item (M and N) to produce per day to maximise the profit, and the maximum profit.

---

Solution:

Let $\mathbf{x}$ be the number of units of item M produced per day.

Let $\mathbf{y}$ be the number of units of item N produced per day.

Since the number of units cannot be negative, we have the non-negativity constraints:

The constraints based on the machine hours are:

Machine I constraint (at most 12 hours):

Machine II constraint (at most 12 hours):

Machine III constraint (at least 5 hours):

$1x + 1.25y \ge 5$

Multiplying by 4 to remove decimal:

The objective is to maximise the total profit. The profit function $Z$ is given by:

$Z = 600x + 400y$

The mathematical formulation of the LPP is:

Maximise $Z = 600x + 400y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $x + 2y = 12$

Points for Line 1: $(12, 0)$ and $(0, 6)$.

Line 2: $2x + y = 12$

Points for Line 2: $(6, 0)$ and $(0, 12)$.

Line 3: $4x + 5y = 20$

Points for Line 3: $(5, 0)$ and $(0, 4)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$). It is the area below or on Line 1 ($x+2y \le 12$), below or on Line 2 ($2x+y \le 12$), and above or on Line 3 ($4x+5y \ge 20$).

The corner points of the feasible region are the vertices formed by the intersection of the boundary lines within the feasible area. These points are the origin (for $\le$ constraints combined with non-negativity) unless cut off by a $\ge$ constraint which moves the boundary away from the origin.

The feasible region is the polygon with vertices at the intersection points that satisfy all constraints:

1. Intersection of $y=0$ and $2x+y=12$: $(6, 0)$. Check constraints: $6 \ge 0, 0 \ge 0$ (Ok); $6+2(0)=6 \le 12$ (Ok); $2(6)+0=12 \le 12$ (Ok); $4(6)+5(0)=24 \ge 20$ (Ok). Feasible.

2. Intersection of $x+2y=12$ and $2x+y=12$. Solving the system: $x=4, y=4$. Point: $(4, 4)$. Check constraints: $4 \ge 0, 4 \ge 0$ (Ok); $4+2(4)=12 \le 12$ (Ok); $2(4)+4=12 \le 12$ (Ok); $4(4)+5(4)=16+20=36 \ge 20$ (Ok). Feasible.

3. Intersection of $x=0$ and $x+2y=12$: $(0, 6)$. Check constraints: $0 \ge 0, 6 \ge 0$ (Ok); $0+2(6)=12 \le 12$ (Ok); $2(0)+6=6 \le 12$ (Ok); $4(0)+5(6)=30 \ge 20$ (Ok). Feasible.

4. Intersection of $x=0$ and $4x+5y=20$: $(0, 4)$. Check constraints: $0 \ge 0, 4 \ge 0$ (Ok); $0+2(4)=8 \le 12$ (Ok); $2(0)+4=4 \le 12$ (Ok); $4(0)+5(4)=20 \ge 20$ (Ok). Feasible.

5. Intersection of $y=0$ and $4x+5y=20$: $(5, 0)$. Check constraints: $5 \ge 0, 0 \ge 0$ (Ok); $5+2(0)=5 \le 12$ (Ok); $2(5)+0=10 \le 12$ (Ok); $4(5)+5(0)=20 \ge 20$ (Ok). Feasible.

The corner points of the feasible region are $(6, 0)$, $(4, 4)$, $(0, 6)$, $(0, 4)$, and $(5, 0)$.

Now, we evaluate the objective function $Z = 600x + 400y$ at each corner point to find the maximum profit.

Corner Point $(x, y)$	$Z = 600x + 400y$	Value of $Z$ ($\textsf{₹}$)
$(6, 0)$	$600(6) + 400(0)$	$3600 + 0 = 3600$
$(4, 4)$	$600(4) + 400(4)$	$2400 + 1600 = 4000$
$(0, 6)$	$600(0) + 400(6)$	$0 + 2400 = 2400$
$(0, 4)$	$600(0) + 400(4)$	$0 + 1600 = 1600$
$(5, 0)$	$600(5) + 400(0)$	$3000 + 0 = 3000$

The maximum value of $Z$ is $\textsf{₹}4000$, which occurs at the corner point $(4, 4)$.

This means that the manufacturer should produce 4 units of item M and 4 units of item N per day to maximise profit.

---

Conclusion:

To maximise profit, the manufacturer should produce 4 units of item M and 4 units of item N per day. The maximum profit will be $\textsf{₹}4000$.

This is a Transportation Problem which can be formulated and solved as a Linear Programming Problem (LPP).

---

Given:

Factory capacities: P = 8 units, Q = 6 units.

Depot requirements: A = 5 units, B = 5 units, C = 4 units.

Total supply = 8 + 6 = 14 units.

Total demand = 5 + 5 + 4 = 14 units.

Since total supply equals total demand, this is a balanced transportation problem.

Transportation costs per unit are:

From/To	A (₹)	B (₹)	C (₹)
P	160	100	150
Q	100	120	100

---

To Find:

The optimal transportation plan (number of units from each factory to each depot) that minimises the total transportation cost, and the minimum cost.

---

Solution:

Let $x_1, x_2, x_3$ be the number of units transported from factory P to depots A, B, C respectively.

Let $y_1, y_2, y_3$ be the number of units transported from factory Q to depots A, B, C respectively.

The objective is to minimise the total transportation cost $Z = 160x_1 + 100x_2 + 150x_3 + 100y_1 + 120y_2 + 100y_3$.

The constraints are based on factory capacities and depot requirements:

Non-negativity constraints:

$x_1, x_2, x_3, y_1, y_2, y_3 \ge 0$

The mathematical formulation of the LPP is:

Minimise $Z = 160x_1 + 100x_2 + 150x_3 + 100y_1 + 120y_2 + 100y_3$

Subject to constraints (1) through (5) and non-negativity constraints.

To solve this using a 2D graphical method, we reduce the problem to two variables. Let $x_1 = x$ and $x_2 = y$.

From the equality constraints (1), (3), (4), and (5), we can express other variables in terms of $x$ and $y$:

The non-negativity constraints on all variables ($x_i \ge 0, y_i \ge 0$) translate to constraints on $x$ and $y$. These define the feasible region:

The feasible region for $(x, y)$ is defined by the inequalities: $x \ge 0, y \ge 0, x \le 5, y \le 5, x + y \ge 4, x + y \le 8$.

We express the objective function $Z$ in terms of $x$ and $y$ by substituting the variable expressions:

$Z(x,y) = 160x + 100y + 150(8-x-y) + 100(5-x) + 120(5-y) + 100(x+y-4)$

Simplifying the expression for $Z(x,y)$ gives:

$Z(x,y) = 160x + 100y + 1200 - 150x - 150y + 500 - 100x + 600 - 120y + 100x + 100y - 400$

$Z(x,y) = (160 - 150 - 100 + 100)x + (100 - 150 - 120 + 100)y + (1200 + 500 + 600 - 400)$

$Z(x,y) = 10x - 70y + 1900$

We need to minimise $Z = 10x - 70y + 1900$ subject to the feasible region constraints.

The corner points of this feasible region in the $(x, y)$ plane are:

• Intersection of $y=0$ and $x+y=4$: $(4,0)$
• Intersection of $y=0$ and $x=5$: $(5,0)$
• Intersection of $x=5$ and $x+y=8$: $(5,3)$
• Intersection of $y=5$ and $x+y=8$: $(3,5)$
• Intersection of $x=0$ and $y=5$: $(0,5)$
• Intersection of $x=0$ and $x+y=4$: $(0,4)$

The corner points are $(4, 0)$, $(5, 0)$, $(5, 3)$, $(3, 5)$, $(0, 5)$, and $(0, 4)$.

Now, we evaluate the objective function $Z = 10x - 70y + 1900$ at each corner point:

Corner Point $(x, y)$	$Z = 10x - 70y + 1900$	Value of $Z$ (₹)
$(4, 0)$	$10(4) - 70(0) + 1900$	$40 + 1900 = 1940$
$(5, 0)$	$10(5) - 70(0) + 1900$	$50 + 1900 = 1950$
$(5, 3)$	$10(5) - 70(3) + 1900$	$50 - 210 + 1900 = 1740$
$(3, 5)$	$10(3) - 70(5) + 1900$	$30 - 350 + 1900 = 1580$
$(0, 5)$	$10(0) - 70(5) + 1900$	$0 - 350 + 1900 = 1550$
$(0, 4)$	$10(0) - 70(4) + 1900$	$0 - 280 + 1900 = 1620$

The minimum value of $Z$ is $\textsf{₹}1550$, which occurs at the corner point $(0, 5)$ in the $(x, y)$ plane.

The optimal values are $x=0$ and $y=5$. We use these to find the optimal shipment quantities:

Let's verify the total units shipped from each factory and received by each depot:

From P: $x_1+x_2+x_3 = 0+5+3=8$. Matches P's capacity.

From Q: $y_1+y_2+y_3 = 5+0+1=6$. Matches Q's capacity.

To A: $x_1+y_1 = 0+5=5$. Matches A's requirement.

To B: $x_2+y_2 = 5+0=5$. Matches B's requirement.

To C: $x_3+y_3 = 3+1=4$. Matches C's requirement.

---

Conclusion:

To minimise the transportation cost, the units should be transported as follows:

• From Factory P to Depot A: 0 units
• From Factory P to Depot B: 5 units
• From Factory P to Depot C: 3 units
• From Factory Q to Depot A: 5 units
• From Factory Q to Depot B: 0 units
• From Factory Q to Depot C: 1 unit

The minimum transportation cost is $\textsf{₹}1550$.

This is a Linear Programming Problem (LPP), related to Example 9 but with a different objective function.

---

Given:

The constraints and resources are the same as in Example 9.

Let $\mathbf{x}$ be the number of packets of food P used.

Let $\mathbf{y}$ be the number of packets of food Q used.

The constraints are:

The objective is to maximise the amount of Vitamin A. The Vitamin A content is 6 units per packet of Food P and 3 units per packet of Food Q. The objective function $Z$ (amount of Vitamin A) is:

$Z = 6x + 3y$

---

To Find:

The number of packets of each food (P and Q) to be used to maximise the amount of vitamin A in the diet, and the maximum amount of vitamin A.

---

Solution:

The feasible region is determined by the system of inequalities: $x \ge 0, y \ge 0, 4x + y \ge 80, x + 5y \ge 115, 3x + 2y \le 150$. This is the same feasible region as in Example 9.

The corner points of this feasible region were calculated in Example 9 as:

$(2, 72)$, $(15, 20)$, and $(40, 15)$.

To find the maximum amount of Vitamin A, we evaluate the objective function $Z = 6x + 3y$ at each of these corner points.

Corner Point $(x, y)$	$Z = 6x + 3y$	Value of $Z$ (Vitamin A units)
$(2, 72)$	$6(2) + 3(72)$	$12 + 216 = 228$
$(15, 20)$	$6(15) + 3(20)$	$90 + 60 = 150$
$(40, 15)$	$6(40) + 3(15)$	$240 + 45 = 285$

The maximum value of $Z$ is $285$, which occurs at the corner point $(40, 15)$.

This means that to maximise the amount of Vitamin A while meeting the nutritional requirements and cholesterol limit, 40 packets of food P and 15 packets of food Q should be used.

---

Conclusion:

To maximise the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used. The maximum amount of vitamin A in the diet will be 285 units.

This is a Linear Programming Problem (LPP).

---

Given:

Nutrient content and cost per bag of Brand P and Brand Q:

	Nutrient A (units/bag)	Nutrient B (units/bag)	Nutrient C (units/bag)	Cost (₹/bag)
Brand P	3	2.5	2	250
Brand Q	1.5	11.25	3	200

Minimum requirements:

Nutrient A: At least 18 units

Nutrient B: At least 45 units

Nutrient C: At least 24 units

---

To Find:

The number of bags of each brand (P and Q) to be mixed to produce a mixture having a minimum cost, and the minimum cost of the mixture per bag.

---

Solution:

Let $\mathbf{x}$ be the number of bags of Brand P mixed.

Let $\mathbf{y}$ be the number of bags of Brand Q mixed.

Since the number of bags cannot be negative, we have the non-negativity constraints:

The constraints based on the minimum nutrient requirements are:

Nutrient A requirement (at least 18 units):

$3x + 1.5y \ge 18$

Multiplying by 2 to clear the decimal: $6x + 3y \ge 36$. Dividing by 3:

Nutrient B requirement (at least 45 units):

$2.5x + 11.25y \ge 45$

Multiplying by 100 to clear decimals: $250x + 1125y \ge 4500$. Dividing by 25: $10x + 45y \ge 180$. Dividing by 5:

Nutrient C requirement (at least 24 units):

The objective is to minimise the total cost. The cost function $Z$ is given by:

$Z = 250x + 200y$

The mathematical formulation of the LPP is:

Minimise $Z = 250x + 200y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $2x + y = 12$. Points: $(0, 12)$ and $(6, 0)$.

Line 2: $2x + 9y = 36$. Points: $(0, 4)$ and $(18, 0)$.

Line 3: $2x + 3y = 24$. Points: $(0, 8)$ and $(12, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$). It is the area above or on Line 1 ($2x+y \ge 12$), above or on Line 2 ($2x+9y \ge 36$), and above or on Line 3 ($2x+3y \ge 24$). This results in an unbounded feasible region.

The corner points of the feasible region are the vertices formed by the intersection of the boundary lines:

• Intersection of $y=0$ and $2x + 9y = 36$: $2x = 36 \implies x = 18$. Point $(18, 0)$. This point satisfies $2(18)+0=36 \ge 12$ and $2(18)+0=36 \ge 24$. Feasible.
• Intersection of $x=0$ and $2x + y = 12$: $y = 12$. Point $(0, 12)$. This point satisfies $2(0)+9(12)=108 \ge 36$ and $2(0)+3(12)=36 \ge 24$. Feasible.
• Intersection of $2x + y = 12$ and $2x + 3y = 24$: Subtracting the first from the second: $2y = 12 \implies y = 6$. Substitute into $2x+y=12$: $2x+6=12 \implies 2x=6 \implies x=3$. Point $(3, 6)$. This point satisfies $2(3)+9(6)=6+54=60 \ge 36$. Feasible.
• Intersection of $2x + 3y = 24$ and $2x + 9y = 36$: Subtracting the first from the second: $6y = 12 \implies y = 2$. Substitute into $2x+3y=24$: $2x+3(2)=24 \implies 2x+6=24 \implies 2x=18 \implies x=9$. Point $(9, 2)$. This point satisfies $2(9)+2=18+2=20 \ge 12$. Feasible.

The corner points of the feasible region are $(18, 0)$, $(9, 2)$, $(3, 6)$, and $(0, 12)$.

Now, we evaluate the objective function $Z = 250x + 200y$ at each corner point to find the minimum cost.

Corner Point $(x, y)$	$Z = 250x + 200y$	Value of $Z$ ($\textsf{₹}$)
$(18, 0)$	$250(18) + 200(0)$	$4500$
$(9, 2)$	$250(9) + 200(2)$	$2250 + 400 = 2650$
$(3, 6)$	$250(3) + 200(6)$	$750 + 1200 = 1950$
$(0, 12)$	$250(0) + 200(12)$	$2400$

The minimum value of $Z$ is $\textsf{₹}1950$, which occurs at the corner point $(3, 6)$.

Since the feasible region is unbounded and the coefficients of the objective function (250 and 200) are positive, the minimum value is attained at one of the corner points.

---

Conclusion:

To produce a mixture having a minimum cost per bag while meeting the nutrient requirements, the farmer should mix 3 bags of Brand P and 6 bags of Brand Q. The minimum cost of the mixture per bag is $\textsf{₹}1950$.

This is a Linear Programming Problem (LPP).

---

Given:

Vitamin content per kg of Food X and Food Y:

	Vitamin A (units/kg)	Vitamin B (units/kg)	Vitamin C (units/kg)
Food X	1	2	3
Food Y	2	2	1

Minimum requirements:

Vitamin A: At least 10 units

Vitamin B: At least 12 units

Vitamin C: At least 8 units

Cost of Food X = $\textsf{₹}16$/kg

Cost of Food Y = $\textsf{₹}20$/kg

---

To Find:

The quantities of each kind of food (X and Y) to be mixed to obtain the required diet at the least cost, and the minimum cost.

---

Solution:

Let $\mathbf{x}$ be the quantity of food X mixed (in kg).

Let $\mathbf{y}$ be the quantity of food Y mixed (in kg).

Since the quantities of food cannot be negative, we have the non-negativity constraints:

The constraints based on the minimum vitamin requirements are:

Vitamin A requirement (at least 10 units):

Vitamin B requirement (at least 12 units):

Dividing by 2:

Vitamin C requirement (at least 8 units):

The objective is to minimise the total cost. The cost function $Z$ is given by:

$Z = 16x + 20y$

The mathematical formulation of the LPP is:

Minimise $Z = 16x + 20y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $x + 2y = 10$. Points: $(0, 5)$ and $(10, 0)$.

Line 2: $x + y = 6$. Points: $(0, 6)$ and $(6, 0)$.

Line 3: $3x + y = 8$. Points: $(0, 8)$ and $(8/3, 0) = (2.67, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$). It is the area above or on Line 1 ($x+2y \ge 10$), above or on Line 2 ($x+y \ge 6$), and above or on Line 3 ($3x+y \ge 8$). This results in an unbounded feasible region.

The corner points of the feasible region are the vertices formed by the intersection of the boundary lines:

• Intersection of $y=0$ and $3x + y = 8$: $3x = 8 \implies x = 8/3$. Point $(8/3, 0)$. This satisfies $8/3+0 \ge 6$ (False, $2.67 \ge 6$ is false), $8/3+2(0)=8/3 \ge 10$ (False, $2.67 \ge 10$ is false). This point is not a vertex of the feasible region defined by all constraints. Let's re-evaluate the relevant intersection points.
• Intersection of $y=0$ and $x + y = 6$: $(6, 0)$. Satisfies $6+2(0)=6 \ge 10$ (False, $6 \ge 10$ is false). Not a vertex.
• Intersection of $y=0$ and $x+2y=10$: $(10, 0)$. Satisfies $10+0=10 \ge 6$ (True), $3(10)+0=30 \ge 8$ (True). Feasible. Point $(10, 0)$.
• Intersection of $x=0$ and $x+2y=10$: $(0, 5)$. Satisfies $0+5=5 \ge 6$ (False, $5 \ge 6$ is false). Not a vertex.
• Intersection of $x=0$ and $x+y=6$: $(0, 6)$. Satisfies $0+2(6)=12 \ge 10$ (True), $3(0)+6=6 \ge 8$ (False, $6 \ge 8$ is false). Not a vertex.
• Intersection of $x=0$ and $3x+y=8$: $(0, 8)$. Satisfies $0+2(8)=16 \ge 10$ (True), $0+8=8 \ge 6$ (True). Feasible. Point $(0, 8)$.
• Intersection of $3x + y = 8$ and $x + y = 6$: Subtracting the second from the first: $2x = 2 \implies x = 1$. Substitute into $x+y=6$: $1+y=6 \implies y=5$. Point $(1, 5)$. Satisfies $1+2(5)=11 \ge 10$ (True). Feasible. Point $(1, 5)$.
• Intersection of $x + y = 6$ and $x + 2y = 10$: Subtracting the first from the second: $y = 4$. Substitute into $x+y=6$: $x+4=6 \implies x=2$. Point $(2, 4)$. Satisfies $3(2)+4=6+4=10 \ge 8$ (True). Feasible. Point $(2, 4)$.
• Intersection of $3x + y = 8$ and $x + 2y = 10$: From $3x+y=8$, $y=8-3x$. Substitute into $x+2y=10$: $x+2(8-3x)=10 \implies x+16-6x=10 \implies -5x = 10-16 = -6 \implies x = 6/5 = 1.2$. Substitute into $y=8-3x$: $y=8-3(1.2) = 8-3.6 = 4.4$. Point $(1.2, 4.4)$. Satisfies $1.2+4.4 = 5.6 \ge 6$ (False, $5.6 \ge 6$ is false). Not a vertex.

Let's carefully identify the lines and their intersection points that form the vertices of the feasible region defined by $x \ge 0, y \ge 0, x + 2y \ge 10, x + y \ge 6, 3x + y \ge 8$. Plotting these lines will help visualize the feasible region.

Points for boundaries:

$x + 2y = 10$: $(10,0), (0,5)$

$x + y = 6$: $(6,0), (0,6)$

$3x + y = 8$: $(8/3,0), (0,8)$

The feasible region is in the first quadrant and is above all three lines. The vertices are the intersection points that lie on the boundary of this region.

The relevant intersection points are:

• Intersection of $y=0$ and $3x+y=8$ is $(8/3, 0)$. Does it satisfy $x+y \ge 6$ ($8/3 \ge 6$, False)? No.
• Intersection of $y=0$ and $x+y=6$ is $(6, 0)$. Does it satisfy $3x+y \ge 8$ ($18 \ge 8$, True) and $x+2y \ge 10$ ($6 \ge 10$, False)? No.
• Intersection of $y=0$ and $x+2y=10$ is $(10, 0)$. Satisfies all constraints ($10\ge 6$, $30\ge 8$). Vertex: $(10, 0)$.
• Intersection of $x=0$ and $x+2y=10$ is $(0, 5)$. Satisfies $x+y \ge 6$ ($5 \ge 6$, False). No.
• Intersection of $x=0$ and $x+y=6$ is $(0, 6)$. Satisfies $3x+y \ge 8$ ($6 \ge 8$, False). No.
• Intersection of $x=0$ and $3x+y=8$ is $(0, 8)$. Satisfies $x+y \ge 6$ ($8 \ge 6$, True) and $x+2y \ge 10$ ($16 \ge 10$, True). Vertex: $(0, 8)$.
• Intersection of $3x + y = 8$ and $x + y = 6$: $(1, 5)$. Satisfies $x+2y \ge 10$ ($1+10=11 \ge 10$, True). Vertex: $(1, 5)$.
• Intersection of $x + y = 6$ and $x + 2y = 10$: $(2, 4)$. Satisfies $3x+y \ge 8$ ($6+4=10 \ge 8$, True). Vertex: $(2, 4)$.

The corner points of the feasible region are $(10, 0)$, $(2, 4)$, $(1, 5)$, and $(0, 8)$.

Now, we evaluate the objective function $Z = 16x + 20y$ at each corner point to find the minimum cost.

Corner Point $(x, y)$	$Z = 16x + 20y$	Value of $Z$ ($\textsf{₹}$)
$(10, 0)$	$16(10) + 20(0)$	$160$
$(2, 4)$	$16(2) + 20(4)$	$32 + 80 = 112$
$(1, 5)$	$16(1) + 20(5)$	$16 + 100 = 116$
$(0, 8)$	$16(0) + 20(8)$	$160$

The minimum value of $Z$ is $\textsf{₹}112$, which occurs at the corner point $(2, 4)$.

Since the feasible region is unbounded and the coefficients of the objective function (16 and 20) are positive, the minimum value is attained at one of the corner points.

---

Conclusion:

To obtain the required diet at the least cost, the dietician should mix 2 kg of food X and 4 kg of food Y. The minimum cost of the mixture will be $\textsf{₹}112$.

This is a Linear Programming Problem (LPP).

---

Given:

Time required for one toy of type A:

Machine I: 12 minutes

Machine II: 18 minutes

Machine III: 6 minutes

Time required for one toy of type B:

Machine I: 6 minutes

Machine II: 0 minutes

Machine III: 9 minutes

Maximum daily availability for each machine: 6 hours = $6 \times 60 = 360$ minutes.

Profit per toy of type A: $\textsf{₹}7.50$

Profit per toy of type B: $\textsf{₹}5$

---

To Show:

That manufacturing 15 toys of type A and 30 toys of type B per day yields the maximum profit, and to determine this maximum profit.

---

Solution:

Let $\mathbf{x}$ be the number of toys of type A manufactured per day.

Let $\mathbf{y}$ be the number of toys of type B manufactured per day.

Since the number of toys cannot be negative, we have the non-negativity constraints:

The constraints based on the maximum machine availability (in minutes) are:

Machine I constraint (at most 360 minutes):

$12x + 6y \le 360$

Dividing the inequality by 6:

Machine II constraint (at most 360 minutes):

$18x + 0y \le 360$

Dividing the inequality by 18:

Machine III constraint (at most 360 minutes):

$6x + 9y \le 360$

Dividing the inequality by 3:

The objective is to maximise the total profit. The profit function $Z$ is given by:

$Z = 7.5x + 5y$

The mathematical formulation of the LPP is:

Maximise $Z = 7.5x + 5y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $2x + y = 60$. Points: $(0, 60)$, $(30, 0)$.

Line 2: $x = 20$. A vertical line at $x=20$.

Line 3: $2x + 3y = 120$. Points: $(0, 40)$, $(60, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$) and lies below or on the lines $2x + y = 60$, $x = 20$, and $2x + 3y = 120$. Checking the origin $(0,0)$ in the inequalities confirms the feasible region is towards the origin for the first and third constraints ($0 \le 60$, $0 \le 120$), and on or to the left of the line $x=20$ for the second constraint ($0 \le 20$).

The corner points of the feasible region are the vertices formed by the intersection of the boundary lines within the feasible area. These vertices are:

• Intersection of $x=0$ and $y=0$: $(0, 0)$.
• Intersection of $y=0$ and $x=20$: $(20, 0)$. Satisfies $2(20)+0 = 40 \le 60$ and $2(20)+3(0) = 40 \le 120$. Feasible.
• Intersection of $x=0$ and $2x + 3y = 120$: $3y=120 \implies y=40$. Point $(0, 40)$. Satisfies $0 \le 20$ and $2(0)+40 = 40 \le 60$. Feasible.
• Intersection of $x=20$ and $2x + y = 60$: $2(20) + y = 60 \implies 40 + y = 60 \implies y = 20$. Point $(20, 20)$. Satisfies $2(20)+3(20) = 40+60 = 100 \le 120$. Feasible.
• Intersection of $2x + y = 60$ and $2x + 3y = 120$: Subtracting the first from the second: $2y = 60 \implies y = 30$. Substitute into $2x+y=60$: $2x+30=60 \implies 2x=30 \implies x=15$. Point $(15, 30)$. Satisfies $15 \le 20$. Feasible.

The corner points of the feasible region are $(0, 0)$, $(20, 0)$, $(20, 20)$, $(15, 30)$, and $(0, 40)$.

Now, we evaluate the objective function $Z = 7.5x + 5y$ at each corner point to find the maximum profit.

Corner Point $(x, y)$	$Z = 7.5x + 5y$	Value of $Z$ ($\textsf{₹}$)
$(0, 0)$	$7.5(0) + 5(0)$	$0$
$(20, 0)$	$7.5(20) + 5(0)$	$150 + 0 = 150$
$(20, 20)$	$7.5(20) + 5(20)$	$150 + 100 = 250$
$(15, 30)$	$7.5(15) + 5(30)$	$112.5 + 150 = 262.5$
$(0, 40)$	$7.5(0) + 5(40)$	$0 + 200 = 200$

The maximum value of $Z$ is $\textsf{₹}262.5$, which occurs at the corner point $(15, 30)$.

This shows that manufacturing 15 toys of type A and 30 toys of type B per day yields the maximum profit.

---

Conclusion:

Based on the analysis of the feasible region and the evaluation of the profit function at the corner points, the maximum profit of $\textsf{₹}262.50$ is achieved when the manufacturer produces 15 toys of type A and 30 toys of type B in a day.

This is a Linear Programming Problem (LPP).

---

Given:

Maximum passenger capacity = 200.

Profit per executive ticket = $\textsf{₹}1000$.

Profit per economy ticket = $\textsf{₹}600$.

Minimum executive seats reserved = 20.

Constraint: Number of economy passengers is at least 4 times the number of executive passengers.

---

To Find:

The number of executive and economy tickets to sell to maximise profit, and the maximum profit.

---

Solution:

Let $\mathbf{x}$ be the number of executive class tickets sold.

Let $\mathbf{y}$ be the number of economy class tickets sold.

Since the number of tickets cannot be negative, we have the non-negativity constraints:

The constraints are:

Maximum passenger capacity:

Minimum executive seats reserved:

Economy passengers at least 4 times executive passengers:

$y \ge 4x$

Rearranging the inequality:

The objective is to maximise the total profit. The profit function $Z$ is given by:

$Z = 1000x + 600y$

The mathematical formulation of the LPP is:

Maximise $Z = 1000x + 600y$

Subject to the constraints:

(Note: $x \ge 0$ is implied by $x \ge 20$).

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $x + y = 200$. Points: $(0, 200)$, $(200, 0)$.

Line 2: $x = 20$. A vertical line at $x=20$.

Line 3: $y = 4x$. Points: $(0, 0)$, $(10, 40)$, $(20, 80)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$). It is the area below or on Line 1 ($x+y \le 200$), on or to the right of Line 2 ($x \ge 20$), and above or on Line 3 ($y \ge 4x$).

The corner points of the feasible region are the vertices formed by the intersection of the boundary lines within the feasible area:

• Intersection of $x = 20$ and $y = 4x$: Substitute $x=20$ into $y=4x \implies y = 4(20) = 80$. Point $(20, 80)$. Check $x+y \le 200$: $20+80=100 \le 200$. Feasible. Vertex: $(20, 80)$.
• Intersection of $x = 20$ and $x + y = 200$: Substitute $x=20$ into $x+y=200 \implies 20 + y = 200 \implies y = 180$. Point $(20, 180)$. Check $y \ge 4x$: $180 \ge 4(20) \implies 180 \ge 80$. Feasible. Vertex: $(20, 180)$.
• Intersection of $y = 4x$ and $x + y = 200$: Substitute $y=4x$ into $x+y=200 \implies x + 4x = 200 \implies 5x = 200 \implies x = 40$. Substitute $x=40$ into $y=4x \implies y = 4(40) = 160$. Point $(40, 160)$. Check $x \ge 20$: $40 \ge 20$. Feasible. Vertex: $(40, 160)$.

The corner points of the feasible region are $(20, 80)$, $(20, 180)$, and $(40, 160)$.

Now, we evaluate the objective function $Z = 1000x + 600y$ at each corner point to find the maximum profit.

Corner Point $(x, y)$	$Z = 1000x + 600y$	Value of $Z$ ($\textsf{₹}$)
$(20, 80)$	$1000(20) + 600(80)$	$20000 + 48000 = 68000$
$(20, 180)$	$1000(20) + 600(180)$	$20000 + 108000 = 128000$
$(40, 160)$	$1000(40) + 600(160)$	$40000 + 96000 = 136000$

The maximum value of $Z$ is $\textsf{₹}136000$, which occurs at the corner point $(40, 160)$.

This means that to maximise profit, the airline should sell 40 executive class tickets and 160 economy class tickets.

---

Conclusion:

To maximise the profit for the airline, 40 executive class tickets and 160 economy class tickets must be sold. The maximum profit will be $\textsf{₹}136000$.

This is a Transportation Problem which can be formulated and solved as a Linear Programming Problem (LPP).

---

Given:

Godown capacities: A = 100 quintals, B = 50 quintals.

Shop requirements: D = 60 quintals, E = 50 quintals, F = 40 quintals.

Total supply = 100 + 50 = 150 quintals.

Total demand = 60 + 50 + 40 = 150 quintals.

Since total supply equals total demand, this is a balanced transportation problem.

Transportation costs per quintal are:

From/To	D (₹)	E (₹)	F (₹)
A	6	3	2.50
B	4	2	3

---

To Find:

The optimal transportation plan (number of quintals from each godown to each shop) that minimises the total transportation cost, and the minimum cost.

---

Solution:

Let $x_1, x_2, x_3$ be the number of quintals transported from godown A to shops D, E, F respectively.

Let $y_1, y_2, y_3$ be the number of quintals transported from godown B to shops D, E, F respectively.

The objective is to minimise the total transportation cost $Z = 6x_1 + 3x_2 + 2.5x_3 + 4y_1 + 2y_2 + 3y_3$.

The constraints are based on godown capacities and shop requirements:

Non-negativity constraints:

$x_1, x_2, x_3, y_1, y_2, y_3 \ge 0$

To solve this using a 2D graphical method, we reduce the problem to two variables. Let $x_1 = x$ and $x_2 = y$.

From the equality constraints, we express other variables in terms of $x$ and $y$:

The non-negativity constraints on all variables translate to constraints on $x$ and $y$. These define the feasible region:

The feasible region for $(x, y)$ is defined by: $0 \le x \le 60$, $0 \le y \le 50$, $x + y \ge 60$, $x + y \le 100$.

We express the objective function $Z$ in terms of $x$ and $y$ by substituting the variable expressions:

$Z(x,y) = 6x + 3y + 2.5(100-x-y) + 4(60-x) + 2(50-y) + 3(x+y-60)$

$Z(x,y) = 6x + 3y + 250 - 2.5x - 2.5y + 240 - 4x + 100 - 2y + 3x + 3y - 180$

$Z(x,y) = (6 - 2.5 - 4 + 3)x + (3 - 2.5 - 2 + 3)y + (250 + 240 + 100 - 180)$

$Z(x,y) = 2.5x + 1.5y + 410$

We need to minimise $Z = 2.5x + 1.5y + 410$ subject to the feasible region constraints.

The feasible region is a polygon defined by the intersection of the regions. The boundary lines are $x=0, x=60, y=0, y=50, x+y=60, x+y=100$.

The corner points of the feasible region are:

• Intersection of $y=50$ and $x+y=60$: $x+50=60 \implies x=10$. Point $(10, 50)$. Satisfies $0 \le 10 \le 60$. Feasible.
• Intersection of $x=60$ and $x+y=60$: $60+y=60 \implies y=0$. Point $(60, 0)$. Satisfies $0 \le 0 \le 50$ (False). Not a corner.
• Intersection of $x=60$ and $x+y=100$: $60+y=100 \implies y=40$. Point $(60, 40)$. Satisfies $0 \le 40 \le 50$. Feasible.
• Intersection of $y=50$ and $x+y=100$: $x+50=100 \implies x=50$. Point $(50, 50)$. Satisfies $0 \le 50 \le 60$. Feasible.
• Intersection of $y=0$ and $x+y=60$: $(60, 0)$. Satisfies $0 \le 60 \le 60$ and $0 \le 0 \le 50$. Feasible.
• Intersection of $x=0$ and $x+y=60$: $(0, 60)$. Satisfies $0 \le 0 \le 60$ and $0 \le 60 \le 50$ (False). Not a corner.
• Intersection of $x=0$ and $y=50$: $(0, 50)$. Satisfies $0+50=50 \ge 60$ (False). Not a corner.
• Intersection of $x=60$ and $y=0$: $(60, 0)$. Already found.
• Intersection of $y=0$ and $x+y=100$: $(100, 0)$. $100 \le 60$ (False). Not a corner.

Let's list the vertices of the feasible region: $x \ge 0, x \le 60, y \ge 0, y \le 50, x+y \ge 60, x+y \le 100$.

• Intersection of $y=50$ and $x+y=60$: $(10, 50)$.
• Intersection of $y=50$ and $x+y=100$: $(50, 50)$.
• Intersection of $x=60$ and $x+y=100$: $(60, 40)$.
• Intersection of $x=60$ and $x+y=60$: $(60, 0)$ -> Point $(60,0)$ satisfies $0 \le 0 \le 50$. Vertex $(60,0)$.
• Intersection of $y=0$ and $x+y=60$: $(60, 0)$. Already listed.

The vertices are $(10, 50)$, $(50, 50)$, $(60, 40)$, and $(60, 0)$. Let's re-check the region carefully.

The feasible region is bounded by the lines $x=0, x=60, y=0, y=50, x+y=60, x+y=100$. The region $x \ge 0, y \ge 0, x \le 60, y \le 50$ is a rectangle with vertices $(0,0), (60,0), (60,50), (0,50)$. The conditions $x+y \ge 60$ and $x+y \le 100$ cut this region.

The vertices of the feasible region are the intersection points of these boundary lines that lie within the region.

Vertices are intersection of boundary lines:

• $x+y=60$ and $y=50 \implies x=10$. Point $(10, 50)$. $0 \le 10 \le 60$ (True). Feasible.
• $x+y=60$ and $x=60 \implies y=0$. Point $(60, 0)$. $0 \le 0 \le 50$ (True). Feasible.
• $x+y=100$ and $y=50 \implies x=50$. Point $(50, 50)$. $0 \le 50 \le 60$ (True). Feasible.
• $x+y=100$ and $x=60 \implies y=40$. Point $(60, 40)$. $0 \le 40 \le 50$ (True). Feasible.

The corner points of the feasible region are $(10, 50)$, $(50, 50)$, $(60, 40)$, and $(60, 0)$.

Now, we evaluate the objective function $Z = 2.5x + 1.5y + 410$ at each corner point:

Corner Point $(x, y)$	$Z = 2.5x + 1.5y + 410$	Value of $Z$ ($\textsf{₹}$)
$(10, 50)$	$2.5(10) + 1.5(50) + 410$	$25 + 75 + 410 = 510$
$(50, 50)$	$2.5(50) + 1.5(50) + 410$	$125 + 75 + 410 = 610$
$(60, 40)$	$2.5(60) + 1.5(40) + 410$	$150 + 60 + 410 = 620$
$(60, 0)$	$2.5(60) + 1.5(0) + 410$	$150 + 0 + 410 = 560$

The minimum value of $Z$ is $\textsf{₹}510$, which occurs at the corner point $(10, 50)$ in the $(x, y)$ plane.

The optimal values are $x=10$ and $y=50$. We use these to find the optimal shipment quantities:

Let's verify the total units shipped from each godown and received by each shop:

From A: $x_1+x_2+x_3 = 10+50+40=100$. Matches A's capacity.

From B: $y_1+y_2+y_3 = 50+0+0=50$. Matches B's capacity.

To D: $x_1+y_1 = 10+50=60$. Matches D's requirement.

To E: $x_2+y_2 = 50+0=50$. Matches E's requirement.

To F: $x_3+y_3 = 40+0=40$. Matches F's requirement.

---

Conclusion:

To minimise the transportation cost, the supplies should be transported as follows:

• From Godown A to Shop D: 10 quintals
• From Godown A to Shop E: 50 quintals
• From Godown A to Shop F: 40 quintals
• From Godown B to Shop D: 50 quintals
• From Godown B to Shop E: 0 quintals
• From Godown B to Shop F: 0 quintals

The minimum transportation cost is $\textsf{₹}510$.

This is a Transportation Problem which can be formulated and solved as a Linear Programming Problem (LPP).

---

Given:

Depot capacities: A = 7000 L, B = 4000 L.

Petrol pump requirements: D = 4500 L, E = 3000 L, F = 3500 L.

Total supply = $7000 + 4000 = 11000$ L.

Total demand = $4500 + 3000 + 3500 = 11000$ L.

Since total supply equals total demand, this is a balanced transportation problem.

Transportation cost: $\textsf{₹}1$ per 10 litres per km.

Cost per litre per km = $\textsf{₹}1 / 10 = \textsf{₹}0.1$.

Transportation costs per litre are:

From/To	D (₹/L)	E (₹/L)	F (₹/L)
A	$7 \times 0.1 = 0.7$	$6 \times 0.1 = 0.6$	$3 \times 0.1 = 0.3$
B	$3 \times 0.1 = 0.3$	$4 \times 0.1 = 0.4$	$2 \times 0.1 = 0.2$

---

To Find:

The optimal transportation plan (quantity of oil from each depot to each petrol pump) that minimises the total transportation cost, and the minimum cost.

---

Solution:

Let $x_1, x_2, x_3$ be the quantity of oil (in litres) transported from depot A to petrol pumps D, E, F respectively.

Let $y_1, y_2, y_3$ be the quantity of oil (in litres) transported from depot B to petrol pumps D, E, F respectively.

The objective is to minimise the total transportation cost $Z = 0.7x_1 + 0.6x_2 + 0.3x_3 + 0.3y_1 + 0.4y_2 + 0.2y_3$.

The constraints are based on depot capacities and petrol pump requirements:

Non-negativity constraints:

$x_1, x_2, x_3, y_1, y_2, y_3 \ge 0$

To solve this using a 2D graphical method, we reduce the problem to two variables. Let $x_1 = x$ and $x_2 = y$. We express other variables in terms of $x$ and $y$ using the equality constraints:

The non-negativity constraints on all variables ($x_i \ge 0, y_i \ge 0$) translate to constraints on $x$ and $y$. These define the feasible region:

The feasible region for $(x, y)$ is defined by the inequalities: $0 \le x \le 4500$, $0 \le y \le 3000$, $x + y \ge 3500$, $x + y \le 7000$.

Substitute the variable expressions into the objective function $Z$:

$Z(x,y) = 0.7x + 0.6y + 0.3(7000-x-y) + 0.3(4500-x) + 0.4(3000-y) + 0.2(x+y-3500)$

$Z(x,y) = 0.7x + 0.6y + 2100 - 0.3x - 0.3y + 1350 - 0.3x + 1200 - 0.4y + 0.2x + 0.2y - 700$

$Z(x,y) = (0.7 - 0.3 - 0.3 + 0.2)x + (0.6 - 0.3 - 0.4 + 0.2)y + (2100 + 1350 + 1200 - 700)$

$Z(x,y) = 0.3x + 0.1y + 3950$

We need to minimise $Z = 0.3x + 0.1y + 3950$ subject to the feasible region constraints. The corner points of the feasible region defined by $0 \le x \le 4500$, $0 \le y \le 3000$, $x+y \ge 3500$, $x+y \le 7000$ are found by identifying intersections of the boundary lines within the feasible region.

The corner points are:

• Intersection of $y=3000$ and $x+y=3500$: $x+3000=3500 \implies x=500$. Point $(500, 3000)$.
• Intersection of $x=4500$ and $x+y=7000$: $4500+y=7000 \implies y=2500$. Point $(4500, 2500)$.
• Intersection of $y=3000$ and $x+y=7000$: $x+3000=7000 \implies x=4000$. Point $(4000, 3000)$.
• Intersection of $y=0$ and $x+y=3500$: $x+0=3500 \implies x=3500$. Point $(3500, 0)$.
• Intersection of $y=0$ and $x=4500$: $(4500,0)$. $4500+0=4500 \ge 3500$ (True), $4500 \le 7000$ (True). Feasible. Vertex $(4500,0)$.

The corner points of the feasible region are $(500, 3000)$, $(4000, 3000)$, $(4500, 2500)$, $(3500, 0)$, and $(4500, 0)$. Let's re-verify. The region is a polygon. The vertices are where boundary lines intersect AND are feasible. The boundary lines are $x=0, x=4500, y=0, y=3000, x+y=3500, x+y=7000$. The vertices are $(500, 3000)$, $(4000, 3000)$, $(4500, 2500)$, $(4500, 0)$, $(3500, 0)$.

Now, we evaluate the objective function $Z = 0.3x + 0.1y + 3950$ at each corner point:

Corner Point $(x, y)$	$Z = 0.3x + 0.1y + 3950$	Value of $Z$ ($\textsf{₹}$)
$(500, 3000)$	$0.3(500) + 0.1(3000) + 3950$	$150 + 300 + 3950 = 4400$
$(4000, 3000)$	$0.3(4000) + 0.1(3000) + 3950$	$1200 + 300 + 3950 = 5450$
$(4500, 2500)$	$0.3(4500) + 0.1(2500) + 3950$	$1350 + 250 + 3950 = 5550$
$(4500, 0)$	$0.3(4500) + 0.1(0) + 3950$	$1350 + 3950 = 5300$
$(3500, 0)$	$0.3(3500) + 0.1(0) + 3950$	$1050 + 0 + 3950 = 5000$

The minimum value of $Z$ is $\textsf{₹}4400$, which occurs at the corner point $(500, 3000)$.

The optimal values are $x=500$ and $y=3000$. We use these to find the optimal shipment quantities:

---

Conclusion:

To minimise the transportation cost, the delivery should be scheduled as follows:

• From Depot A to Petrol Pump D: 500 litres
• From Depot A to Petrol Pump E: 3000 litres
• From Depot A to Petrol Pump F: 3500 litres
• From Depot B to Petrol Pump D: 4000 litres
• From Depot B to Petrol Pump E: 0 litres
• From Depot B to Petrol Pump F: 0 litres

The minimum transportation cost is $\textsf{₹}4400$.

This is a Linear Programming Problem (LPP).

---

Given:

Nutrient content per bag of Brand P and Brand Q (in kg):

	Nitrogen (kg/bag)	Phosphoric Acid (kg/bag)	Potash (kg/bag)	Chlorine (kg/bag)
Brand P	3	1	3	1.5
Brand Q	3.5	2	1.5	2

Nutrient requirements and limits:

Phosphoric Acid: At least 240 kg

Potash: At least 270 kg

Chlorine: At most 310 kg

---

To Find:

The number of bags of each brand (P and Q) to be used to minimise the amount of nitrogen added, and the minimum amount of nitrogen.

---

Solution:

Let $\mathbf{x}$ be the number of bags of Brand P used.

Let $\mathbf{y}$ be the number of bags of Brand Q used.

Since the number of bags cannot be negative, we have the non-negativity constraints:

The constraints based on the nutrient requirements and limits are:

Phosphoric Acid constraint (at least 240 kg):

Potash constraint (at least 270 kg):

$3x + 1.5y \ge 270$

Multiplying by 2 to remove decimal: $6x + 3y \ge 540$. Dividing by 3:

Chlorine constraint (at most 310 kg):

$1.5x + 2y \le 310$

Multiplying by 2 to remove decimal: $3x + 4y \le 620$.

The objective is to minimise the total amount of Nitrogen. The amount of Nitrogen $Z$ is given by:

$Z = 3x + 3.5y$

The mathematical formulation of the LPP is:

Minimise $Z = 3x + 3.5y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $x + 2y = 240$. Points: $(0, 120)$, $(240, 0)$.

Line 2: $2x + y = 180$. Points: $(0, 180)$, $(90, 0)$.

Line 3: $3x + 4y = 620$. Points: $(0, 155)$, $(620/3, 0) \approx (206.67, 0)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$). It is the area above or on Line 1 ($x+2y \ge 240$), above or on Line 2 ($2x+y \ge 180$), and below or on Line 3 ($3x+4y \le 620$).

The corner points of the feasible region are the vertices formed by the intersection of the boundary lines within the feasible area. These vertices are:

• Intersection of $x + 2y = 240$ and $2x + y = 180$. Multiply first by 2: $2x + 4y = 480$. Subtract $2x + y = 180$: $3y = 300 \implies y = 100$. Substitute into $2x+y=180$: $2x+100=180 \implies 2x=80 \implies x=40$. Point $(40, 100)$. Check $3x+4y \le 620$: $3(40)+4(100) = 120+400 = 520 \le 620$. Feasible. Vertex: $(40, 100)$.
• Intersection of $2x + y = 180$ and $3x + 4y = 620$. From $2x+y=180$, $y=180-2x$. Substitute into $3x+4y=620$: $3x+4(180-2x)=620 \implies 3x+720-8x=620 \implies -5x = 620-720 = -100 \implies x=20$. Substitute into $y=180-2x$: $y=180-2(20) = 180-40 = 140$. Point $(20, 140)$. Check $x+2y \ge 240$: $20+2(140) = 20+280 = 300 \ge 240$. Feasible. Vertex: $(20, 140)$.
• Intersection of $x + 2y = 240$ and $3x + 4y = 620$. Multiply first by 2: $2x + 4y = 480$. Subtract from $3x+4y=620$: $(3x+4y)-(2x+4y)=620-480 \implies x=140$. Substitute into $x+2y=240$: $140+2y=240 \implies 2y=100 \implies y=50$. Point $(140, 50)$. Check $2x+y \ge 180$: $2(140)+50 = 280+50 = 330 \ge 180$. Feasible. Vertex: $(140, 50)$.

The corner points of the feasible region are $(40, 100)$, $(20, 140)$, and $(140, 50)$.

Now, we evaluate the objective function $Z = 3x + 3.5y$ at each corner point to find the minimum amount of Nitrogen.

Corner Point $(x, y)$	$Z = 3x + 3.5y$	Value of $Z$ (Nitrogen kg)
$(40, 100)$	$3(40) + 3.5(100)$	$120 + 350 = 470$
$(20, 140)$	$3(20) + 3.5(140)$	$60 + 490 = 550$
$(140, 50)$	$3(140) + 3.5(50)$	$420 + 175 = 595$

The minimum value of $Z$ is $470$, which occurs at the corner point $(40, 100)$.

---

Conclusion:

To minimise the amount of nitrogen added to the garden while meeting the nutrient requirements and chlorine limit, the fruit grower should use 40 bags of brand P and 100 bags of brand Q. The minimum amount of nitrogen added in the garden will be 470 kg.

This is a Linear Programming Problem (LPP), related to Question 8 but with a different objective.

---

Given:

The constraints are the same as in Question 8.

Let $\mathbf{x}$ be the number of bags of Brand P used.

Let $\mathbf{y}$ be the number of bags of Brand Q used.

The constraints are:

The objective is to maximise the total amount of Nitrogen. The Nitrogen content is 3 kg per bag of Brand P and 3.5 kg per bag of Brand Q. The objective function $Z$ (amount of Nitrogen) is:

$Z = 3x + 3.5y$

---

To Find:

The number of bags of each brand (P and Q) to be used to maximise the amount of nitrogen added to the garden, and the maximum amount of nitrogen.

---

Solution:

The feasible region is determined by the system of inequalities: $x \ge 0, y \ge 0, x + 2y \ge 240, 2x + y \ge 180, 3x + 4y \le 620$. This is the same feasible region as in Question 8.

The corner points of this feasible region were calculated in Question 8 as:

$(40, 100)$, $(20, 140)$, and $(140, 50)$.

To find the maximum amount of Nitrogen, we evaluate the objective function $Z = 3x + 3.5y$ at each of these corner points.

Corner Point $(x, y)$	$Z = 3x + 3.5y$	Value of $Z$ (Nitrogen kg)
$(40, 100)$	$3(40) + 3.5(100)$	$120 + 350 = 470$
$(20, 140)$	$3(20) + 3.5(140)$	$60 + 490 = 550$
$(140, 50)$	$3(140) + 3.5(50)$	$420 + 175 = 595$

The maximum value of $Z$ is $595$, which occurs at the corner point $(140, 50)$.

This means that to maximise the amount of Nitrogen added while meeting the nutrient requirements and chlorine limit, 140 bags of brand P and 50 bags of brand Q should be used.

---

Conclusion:

To maximise the amount of nitrogen added to the garden, the fruit grower should use 140 bags of brand P and 50 bags of brand Q. The maximum amount of nitrogen added in the garden will be 595 kg.

This is a Linear Programming Problem (LPP).

---

Given:

Maximum combined production level = 1200 dolls per week.

Demand for dolls of type B is at most half of that for dolls of type A.

Production level of dolls of type A can exceed three times the production of dolls of type B by at most 600 units.

Profit per doll of type A = $\textsf{₹}12$.

Profit per doll of type B = $\textsf{₹}16$.

---

To Find:

The number of dolls of each type (A and B) to produce weekly to maximise the profit, and the maximum profit.

---

Solution:

Let $\mathbf{x}$ be the number of dolls of type A produced per week.

Let $\mathbf{y}$ be the number of dolls of type B produced per week.

Since the number of dolls cannot be negative, we have the non-negativity constraints:

The constraints based on the given conditions are:

Combined production level (at most 1200):

Demand for dolls of type B is at most half of that for dolls of type A:

$y \le \frac{1}{2}x$

Multiplying by 2:

$2y \le x$

Rearranging:

Production level of dolls of type A can exceed three times the production of dolls of type B by at most 600 units:

$x - 3y \le 600$

The objective is to maximise the total profit. The profit function $Z$ is given by:

$Z = 12x + 16y$

The mathematical formulation of the LPP is:

Maximise $Z = 12x + 16y$

Subject to the constraints:

---

We solve this LPP graphically. First, we plot the lines corresponding to the equality forms of the constraints:

Line 1: $x + y = 1200$. Points: $(0, 1200)$, $(1200, 0)$.

Line 2: $x - 2y = 0$ (or $x = 2y$). Points: $(0, 0)$, $(100, 50)$, $(1200, 600)$.

Line 3: $x - 3y = 600$. Points: $(600, 0)$, $(0, -200)$ (not in first quadrant). If $y=100$, $x=900$. Point $(900, 100)$.

The feasible region is in the first quadrant ($x \ge 0, y \ge 0$). It is the area below or on Line 1 ($x+y \le 1200$), above or on Line 2 ($x-2y \ge 0$), and below or on Line 3 ($x-3y \le 600$).

The corner points of the feasible region are the vertices formed by the intersection of the boundary lines within the feasible area. These vertices are:

• Intersection of $y=0$ and $x-2y=0$: $x-0=0 \implies x=0$. Point $(0, 0)$. Satisfies $0+0 \le 1200$, $0-0 \le 600$. Feasible. Vertex: $(0, 0)$.
• Intersection of $y=0$ and $x - 3y = 600$: $x - 0 = 600 \implies x = 600$. Point $(600, 0)$. Satisfies $600+0 = 600 \le 1200$, $600-0 = 600 \ge 0$. Feasible. Vertex: $(600, 0)$.
• Intersection of $x-3y=600$ and $x+y=1200$. From second equation, $x=1200-y$. Substitute into first: $(1200-y)-3y=600 \implies 1200-4y=600 \implies 4y=600 \implies y=150$. Substitute into $x=1200-y$: $x=1200-150=1050$. Point $(1050, 150)$. Check $x-2y \ge 0$: $1050-2(150)=1050-300=750 \ge 0$. Feasible. Vertex: $(1050, 150)$.
• Intersection of $x-2y=0$ and $x+y=1200$. From first equation, $x=2y$. Substitute into second: $2y+y=1200 \implies 3y=1200 \implies y=400$. Substitute into $x=2y$: $x=2(400)=800$. Point $(800, 400)$. Check $x-3y \le 600$: $800-3(400)=800-1200 = -400 \le 600$. Feasible. Vertex: $(800, 400)$.
• Intersection of $x-2y=0$ and $x-3y=600$: From first equation, $x=2y$. Substitute into second: $2y-3y=600 \implies -y=600 \implies y=-600$. Not in first quadrant. Not a vertex of the feasible region.

The corner points of the feasible region are $(0, 0)$, $(600, 0)$, $(1050, 150)$, and $(800, 400)$.

Now, we evaluate the objective function $Z = 12x + 16y$ at each corner point to find the maximum profit.

Corner Point $(x, y)$	$Z = 12x + 16y$	Value of $Z$ ($\textsf{₹}$)
$(0, 0)$	$12(0) + 16(0)$	$0$
$(600, 0)$	$12(600) + 16(0)$	$7200$
$(1050, 150)$	$12(1050) + 16(150)$	$12600 + 2400 = 15000$
$(800, 400)$	$12(800) + 16(400)$	$9600 + 6400 = 16000$

The maximum value of $Z$ is $\textsf{₹}16000$, which occurs at the corner point $(800, 400)$.

This means that to maximise profit, the company should produce 800 dolls of type A and 400 dolls of type B weekly.

---

Conclusion:

To maximise the profit, the company should produce 800 dolls of type A and 400 dolls of type B weekly. The maximum profit is $\textsf{₹}16000$.