Chapter 12 Linear Programming

According to the Fundamental Theorem of Linear Programming, the optimal solution (maximum or minimum) for a linear programming problem, if it exists, will occur at a corner point (vertex) of the feasible region.

From the given figure, the feasible region is the polygon OABC. The corner points (vertices) of this feasible region are:

• O = (0, 0)
• A = (25, 0)
• B = (15, 15)
• C = (0, 20)

The objective function is given by $Z = 4x + 3y$. We need to evaluate the value of Z at each of the corner points.

Evaluating Z at the corner points:

At O (0, 0): $Z = 4(0) + 3(0) = 0$

At A (25, 0): $Z = 4(25) + 3(0) = 100 + 0 = 100$

At B (15, 15): $Z = 4(15) + 3(15) = 60 + 45 = 105$

At C (0, 20): $Z = 4(0) + 3(20) = 0 + 60 = 60$

We can summarize the values of Z in a table:

Comparing the values of Z obtained at the corner points, we find that the maximum value is 105.

The maximum value of the objective function $Z = 4x + 3y$ is $\mathbf{105}$, and it occurs at the corner point (15, 15).

The feasible region is given in the figure. It is an unbounded region.

The corner points (vertices) of the feasible region are the points where the boundary lines intersect. From the figure, the corner points are:

• A = (0, 10)
• B = (5, 5)
• C = (15, 0)

The objective function is given by $Z = 3x + 2y$. We need to evaluate the value of Z at each of the corner points.

Evaluating Z at the corner points:

At A (0, 10): $Z = 3(0) + 2(10) = 0 + 20 = 20$

At B (5, 5): $Z = 3(5) + 2(5) = 15 + 10 = 25$

At C (15, 0): $Z = 3(15) + 2(0) = 45 + 0 = 45$

We can summarize the values of Z in a table:

The minimum value of Z at the corner points is 20, which occurs at (0, 10).

Since the feasible region is unbounded, we need to check if the minimum value of Z exists.

Consider the inequality $3x + 2y < 20$. We draw the graph of the line $3x + 2y = 20$. This line passes through (0, 10) and $(20/3, 0)$.

The open half-plane represented by $3x + 2y < 20$ is the region below the line $3x + 2y = 20$.

We need to determine if this open half-plane $3x + 2y < 20$ has any point in common with the feasible region.

Observing the figure, the feasible region is above the lines segment connecting (0,10), (5,5), and (15,0). The line $3x + 2y = 20$ passes through the point (0, 10).

Consider any point $(x, y)$ in the feasible region. The feasible region is defined by the constraints whose boundary lines pass through (0,10), (5,5), and (15,0). These lines are $x+y=10$ (through (0,10) and (5,5)) and $x+2y=15$ (through (5,5) and (15,0)), along with $x \ge 0$ and $y \ge 0$. The feasible region satisfies $x+y \ge 10$ and $x+2y \ge 15$.

For any point $(x, y)$ in the feasible region, we have $x \ge 0, y \ge 0, x+y \ge 10$, and $x+2y \ge 15$.

If we take a point $(x,y)$ in the feasible region other than (0,10), either $x > 0$ or $y > 10$ or both conditions might apply, subject to the constraints.

Consider the line $3x+2y=20$. This line passes through the corner point (0,10). All other points in the feasible region lie such that $3x+2y \ge 20$. This is because the feasible region lies "above" the line $3x+2y=20$ at points other than (0,10) which is on the line.

For example, consider point (5,5). $3(5)+2(5)=25 > 20$. Consider point (15,0). $3(15)+2(0)=45 > 20$.

Since the open half-plane $3x + 2y < 20$ does not contain any point of the feasible region, the minimum value of Z exists and is equal to the minimum value at the corner point (0, 10).

Therefore, the minimum value of the objective function $Z = 3x + 2y$ is $\mathbf{20}$, and it occurs at the corner point (0, 10).

The given Linear Programming Problem is:

Maximise $Z = 2x + 3y$

Subject to the constraints:

$x + y \le 4$

$x \ge 0$

$y \ge 0$

To find the feasible region, we first graph the boundary lines of the constraints.

• For $x+y \le 4$, the boundary line is $x+y = 4$. We find two points on this line:
  • If $x = 0$, then $0 + y = 4 \implies y = 4$. Point is (0, 4).
  • If $y = 0$, then $x + 0 = 4 \implies x = 4$. Point is (4, 0).
  We plot these points and draw the line passing through them. Since the inequality is $x+y \le 4$, the region satisfying this constraint is the half-plane containing the origin (0,0) (since $0+0 \le 4$ is true).
• For $x \ge 0$, this represents the region on or to the right of the y-axis.
• For $y \ge 0$, this represents the region on or above the x-axis.

The feasible region is the intersection of these three regions. It is the region in the first quadrant ($x \ge 0, y \ge 0$) below or on the line $x+y=4$. This forms a triangular region.

The corner points (vertices) of the feasible region are the points of intersection of the boundary lines. These are:

• Intersection of $x=0$ and $y=0$: (0, 0)
• Intersection of $x=0$ and $x+y=4$: Substitute $x=0$ into $x+y=4 \implies 0+y=4 \implies y=4$. Point is (0, 4).
• Intersection of $y=0$ and $x+y=4$: Substitute $y=0$ into $x+y=4 \implies x+0=4 \implies x=4$. Point is (4, 0).

The corner points are (0, 0), (0, 4), and (4, 0).

Now we evaluate the objective function $Z = 2x + 3y$ at each corner point to find the maximum value.

At corner point (0, 0):

$Z = 2(0) + 3(0) = 0$

At corner point (0, 4):

$Z = 2(0) + 3(4) = 0 + 12 = 12$

At corner point (4, 0):

$Z = 2(4) + 3(0) = 8 + 0 = 8$

We can list the values of Z at the corner points in a table:

Comparing the values of Z, the maximum value is 12.

Therefore, the maximum value of $Z = 2x + 3y$ is $\mathbf{12}$, which occurs at the point (0, 4).

Let us formulate the given problem as a Linear Programming Problem.

Decision Variables:

Let $x$ be the number of black and white television sets produced per week.

Let $y$ be the number of coloured television sets produced per week.

Objective Function (Maximise Profit):

The profit from one black and white set is $\textsf{₹} 510$.

The profit from one coloured set is $\textsf{₹} 675$.

Total profit, $Z$, is the sum of profits from both types of sets.

$Z = 510x + 675y$

We want to maximise $Z$.

Constraints:

1. Resource Constraint (Maximum number of sets):

The total number of sets produced per week cannot exceed 300.

So, $x + y \le 300$.

2. Budget Constraint (Maximum expenditure):

The cost of making one black and white set is $\textsf{₹} 1800$.

The cost of making one coloured set is $\textsf{₹} 2700$.

The total weekly expenditure cannot exceed $\textsf{₹} 648000$.

So, the total cost is $1800x + 2700y$.

Thus, $1800x + 2700y \le 648000$.

We can simplify this inequality by dividing by 900:

$\frac{\cancel{1800}^{2}x}{\cancel{900}} + \frac{\cancel{2700}^{3}y}{\cancel{900}} \le \frac{\cancel{648000}^{720}}{\cancel{900}}$

$2x + 3y \le 720$.

3. Non-negativity Constraints:

The number of television sets produced cannot be negative.

So, $x \ge 0$ and $y \ge 0$.

Formulation of the LPP:

Maximise $Z = 510x + 675y$

Subject to the constraints:

$x + y \le 300$

$2x + 3y \le 720$

$x \ge 0$

$y \ge 0$

From Example 4, the Linear Programming Problem formulation is:

Maximise $Z = 510x + 675y$

Subject to:

$x + y \le 300$

$2x + 3y \le 720$

$x \ge 0$

$y \ge 0$

where $x$ is the number of black and white television sets and $y$ is the number of coloured television sets.

To solve this LPP graphically, we first determine the feasible region defined by the constraints.

We draw the boundary lines for each inequality:

• Line 1: $x + y = 300$
  • If $x=0$, $y=300$. Point (0, 300).
  • If $y=0$, $x=300$. Point (300, 0).
  The inequality $x + y \le 300$ is the region including the origin.
• Line 2: $2x + 3y = 720$
  • If $x=0$, $3y=720 \implies y=240$. Point (0, 240).
  • If $y=0$, $2x=720 \implies x=360$. Point (360, 0).
  The inequality $2x + 3y \le 720$ is the region including the origin.
• $x \ge 0$: The region on or to the right of the y-axis.
• $y \ge 0$: The region on or above the x-axis.

The feasible region is the intersection of the regions satisfying all the constraints. It is a bounded region in the first quadrant.

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

• Intersection of $x=0$ and $y=0$: (0, 0)
• Intersection of $y=0$ and $x+y=300$: Substitute $y=0$ into $x+y=300 \implies x+0=300 \implies x=300$. Point: (300, 0). (This point satisfies $2x+3y \le 720$ as $2(300)+3(0) = 600 \le 720$).
• Intersection of $x=0$ and $2x+3y=720$: Substitute $x=0$ into $2x+3y=720 \implies 2(0)+3y=720 \implies 3y=720 \implies y=240$. Point: (0, 240). (This point satisfies $x+y \le 300$ as $0+240 = 240 \le 300$).
• Intersection of $x+y=300$ and $2x+3y=720$:

  From $x+y=300$, we get $y = 300 - x$. Substitute this into the second equation:

  $2x + 3(300 - x) = 720$

  $2x + 900 - 3x = 720$

  $-x = 720 - 900$

  $-x = -180$

  $x = 180$

  Now substitute $x=180$ back into $y = 300 - x$:

  $y = 300 - 180 = 120$

  Point: (180, 120). (This point satisfies $x \ge 0$ and $y \ge 0$ as $180 \ge 0$ and $120 \ge 0$).

The corner points of the feasible region are O(0, 0), A(300, 0), B(180, 120), and C(0, 240).

Next, we evaluate the objective function $Z = 510x + 675y$ at each corner point.

• At O (0, 0):

  $Z = 510(0) + 675(0) = 0$

• At A (300, 0):

  $Z = 510(300) + 675(0) = 153000 + 0 = 153000$

• At B (180, 120):

  $Z = 510(180) + 675(120)$

  $Z = 91800 + 81000 = 172800$

• At C (0, 240):

  $Z = 510(0) + 675(240)$

  $Z = 0 + 162000 = 162000$

We list the values of Z in the following table:

Comparing the values of Z, the maximum value is 172800.

The maximum profit of $\textsf{₹} 172800$ is attained when $x = 180$ and $y = 120$.

Therefore, the company should produce 180 black and white television sets and 120 coloured television sets to achieve the maximum profit.

The given Linear Programming Problem is:

Minimise $Z = 3x + 5y$

Subject to the constraints:

$x + 2y \ge 10$

$x + y \ge 6$

$3x + y \ge 8$

$x \ge 0, y \ge 0$

To find the feasible region, we graph the boundary lines corresponding to the inequality constraints:

• Line 1: $x + 2y = 10$
  • If $x = 0$, $2y = 10 \implies y = 5$. Point: (0, 5).
  • If $y = 0$, $x = 10$. Point: (10, 0).
  The region $x + 2y \ge 10$ is the half-plane not containing the origin.
• Line 2: $x + y = 6$
  • If $x = 0$, $y = 6$. Point: (0, 6).
  • If $y = 0$, $x = 6$. Point: (6, 0).
  The region $x + y \ge 6$ is the half-plane not containing the origin.
• Line 3: $3x + y = 8$
  • If $x = 0$, $y = 8$. Point: (0, 8).
  • If $y = 0$, $3x = 8 \implies x = 8/3$. Point: (8/3, 0).
  The region $3x + y \ge 8$ is the half-plane not containing the origin.
• $x \ge 0, y \ge 0$: This restricts the feasible region to the first quadrant.

The feasible region is the intersection of these half-planes in the first quadrant. Graphing these lines and inequalities shows that the feasible region is an unbounded region.

The corner points (vertices) of the feasible region are the points where the boundary lines intersect and satisfy all constraints. We find the intersection points:

• Intersection of $x = 0$ and $3x + y = 8$: Substitute $x=0$ into $3x+y=8 \implies 3(0)+y=8 \implies y=8$. Point (0, 8). Check with $x+2y \ge 10$: $0+2(8)=16 \ge 10$ (True). Check with $x+y \ge 6$: $0+8=8 \ge 6$ (True). This is a corner point.
• Intersection of $3x + y = 8$ and $x + y = 6$: Subtract $x+y=6$ from $3x+y=8$: $(3x+y) - (x+y) = 8 - 6$ $2x = 2 \implies x = 1$. Substitute $x=1$ into $x+y=6$: $1+y=6 \implies y=5$. Point (1, 5). Check with $x+2y \ge 10$: $1+2(5)=1+10=11 \ge 10$ (True). This is a corner point.
• Intersection of $x + y = 6$ and $x + 2y = 10$: Subtract $x+y=6$ from $x+2y=10$: $(x+2y) - (x+y) = 10 - 6$ $y = 4$. Substitute $y=4$ into $x+y=6$: $x+4=6 \implies x=2$. Point (2, 4). Check with $3x+y \ge 8$: $3(2)+4=6+4=10 \ge 8$ (True). This is a corner point.
• Intersection of $x + 2y = 10$ and $y = 0$: Substitute $y=0$ into $x+2y=10 \implies x+2(0)=10 \implies x=10$. Point (10, 0). Check with $x+y \ge 6$: $10+0=10 \ge 6$ (True). Check with $3x+y \ge 8$: $3(10)+0=30 \ge 8$ (True). This is a corner point.

The corner points of the feasible region are (0, 8), (1, 5), (2, 4), and (10, 0).

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

• At (0, 8): $Z = 3(0) + 5(8) = 0 + 40 = 40$.
• At (1, 5): $Z = 3(1) + 5(5) = 3 + 25 = 28$.
• At (2, 4): $Z = 3(2) + 5(4) = 6 + 20 = 26$.
• At (10, 0): $Z = 3(10) + 5(0) = 30 + 0 = 30$.

We can list these values in a table:

The minimum value of Z at the corner points is 26, which occurs at the point (2, 4).

Since the feasible region is unbounded, we need to check if this minimum value is the actual minimum.

Consider the inequality $3x + 5y < 26$. This represents an open half-plane. We draw the line $3x + 5y = 26$. This line passes through the point (2, 4).

We need to check if the open half-plane $3x + 5y < 26$ has any point in common with the feasible region.

Since all the corner points (0, 8), (1, 5), (2, 4), (10, 0) give values of Z which are 26 or greater than 26, and the feasible region lies above or to the right of the boundary lines, the region $3x + 5y < 26$ does not overlap with the feasible region.

Therefore, the minimum value of $Z = 3x + 5y$ exists and is equal to the smallest value found at the corner points.

The minimum value of Z is $\mathbf{26}$, which occurs at the point (2, 4).

The given corner points of the feasible region are (0, 10), (5, 5), (15, 15), and (0, 20).

The objective function is $Z = px + qy$, where $p > 0$ and $q > 0$.

We are given that the maximum value of Z occurs at both the points (15, 15) and (0, 20).

This implies that the value of the objective function $Z$ is the same at these two points, and this value is the maximum value.

Let's evaluate Z at point (15, 15):

$Z_{(15, 15)} = p(15) + q(15) = 15p + 15q$

Let's evaluate Z at point (0, 20):

$Z_{(0, 20)} = p(0) + q(20) = 0 + 20q = 20q$

Since the maximum occurs at both points, the values of Z must be equal:

$15p + 15q = 20q$

Now, we solve for the relationship between p and q:

$15p = 20q - 15q$

$15p = 5q$

Divide both sides by 5:

$\frac{15p}{5} = \frac{5q}{5}$

$3p = q$

or

$q = 3p$

This is the condition on p and q for the maximum of Z to occur at both (15, 15) and (0, 20).

Comparing this condition with the given options:

(A) $p = q$

(B) $p = 2q$

(C) $q = 2p$

(D) $q = 3p$

The condition $q = 3p$ matches option (D).

The correct option is $\mathbf{(D)}$.

The objective function to be minimised is $Z = 4x + 3y$.

The feasible region is shown in the given figure. The corner points (vertices) of this feasible region are explicitly marked in the figure or can be identified as the intersection points of the boundary lines. The corner points are:

• (0, 8)
• (2, 5)
• (4, 3)
• (9, 0)

The feasible region appears to be the quadrilateral formed by these points in the first quadrant, along with the constraints $x \ge 0$ and $y \ge 0$. Since the region is bounded, the minimum value of Z will occur at one of these corner points.

We evaluate the objective function $Z = 4x + 3y$ at each of these corner points:

At the point (0, 8):

$Z = 4(0) + 3(8) = 0 + 24 = 24$

At the point (2, 5):

$Z = 4(2) + 3(5) = 8 + 15 = 23$

At the point (4, 3):

$Z = 4(4) + 3(3) = 16 + 9 = 25$

At the point (9, 0):

$Z = 4(9) + 3(0) = 36 + 0 = 36$

We can list the values of Z in a table:

Comparing the values of Z, the minimum value is 23.

The minimum value of Z occurs at the point (2, 5).

Therefore, the correct option is (B).

The correct option is $\mathbf{(B)}$.

In a LPP, the linear function which has to be maximised or minimised is called a linear objective function.

The blank should be filled with the word "objective".

The common region determined by all the linear constraints of a LPP is called the feasible region.

The blank should be filled with the word "feasible".

The statement is True.

According to the Fundamental Theorem of Linear Programming, if the feasible region for a linear programming problem is a bounded convex polygon, then the objective function has both a maximum and a minimum value, and these occur at the corner points (vertices) of the feasible region.

Here, R represents the feasible region. If the feasible region is bounded, then the objective function $Z = ax + by$ is guaranteed to have both a maximum and a minimum value within that region.

The statement is False.

While the optimal value (maximum or minimum) of the objective function in a Linear Programming Problem always occurs at a corner point of the feasible region (if an optimal solution exists), it is not always true that it occurs at only one corner point.

If the objective function is parallel to one of the boundary lines of the feasible region, and that boundary line contains multiple corner points, then the optimal value can occur at more than one corner point and at every point on the line segment connecting these corner points.

Given:

Objective function to maximize: $Z = 11x + 7y$

Constraints:

$2x + y \leq 6$

$x \leq 2$

$x \geq 0$

$y \geq 0$

To Find:

The maximum value of Z subject to the given constraints.

Solution:

The given constraints are linear inequalities. The region satisfying all constraints simultaneously is the feasible region.

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

Consider the boundary lines corresponding to the other inequalities:

$2x + y = 6$

$x = 2$

We find the vertices (corner points) of the feasible region formed by the intersection of these boundary lines and the axes ($x=0, y=0$).

1. Intersection of $x=0$ and $y=0$: Point (0, 0).

2. Intersection of $y=0$ and $x=2$: Point (2, 0).

3. Intersection of $x=2$ and $2x + y = 6$. Substitute $x=2$ into the equation $2x + y = 6$:

$2(2) + y = 6$

$4 + y = 6$

$y = 6 - 4$

$y = 2$

Point: (2, 2).

4. Intersection of $x=0$ and $2x + y = 6$. Substitute $x=0$ into the equation $2x + y = 6$:

$2(0) + y = 6$

$0 + y = 6$

$y = 6$

Point: (0, 6).

The vertices of the feasible region are (0, 0), (2, 0), (2, 2), and (0, 6).

Now, we evaluate the objective function $Z = 11x + 7y$ at each vertex:

Comparing the values of Z at the vertices, the maximum value is 42.

Conclusion:

The maximum value of Z is 42, which occurs at the point (0, 6).

Given:

Objective function to maximize: $Z = 3x + 4y$

Constraints:

$x + y \leq 1$

$x \geq 0$

$y \geq 0$

To Find:

The maximum value of Z subject to the given constraints.

Solution:

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

The inequality $x + y \leq 1$ is bounded by the line $x + y = 1$.

To find the feasible region, we consider the boundary line $x + y = 1$ and the axes $x = 0, y = 0$.

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

1. Intersection of $x=0$ and $y=0$: Point (0, 0).

2. Intersection of $y=0$ and $x + y = 1$. Substitute $y=0$ into $x + y = 1$:

$x + 0 = 1$

$x = 1$

Point: (1, 0).

3. Intersection of $x=0$ and $x + y = 1$. Substitute $x=0$ into $x + y = 1$:

$0 + y = 1$

$y = 1$

Point: (0, 1).

The vertices of the feasible region are (0, 0), (1, 0), and (0, 1).

Now, we evaluate the objective function $Z = 3x + 4y$ at each vertex:

Comparing the values of Z at the vertices, the maximum value is 4.

Conclusion:

The maximum value of Z is 4, which occurs at the point (0, 1).

Given:

Objective function to maximize: $Z = 11x + 7y$

Constraints:

$x \leq 3$

$y \leq 2$

$x \geq 0$

$y \geq 0$

To Find:

The maximum value of Z subject to the given constraints.

Solution:

The given constraints define the feasible region. The constraints $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

The constraint $x \leq 3$ is the region to the left of or on the vertical line $x = 3$.

The constraint $y \leq 2$ is the region below or on the horizontal line $y = 2$.

The feasible region is the rectangle bounded by the lines $x=0$, $y=0$, $x=3$, and $y=2$.

The vertices (corner points) of this feasible region are the intersections of these boundary lines:

1. Intersection of $x=0$ and $y=0$: Point (0, 0).

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

3. Intersection of $x=3$ and $y=2$: Point (3, 2).

4. Intersection of $y=2$ and $x=0$: Point (0, 2).

The vertices of the feasible region are (0, 0), (3, 0), (3, 2), and (0, 2).

Now, we evaluate the objective function $Z = 11x + 7y$ at each vertex:

Comparing the values of Z at the vertices, the maximum value is 47.

Conclusion:

The maximum value of Z is 47, which occurs at the point (3, 2).

Given:

Objective function to minimize: $Z = 13x – 15y$

Constraints:

$x + y \leq 7$

$2x – 3y + 6 \geq 0$

$x \geq 0$

$y \geq 0$

To Find:

The minimum value of Z subject to the given constraints.

Solution:

The constraints define the feasible region. The conditions $x \geq 0$ and $y \geq 0$ restrict the feasible region to the first quadrant.

We consider the boundary lines corresponding to the inequalities:

1. $x + y = 7$

2. $2x – 3y + 6 = 0 \implies 2x - 3y = -6$

We identify the vertices (corner points) of the feasible region, which are the intersection points of these boundary lines within the first quadrant.

1. Intersection of $x=0$ and $y=0$: Point (0, 0). This point satisfies $0+0 \leq 7$ and $2(0)-3(0)+6 \geq 0$. So, (0, 0) is a vertex.

2. Intersection of $y=0$ and $x+y=7$: Setting $y=0$ in $x+y=7$, we get $x=7$. Point (7, 0). This point satisfies $2(7)-3(0)+6 = 14+6 = 20 \geq 0$. So, (7, 0) is a vertex.

3. Intersection of $x=0$ and $2x-3y=-6$: Setting $x=0$ in $2x-3y=-6$, we get $-3y=-6 \implies y=2$. Point (0, 2). This point satisfies $0+2 \leq 7$. So, (0, 2) is a vertex.

4. Intersection of $x+y=7$ and $2x-3y=-6$.

From $x+y=7$, we have $y = 7-x$. Substitute this into the second equation:

$2x - 3(7-x) = -6$

$2x - 21 + 3x = -6$

$5x = 21 - 6$

$5x = 15$

$x = 3$

Substitute $x=3$ back into $y=7-x$:

$y = 7 - 3$

$y = 4$

Point: (3, 4). This point is in the first quadrant and satisfies both inequalities as shown in the vertex finding step above. So, (3, 4) is a vertex.

The vertices of the feasible region are (0, 0), (7, 0), (0, 2), and (3, 4).

Now, we evaluate the objective function $Z = 13x - 15y$ at each vertex:

Comparing the values of Z at the vertices, the minimum value is -30.

Conclusion:

The minimum value of Z is -30, which occurs at the point (0, 2).

Given:

Objective function to maximize: $Z = 3x + 4y$

The feasible region is shown in Fig. 12.7.

To Find:

The maximum value of Z over the given feasible region.

Solution:

From the provided figure (Fig. 12.7), the vertices (corner points) of the feasible region are clearly visible. These vertices are where the boundary lines of the inequalities intersect.

The vertices of the feasible region are:

(0, 0)

(1, 0)

(0, 1)

To find the maximum value of the objective function $Z = 3x + 4y$, we evaluate Z at each of these vertices.

By comparing the values of Z at these vertices, we find the maximum value.

The values obtained are 0, 3, and 4.

The maximum value among these is 4.

Conclusion:

The maximum value of Z is 4, which occurs at the vertex (0, 1).

Given:

Objective function to maximize: $Z = 5x + 7y$

The feasible region is shown in Fig. 12.8.

To Find:

The maximum value of Z over the given feasible region.

Solution:

The vertices (corner points) of the feasible region are the points where the boundary lines intersect. From the provided figure (Fig. 12.8), the vertices of the shaded feasible region are clearly identified.

The vertices are:

(0, 0)

(5, 0)

(3, 4)

(0, 5)

According to the fundamental theorem of linear programming, the optimal solution (maximum or minimum) of the objective function, if it exists, occurs at one of the vertices of the feasible region.

We evaluate the objective function $Z = 5x + 7y$ at each of these vertices:

Comparing the values of Z calculated at the vertices, we can determine the maximum value.

The values are 0, 25, 43, and 35.

The maximum value among these is 43.

Conclusion:

The maximum value of Z is 43, which occurs at the vertex (3, 4).

Given:

Objective function to minimize: $Z = 11x + 7y$

The feasible region is shown in Fig. 12.9.

To Find:

The minimum value of Z over the given feasible region.

Solution:

The vertices (corner points) of the feasible region shown in Fig. 12.9 are the points where the boundary lines intersect. These vertices are:

From the figure, we can identify the vertices:

(0, 0)

(6, 0)

(0, 9)

The intersection point of the two lines forming the boundary of the feasible region is also a vertex. By observing the graph, this intersection point appears to be (2, 4).

Let's confirm the equations of the lines from the points they pass through:

Line 1 passes through (6, 0) and (2, 4). The equation is $\frac{y - 0}{x - 6} = \frac{4 - 0}{2 - 6} = \frac{4}{-4} = -1$.

$y = -1(x - 6)$

$y = -x + 6 \implies x + y = 6$.

Line 2 passes through (0, 9) and (2, 4). The equation is $\frac{y - 9}{x - 0} = \frac{4 - 9}{2 - 0} = \frac{-5}{2}$.

$2(y - 9) = -5x$

$2y - 18 = -5x \implies 5x + 2y = 18$.

The intersection of $x + y = 6$ and $5x + 2y = 18$ can be found by solving the system of equations.

From $x+y=6$, $y = 6-x$. Substituting into the second equation:

$5x + 2(6-x) = 18$

$5x + 12 - 2x = 18$

$3x = 18 - 12$

$3x = 6$

$x = 2$

Substituting $x=2$ into $y=6-x$:

$y = 6 - 2 = 4$.

So the intersection point is indeed (2, 4).

The vertices of the feasible region are (0, 0), (6, 0), (2, 4), and (0, 9).

Now, we evaluate the objective function $Z = 11x + 7y$ at each vertex:

Comparing the values of Z at the vertices, we find the minimum value.

The values are 0, 66, 50, and 63.

The minimum value among these is 0.

Conclusion:

The minimum value of Z is 0, which occurs at the vertex (0, 0).

Given:

Objective function to maximize: $Z = 11x + 7y$

The feasible region is the same as in Exercise 7, shown in Fig. 12.9.

To Find:

The maximum value of Z over the feasible region from Fig. 12.9.

Solution:

As determined in the solution to Exercise 7, the vertices (corner points) of the feasible region shown in Fig. 12.9 are:

(0, 0)

(6, 0)

(2, 4)

(0, 9)

To find the maximum value of the objective function $Z = 11x + 7y$, we evaluate Z at each of these vertices:

Comparing the values of Z calculated at the vertices, we identify the maximum value.

The values are 0, 66, 50, and 63.

The maximum value among these is 66.

Conclusion:

The maximum value of Z is 66, which occurs at the vertex (6, 0).

Given:

Objective function to minimize: $Z = 4x + y$

The feasible region is shown in Fig. 12.10.

To Find:

The minimum value of Z over the given feasible region, if it exists.

Solution:

From the provided figure (Fig. 12.10), we can identify the corner points (vertices) of the feasible region. These are the points where the boundary lines intersect.

The vertices of the feasible region are:

(0, 3)

(1, 2)

(6, 0)

We evaluate the objective function $Z = 4x + y$ at each of these vertices:

The values of Z at the corner points are 3, 6, and 24.

The smallest value among these is 3.

Since the feasible region is unbounded, we need to check if the minimum value actually exists.

The feasible region is defined by constraints which are '$\geq$' type inequalities ($x+y \geq 3$, $2x+5y \geq 12$, $x \geq 0$, $y \geq 0$). The coefficients of x and y in the objective function $Z = 4x + y$ (which are 4 and 1) are both positive. In such cases, for an unbounded feasible region, the minimum value of the objective function always exists and occurs at a corner point.

Alternatively, consider the inequality $4x + y < 3$. This represents the region below the line $4x + y = 3$. The line $4x + y = 3$ passes through the vertex (0, 3).

We check if there is any point in the feasible region (other than (0, 3)) that also lies in the region $4x + y < 3$.

The feasible region lies above or on the boundary lines $x+y=3$ and $2x+5y=12$. The slope of the line $4x+y=3$ (slope -4) is steeper than the slopes of the boundary lines ($x+y=3$ has slope -1, $2x+5y=12$ has slope -0.4).

As we move away from the origin within the feasible region, the value of $4x+y$ increases because the coefficients 4 and 1 are positive.

Since the feasible region is above or on the line $4x+y=3$ (at point (0,3)), there are no points in the feasible region for which $4x+y < 3$.

Therefore, the minimum value of Z exists and is the smallest value found at the vertices.

Conclusion:

The minimum value of Z is 3, which occurs at the vertex (0, 3).

Given:

Objective function: $Z = x + 2y$

The feasible region is the shaded area shown in Fig. 12.11.

To Find:

The maximum and minimum values of Z over the given feasible region.

Solution:

The feasible region in the figure is a polygon. The vertices (corner points) of this feasible region are clearly marked in the figure. These vertices are:

(0, 0)

(10, 0)

(6, 4)

(4, 5)

(0, 5)

According to the fundamental theorem of linear programming, for a closed and bounded feasible region, the maximum and minimum values of the objective function occur at the vertices of the feasible region.

We evaluate the objective function $Z = x + 2y$ at each of these vertices:

Comparing the values of Z calculated at the vertices, we find the minimum and maximum values.

The values are 0, 10, 14, 14, and 10.

The minimum value among these is 0.

The maximum value among these is 14.

Conclusion:

The minimum value of Z is 0, which occurs at the vertex (0, 0).

The maximum value of Z is 14, which occurs at the vertices (6, 4) and (4, 5).

Given:

Stock of components:

Resistors: 200

Transistors: 120

Capacitors: 150

Requirements per circuit:

Type A: 20 resistors, 10 transistors, 10 capacitors

Type B: 10 resistors, 20 transistors, 30 capacitors

Profit per circuit:

Type A: $\textsf{₹}50$

Type B: $\textsf{₹}60$

To Formulate:

Formulate the problem as a Linear Programming Problem (LPP) to maximize the manufacturer's profit.

Formulation:

Let $x$ be the number of Type A circuits produced.

Let $y$ be the number of Type B circuits produced.

These are our decision variables.

The manufacturer wants to maximize the total profit. The profit from Type A circuits is $50x$, and the profit from Type B circuits is $60y$.

The objective function to maximize is:

$Z = 50x + 60y$

The production is subject to the availability of components. We have constraints based on the total number of resistors, transistors, and capacitors used.

Constraints:

1. Constraint on Resistors:

Total resistors used = (Resistors for Type A) + (Resistors for Type B)

Total resistors used = $20x + 10y$

Since the total resistors available are 200:

$20x + 10y \leq 200$

Dividing by 10, we get:

$2x + y \leq 20$

2. Constraint on Transistors:

Total transistors used = (Transistors for Type A) + (Transistors for Type B)

Total transistors used = $10x + 20y$

Since the total transistors available are 120:

$10x + 20y \leq 120$

Dividing by 10, we get:

$x + 2y \leq 12$

3. Constraint on Capacitors:

Total capacitors used = (Capacitors for Type A) + (Capacitors for Type B)

Total capacitors used = $10x + 30y$

Since the total capacitors available are 150:

$10x + 30y \leq 150$

Dividing by 10, we get:

$x + 3y \leq 15$

Additionally, the number of circuits produced cannot be negative.

Non-negativity constraints:

$x \geq 0$

$y \geq 0$

The Linear Programming Problem is:

Maximize $Z = 50x + 60y$

Subject to the constraints:

$2x + y \leq 20$

$x + 2y \leq 12$

$x + 3y \leq 15$

$x \geq 0$

$y \geq 0$

Given:

Total packages to transport: 1200

Large vans:

Capacity: 200 packages

Cost: $\textsf{₹}400$ per van

Small vans:

Capacity: 80 packages

Cost: $\textsf{₹}200$ per van

Total budget for the job: Not more than $\textsf{₹}3000$

Constraint on number of vans: Number of large vans cannot exceed the number of small vans.

Objective: Minimise the total cost.

To Formulate:

Formulate the problem as a Linear Programming Problem (LPP) to minimize the transportation cost.

Formulation:

Let $x$ be the number of large vans used.

Let $y$ be the number of small vans used.

These are our decision variables.

The firm wants to minimize the total cost of engaging vans. The cost for large vans is $400x$, and the cost for small vans is $200y$.

The objective function to minimize is:

$Z = 400x + 200y$

The transportation must satisfy the given conditions, which form the constraints.

Constraints:

1. Constraint on total packages transported:

Total packages carried = (Packages by large vans) + (Packages by small vans)

Total packages carried = $200x + 80y$

The firm must transport at least 1200 packages:

$200x + 80y \geq 1200$

Dividing by 40, we get:

$5x + 2y \geq 30$

2. Constraint on total cost:

Total cost = (Cost of large vans) + (Cost of small vans)

Total cost = $400x + 200y$

The total cost must not exceed $\textsf{₹}3000$:

$400x + 200y \leq 3000$

Dividing by 100, we get:

$4x + 2y \leq 30$

Dividing by 2, we get:

$2x + y \leq 15$

3. Constraint on the number of large and small vans:

The number of large vans cannot exceed the number of small vans:

$x \leq y$

This can be written as:

$x - y \leq 0$

Also, the number of vans must be non-negative.

Non-negativity constraints:

$x \geq 0$

$y \geq 0$

Since the number of vans must be whole numbers, $x$ and $y$ are also restricted to be integers (although this is not explicitly required in standard LPP formulation unless specified as Integer Programming). For a general LPP formulation, non-negativity is sufficient.

The Linear Programming Problem is:

Minimize $Z = 400x + 200y$

Subject to the constraints:

$5x + 2y \geq 30$

$2x + y \leq 15$

$x - y \leq 0$

$x \geq 0$

$y \geq 0$

Given:

Machine availability per week: 60 hours for each machine.

1 hour = 60 minutes, so 60 hours = $60 \times 60 = 3600$ minutes.

Machine requirements per box:

Type A: 2 min (threading), 3 min (slotting)

Type B: 8 min (threading), 2 min (slotting)

Profit per box:

Type A: $\textsf{₹}100$

Type B: $\textsf{₹}170$

To Formulate:

Formulate the problem as a Linear Programming Problem (LPP) to maximize the total profit.

Formulation:

Let $x$ be the number of boxes of Type A screws manufactured per week.

Let $y$ be the number of boxes of Type B screws manufactured per week.

These are the decision variables.

The objective is to maximize the total profit. The profit is given by the number of boxes of each type multiplied by their respective profits per box.

The objective function to maximize is:

$Z = 100x + 170y$

The constraints are based on the available time for each machine.

Constraints:

1. Constraint on Threading Machine time:

Total threading time used = (Time for $x$ boxes of A) + (Time for $y$ boxes of B)

Total threading time used = $2x + 8y$ minutes

Available threading time = 3600 minutes

The total time used cannot exceed the available time:

$2x + 8y \leq 3600$

This inequality can be simplified by dividing by 2:

$x + 4y \leq 1800$

2. Constraint on Slotting Machine time:

Total slotting time used = (Time for $x$ boxes of A) + (Time for $y$ boxes of B)

Total slotting time used = $3x + 2y$ minutes

Available slotting time = 3600 minutes

The total time used cannot exceed the available time:

$3x + 2y \leq 3600$

Finally, the number of boxes manufactured cannot be negative.

Non-negativity constraints:

$x \geq 0$

$y \geq 0$

The Linear Programming Problem is:

Maximize $Z = 100x + 170y$

Subject to the constraints:

$x + 4y \leq 1800$

$3x + 2y \leq 3600$

$x \geq 0$

$y \geq 0$

Given:

Cost of making sweaters:

Type A: $\textsf{₹}360$ per sweater

Type B: $\textsf{₹}120$ per sweater

Maximum total sweaters per day: 300

Maximum total cost per day: $\textsf{₹}72000$

Constraint on numbers: Number of type B sweaters cannot exceed number of type A sweaters by more than 100.

Profit per sweater:

Type A: $\textsf{₹}200$

Type B: $\textsf{₹}120$

Objective: Maximise the total profit.

To Formulate:

Formulate the problem as a Linear Programming Problem (LPP) to maximize the company's profit.

Formulation:

Let $x$ be the number of type A sweaters manufactured per day.

Let $y$ be the number of type B sweaters manufactured per day.

These are our decision variables.

The objective is to maximize the total profit. The profit from type A sweaters is $200x$, and the profit from type B sweaters is $120y$.

The objective function to maximize is:

$Z = 200x + 120y$

The production is subject to several constraints.

Constraints:

1. Constraint on total number of sweaters:

The total number of sweaters cannot exceed 300:

$x + y \leq 300$

2. Constraint on total cost:

Total cost = (Cost of A sweaters) + (Cost of B sweaters)

Total cost = $360x + 120y$

The total cost must not exceed $\textsf{₹}72000$:

$360x + 120y \leq 72000$

This inequality can be simplified by dividing by 120:

$\frac{360x}{120} + \frac{120y}{120} \leq \frac{72000}{120}$

$3x + y \leq 600$

3. Constraint on the relative number of sweaters:

The number of type B sweaters ($y$) cannot exceed the number of type A sweaters ($x$) by more than 100.

$y \leq x + 100$

This can be rewritten as:

$y - x \leq 100$

Or:

$-x + y \leq 100$

Also, the number of sweaters manufactured cannot be negative.

Non-negativity constraints:

$x \geq 0$

$y \geq 0$

The Linear Programming Problem is:

Maximize $Z = 200x + 120y$

Subject to the constraints:

$x + y \leq 300$

$3x + y \leq 600$

$-x + y \leq 100$

$x \geq 0$

$y \geq 0$

Given:

Speed 1 ($v_1$): 50 km/hour, Cost 1 ($c_1$): $\textsf{₹}2$/km

Speed 2 ($v_2$): 80 km/hour, Cost 2 ($c_2$): $\textsf{₹}3$/km

Maximum petrol cost: $\textsf{₹}120$

Maximum time: 1 hour

Objective: Maximize total distance travelled.

To Formulate:

Formulate the problem as a Linear Programming Problem (LPP) to maximize the distance travelled.

Formulation:

Let $x$ be the distance (in km) travelled at the speed of 50 km/hour.

Let $y$ be the distance (in km) travelled at the speed of 80 km/hour.

These are our decision variables.

The objective is to maximize the total distance travelled, which is the sum of the distances travelled at each speed.

The objective function to maximize is:

$Z = x + y$

The constraints are based on the total cost of petrol and the total time taken for the journey.

Constraints:

1. Constraint on total petrol cost:

Cost for distance $x$ at $\textsf{₹}2$/km is $2x$.

Cost for distance $y$ at $\textsf{₹}3$/km is $3y$.

Total cost = $2x + 3y$.

The total cost must be at most $\textsf{₹}120$:

$2x + 3y \leq 120$

2. Constraint on total time taken:

Time = $\frac{\text{Distance}}{\text{Speed}}$.

Time taken for distance $x$ at 50 km/hour is $\frac{x}{50}$ hours.

Time taken for distance $y$ at 80 km/hour is $\frac{y}{80}$ hours.

Total time = $\frac{x}{50} + \frac{y}{80}$.

The total time must be at most 1 hour:

$\frac{x}{50} + \frac{y}{80} \leq 1$

To clear the fractions, multiply the inequality by the least common multiple of 50 and 80, which is 400:

$400 \left(\frac{x}{50}\right) + 400 \left(\frac{y}{80}\right) \leq 400(1)$

$8x + 5y \leq 400$

Also, the distances travelled cannot be negative.

Non-negativity constraints:

$x \geq 0$

$y \geq 0$

The Linear Programming Problem is:

Maximize $Z = x + y$

Subject to the constraints:

$2x + 3y \leq 120$

$8x + 5y \leq 400$

$x \geq 0$

$y \geq 0$

Given:

The Linear Programming Problem formulated in Exercise 11 is:

Maximize $Z = 50x + 60y$

Subject to the constraints:

$2x + y \leq 20$

$x + 2y \leq 12$

$x + 3y \leq 15$

$x \geq 0, y \geq 0$

where $x$ is the number of Type A circuits and $y$ is the number of Type B circuits.

To Find:

The number of Type A and Type B circuits to produce to maximize profit, and the maximum profit.

Solution:

To solve this LPP using the graphical method, we first determine the feasible region defined by the given constraints. The constraints $x \geq 0$ and $y \geq 0$ restrict the feasible region to the first quadrant.

We consider the boundary lines of the other inequalities:

1. $2x + y = 20$

2. $x + 2y = 12$

3. $x + 3y = 15$

We find the vertices (corner points) of the feasible region formed by the intersection of these boundary lines and the axes within the first quadrant.

The vertices of the feasible region are:

A: Intersection of $x=0$ and $y=0$ $\implies$ (0, 0).

B: Intersection of $y=0$ and $2x + y = 20$. Setting $y=0$, $2x=20 \implies x=10$. Point (10, 0). This point satisfies $10+2(0)=10 \leq 12$ and $10+3(0)=10 \leq 15$. Feasible.

C: Intersection of $x=0$ and $x + 3y = 15$. Setting $x=0$, $3y=15 \implies y=5$. Point (0, 5). This point satisfies $2(0)+5=5 \leq 20$ and $0+2(5)=10 \leq 12$. Feasible.

D: Intersection of $x + 2y = 12$ and $x + 3y = 15$. Subtracting the first from the second gives $y=3$. Substituting $y=3$ into $x+2y=12$ gives $x+6=12 \implies x=6$. Point (6, 3). This point satisfies $2(6)+3=15 \leq 20$. Feasible.

E: Intersection of $2x + y = 20$ and $x + 2y = 12$. Multiply the second equation by 2: $2x + 4y = 24$. Subtract the first equation ($2x+y=20$) from this: $(2x+4y) - (2x+y) = 24-20 \implies 3y=4 \implies y=4/3$. Substitute $y=4/3$ into $2x+y=20$: $2x + 4/3 = 20 \implies 2x = 20 - 4/3 = 56/3 \implies x = 28/3$. Point (28/3, 4/3). This point satisfies $28/3 + 3(4/3) = 28/3 + 4 = (28+12)/3 = 40/3 = 13.33... \leq 15$. Feasible.

The vertices of the feasible region are (0, 0), (10, 0), (28/3, 4/3), (6, 3), and (0, 5).

We now evaluate the objective function $Z = 50x + 60y$ at each vertex:

Comparing the values of Z at the vertices (0, 500, 1640/3, 480, 300), the maximum value is 1640/3.

The maximum profit occurs at the vertex ($\frac{28}{3}, \frac{4}{3}$).

Conclusion:

To maximize profit, the manufacturer should produce $\frac{28}{3}$ circuits of Type A and $\frac{4}{3}$ circuits of Type B.

The maximum profit is $\textsf{₹}\frac{1640}{3}$.

Given:

The Linear Programming Problem (LPP) formulated in Exercise 12 is:

Minimize $Z = 400x + 200y$

Subject to the constraints:

$5x + 2y \geq 30$

$2x + y \leq 15$

$x - y \leq 0 \implies x \leq y$

$x \geq 0, y \geq 0$

where $x$ is the number of large vans and $y$ is the number of small vans.

To Find:

The minimum value of the cost function Z.

Solution:

To find the minimum value of Z, we first determine the feasible region defined by the constraints. The constraints $x \geq 0$ and $y \geq 0$ mean the feasible region lies in the first quadrant.

We consider the boundary lines of the inequalities:

1. $5x + 2y = 30$

2. $2x + y = 15$

3. $x = y$

We find the vertices (corner points) of the feasible region by identifying the intersection points of these boundary lines that satisfy all the given constraints.

Let's find the intersections of the boundary lines in the first quadrant:

a) Intersection of $x=0$ and $2x+y=15$: $2(0)+y=15 \implies y=15$. Point (0, 15). Check constraints: $5(0)+2(15)=30 \geq 30$ (True), $0 \leq 15$ (True). This point is feasible.

b) Intersection of $x=0$ and $5x+2y=30$: $5(0)+2y=30 \implies 2y=30 \implies y=15$. Point (0, 15). Same as above.

c) Intersection of $y=x$ and $2x+y=15$: Substitute $y=x$ into $2x+y=15 \implies 2x+x=15 \implies 3x=15 \implies x=5$. Since $y=x$, $y=5$. Point (5, 5). Check constraints: $5(5)+2(5)=25+10=35 \geq 30$ (True). This point is feasible.

d) Intersection of $y=x$ and $5x+2y=30$: Substitute $y=x$ into $5x+2y=30 \implies 5x+2x=30 \implies 7x=30 \implies x=\frac{30}{7}$. Since $y=x$, $y=\frac{30}{7}$. Point ($\frac{30}{7}$, $\frac{30}{7}$). Check constraints: $2(\frac{30}{7})+\frac{30}{7}=\frac{60}{7}+\frac{30}{7}=\frac{90}{7} \approx 12.86 \leq 15$ (True). This point is feasible.

e) Intersection of $2x+y=15$ and $5x+2y=30$: Multiply the first equation by 2: $4x+2y=30$. Subtract this from the second equation: $(5x+2y)-(4x+2y) = 30-30 \implies x=0$. Substitute $x=0$ into $2x+y=15 \implies 2(0)+y=15 \implies y=15$. Point (0, 15). Same as a) and b).

The vertices of the feasible region are the points (0, 15), (5, 5), and ($\frac{30}{7}$, $\frac{30}{7}$).

We evaluate the objective function $Z = 400x + 200y$ at each of these vertices:

Comparing the values of Z at the vertices: $3000$, $3000$, and $\frac{18000}{7}$.

We calculate $\frac{18000}{7} \approx 2571.43$.

The minimum value is the smallest among 3000, 3000, and $\frac{18000}{7}$.

The minimum value is $\frac{18000}{7}$.

Conclusion:

The minimum cost is $\textsf{₹}\frac{18000}{7}$, which occurs when using $\frac{30}{7}$ large vans and $\frac{30}{7}$ small vans.

Given:

The Linear Programming Problem (LPP) formulated in Exercise 13 is:

Maximize $Z = 100x + 170y$

Subject to the constraints:

$x + 4y \leq 1800$

$3x + 2y \leq 3600$

$x \geq 0$

$y \geq 0$

where $x$ is the number of boxes of Type A screws and $y$ is the number of boxes of Type B screws.

To Find:

The number of boxes of Type A and Type B screws to produce to maximize profit, and the maximum profit.

Solution:

To find the maximum profit, we need to find the optimal solution to the LPP by determining the feasible region and evaluating the objective function at its vertices.

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

We consider the boundary lines:

1. $x + 4y = 1800$

2. $3x + 2y = 3600$

We find the vertices (corner points) of the feasible region:

a) Intersection of $x=0$ and $y=0$: Point (0, 0).

b) Intersection of $y=0$ and $3x + 2y = 3600$: Setting $y=0$, $3x = 3600 \implies x = 1200$. Point (1200, 0). Check constraint 1: $1200 + 4(0) = 1200 \leq 1800$. Feasible.

c) Intersection of $x=0$ and $x + 4y = 1800$: Setting $x=0$, $4y = 1800 \implies y = 450$. Point (0, 450). Check constraint 2: $3(0) + 2(450) = 900 \leq 3600$. Feasible.

d) Intersection of $x + 4y = 1800$ and $3x + 2y = 3600$.

Multiply the first equation by 3:

$3(x + 4y) = 3(1800)$

$3x + 12y = 5400$ ... (i)

The second equation is:

$3x + 2y = 3600$ ... (ii)

Subtract equation (ii) from equation (i):

$(3x + 12y) - (3x + 2y) = 5400 - 3600$

$10y = 1800$

$y = 180$

Substitute $y = 180$ into $x + 4y = 1800$:

$x + 4(180) = 1800$

$x + 720 = 1800$

$x = 1800 - 720$

$x = 1080$

Point: (1080, 180).

The vertices of the feasible region are (0, 0), (1200, 0), (1080, 180), and (0, 450).

We evaluate the objective function $Z = 100x + 170y$ at each vertex:

Comparing the values of Z at the vertices, the maximum value is 138600.

This maximum profit occurs at the point (1080, 180).

Conclusion:

To maximize his profit, the manufacturer should produce 1080 boxes of Type A circuits and 180 boxes of Type B circuits.

The maximum profit will be $\textsf{₹}138600$.

Given:

The Linear Programming Problem (LPP) formulated in Exercise 14 is:

Maximize $Z = 200x + 120y$

Subject to the constraints:

$x + y \leq 300$

$3x + y \leq 600$

$-x + y \leq 100$

$x \geq 0$

$y \geq 0$

where $x$ is the number of type A sweaters and $y$ is the number of type B sweaters manufactured per day.

To Find:

The number of sweaters of each type ($x$ and $y$) that maximize the profit, and the maximum profit value (Z).

Solution:

To solve this LPP, we determine the feasible region defined by the constraints. The constraints $x \geq 0$ and $y \geq 0$ mean the feasible region is in the first quadrant.

We consider the boundary lines of the inequalities:

1. $x + y = 300$

2. $3x + y = 600$

3. $-x + y = 100$

We identify the vertices (corner points) of the feasible region by finding the intersection points of these boundary lines within the first quadrant and satisfying all inequalities.

The vertices of the feasible region are:

a) Intersection of $x=0$ and $y=0$: Point (0, 0).

b) Intersection of $y=0$ and $3x+y=600$: $3x=600 \implies x=200$. Point (200, 0). Check constraints: $200+0 \leq 300$ (True), $-200+0 \leq 100$ (True). Feasible.

c) Intersection of $x=0$ and $-x+y=100$: $y=100$. Point (0, 100). Check constraints: $0+100 \leq 300$ (True), $3(0)+100 \leq 600$ (True). Feasible.

d) Intersection of $x+y=300$ and $3x+y=600$. Subtracting the first equation from the second gives: $(3x+y) - (x+y) = 600 - 300 \implies 2x = 300 \implies x=150$. Substituting $x=150$ into $x+y=300$: $150+y=300 \implies y=150$. Point (150, 150). Check constraint 3: $-150+150=0 \leq 100$ (True). Feasible.

e) Intersection of $x+y=300$ and $-x+y=100$. Adding the two equations gives: $(x+y) + (-x+y) = 300 + 100 \implies 2y = 400 \implies y=200$. Substituting $y=200$ into $x+y=300$: $x+200=300 \implies x=100$. Point (100, 200). Check constraint 2: $3(100)+200 = 300+200 = 500 \leq 600$ (True). Feasible.

The vertices of the feasible region are (0, 0), (200, 0), (0, 100), (150, 150), and (100, 200).

We evaluate the objective function $Z = 200x + 120y$ at each vertex:

Comparing the values of Z at the vertices (0, 40000, 12000, 48000, 44000), the maximum value is 48000.

This maximum profit occurs at the vertex (150, 150).

Conclusion:

To maximize the profit, the company should make 150 sweaters of type A and 150 sweaters of type B per day.

The maximum profit will be $\textsf{₹}48000$.

Given:

The Linear Programming Problem (LPP) formulated in Exercise 15 is:

Maximize $Z = x + y$

Subject to the constraints:

$2x + 3y \leq 120$

$8x + 5y \leq 400$

$x \geq 0$

$y \geq 0$

where $x$ is the distance travelled at 50 km/h and $y$ is the distance travelled at 80 km/h. The objective function $Z$ represents the total distance travelled.

To Find:

The maximum distance that the man can travel, which is the maximum value of Z.

Solution:

To find the maximum value of the objective function, we use the graphical method. We determine the feasible region defined by the given constraints. The non-negativity constraints $x \geq 0$ and $y \geq 0$ indicate that the feasible region lies in the first quadrant.

We consider the boundary lines for the other constraints:

1. $2x + 3y = 120$

2. $8x + 5y = 400$

We find the vertices (corner points) of the feasible region formed by the intersection of these boundary lines and the axes within the first quadrant.

a) Intersection of $x=0$ and $y=0$: Point (0, 0).

b) Intersection of $y=0$ and $8x + 5y = 400$: Setting $y=0$, $8x = 400 \implies x = 50$. Point (50, 0). Check constraint 1: $2(50) + 3(0) = 100 \leq 120$ (True). Feasible.

c) Intersection of $x=0$ and $2x + 3y = 120$: Setting $x=0$, $3y = 120 \implies y = 40$. Point (0, 40). Check constraint 2: $8(0) + 5(40) = 200 \leq 400$ (True). Feasible.

d) Intersection of $2x + 3y = 120$ and $8x + 5y = 400$.

We can solve this system of equations. Multiply the first equation by 4:

$4(2x + 3y) = 4(120)$

$8x + 12y = 480$

Now subtract the second equation ($8x + 5y = 400$) from this result:

$(8x + 12y) - (8x + 5y) = 480 - 400$

$7y = 80$

$y = \frac{80}{7}$

Substitute the value of $y$ into the equation $2x + 3y = 120$:

$2x + 3\left(\frac{80}{7}\right) = 120$

$2x + \frac{240}{7} = 120$

$2x = 120 - \frac{240}{7} = \frac{120 \times 7 - 240}{7} = \frac{840 - 240}{7} = \frac{600}{7}$

$x = \frac{600}{7 \times 2} = \frac{300}{7}$

Point: ($\frac{300}{7}$, $\frac{80}{7}$). Since $x \geq 0$ and $y \geq 0$, this point is in the first quadrant and is feasible.

The vertices of the feasible region are (0, 0), (50, 0), ($\frac{300}{7}$, $\frac{80}{7}$), and (0, 40).

We evaluate the objective function $Z = x + y$ at each of these vertices:

Comparing the values of Z at the vertices:

$0$

$50$

$\frac{380}{7} \approx 54.2857$

$40$

The maximum value is $\frac{380}{7}$.

Conclusion:

The maximum distance that the man can travel is $\frac{380}{7}$ km (approximately 54.29 km).

This occurs when he travels $\frac{300}{7}$ km at 50 km/h and $\frac{80}{7}$ km at 80 km/h.

Given:

Objective function to maximize: $Z = x + y$

Constraints:

$x + 4y \leq 8$

$2x + 3y \leq 12$

$3x + y \leq 9$

$x \geq 0$

$y \geq 0$

To Find:

The maximum value of Z subject to the given constraints.

Solution:

The given constraints define the feasible region. The constraints $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

We consider the boundary lines corresponding to the inequalities:

1. $x + 4y = 8$

2. $2x + 3y = 12$

3. $3x + y = 9$

We find the vertices (corner points) of the feasible region, which are the intersection points of these boundary lines within the first quadrant and satisfying all the inequalities.

The vertices are found by considering the intersection of pairs of boundary lines and the axes ($x=0, y=0$).

a) Intersection of $x=0$ and $y=0$: Point (0, 0).

b) Intersection of $y=0$ and $3x + y = 9$: Setting $y=0$, we get $3x = 9 \implies x = 3$. Point (3, 0). Check against $x+4y \leq 8$: $3+4(0)=3 \leq 8$ (True). Check against $2x+3y \leq 12$: $2(3)+3(0)=6 \leq 12$ (True). Feasible. Vertex: (3, 0).

c) Intersection of $x=0$ and $x + 4y = 8$: Setting $x=0$, we get $4y = 8 \implies y = 2$. Point (0, 2). Check against $2x+3y \leq 12$: $2(0)+3(2)=6 \leq 12$ (True). Check against $3x+y \leq 9$: $3(0)+2=2 \leq 9$ (True). Feasible. Vertex: (0, 2).

d) Intersection of $x + 4y = 8$ and $3x + y = 9$. From the second equation, $y = 9 - 3x$. Substituting into the first equation:

$x + 4(9 - 3x) = 8$

$x + 36 - 12x = 8$

$-11x = 8 - 36$

$-11x = -28 \implies x = \frac{28}{11}$

Substituting $x = \frac{28}{11}$ into $y = 9 - 3x$:

$y = 9 - 3\left(\frac{28}{11}\right) = 9 - \frac{84}{11} = \frac{99 - 84}{11} = \frac{15}{11}$

Point ($\frac{28}{11}$, $\frac{15}{11}$). Check against $2x+3y \leq 12$: $2\left(\frac{28}{11}\right) + 3\left(\frac{15}{11}\right) = \frac{56}{11} + \frac{45}{11} = \frac{101}{11}$. Since $\frac{101}{11} \approx 9.18 < 12$, this point is feasible. Vertex: ($\frac{28}{11}$, $\frac{15}{11}$).

e) Intersection of $2x+3y=12$ and $3x+y=9$. (As calculated in thought process, this point (15/7, 18/7) is not feasible with $x+4y \leq 8$).

f) Intersection of $x+4y=8$ and $2x+3y=12$. (As calculated in thought process, this point (24/5, 4/5) is not feasible with $3x+y \leq 9$).

The vertices of the feasible region are (0, 0), (3, 0), ($\frac{28}{11}$, $\frac{15}{11}$), and (0, 2).

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

Comparing the values of Z at the vertices: $0, 3, \frac{43}{11}, 2$.

Since $\frac{43}{11} = 3 + \frac{10}{11} \approx 3.91$, it is the largest value.

Conclusion:

The maximum value of Z is $\frac{43}{11}$, which occurs at the point ($\frac{28}{11}$, $\frac{15}{11}$).

Given:

Production time per unit:

Model X: 6 man-hours

Model Y: 10 man-hours

Total available man-hours per week: 450

Handling and Marketing cost per unit:

Model X: $\textsf{₹}2000$

Model Y: $\textsf{₹}1000$

Total funds available for these purposes per week: $\textsf{₹}80000$

Profit per unit:

Model X: $\textsf{₹}1000$

Model Y: $\textsf{₹}500$

To Find:

The number of bikes of each model (Model X and Model Y) that should be produced per week to maximize the profit, and the maximum profit.

Formulation as LPP:

Let $x$ be the number of units of Model X produced per week.

Let $y$ be the number of units of Model Y produced per week.

These are our decision variables.

The objective is to maximize the total profit.

The objective function to maximize is:

$Z = 1000x + 500y$

The constraints are based on the available man-hours and the available funds for handling and marketing.

Constraints:

1. Constraint on total man-hours:

Total man-hours used = (Man-hours for $x$ units of X) + (Man-hours for $y$ units of Y)

Total man-hours used = $6x + 10y$

Total available man-hours = 450

$6x + 10y \leq 450$

Dividing by 2, we get:

$3x + 5y \leq 225$

2. Constraint on handling and marketing funds:

Total cost for handling and marketing = (Cost for $x$ units of X) + (Cost for $y$ units of Y)

Total cost = $2000x + 1000y$

Total funds available = $\textsf{₹}80000$

$2000x + 1000y \leq 80000$

Dividing by 1000, we get:

$2x + y \leq 80$

Also, the number of units produced cannot be negative.

Non-negativity constraints:

$x \geq 0$

$y \geq 0$

Solving the LPP:

Maximize $Z = 1000x + 500y$

Subject to:

$3x + 5y \leq 225$

$2x + y \leq 80$

$x \geq 0, y \geq 0$

We find the vertices of the feasible region:

a) Intersection of $x=0$ and $y=0$: Point (0, 0).

b) Intersection of $y=0$ and $2x + y = 80$: Setting $y=0$, $2x=80 \implies x=40$. Point (40, 0). Check $3x+5y \leq 225$: $3(40)+5(0)=120 \leq 225$ (True). Feasible.

c) Intersection of $x=0$ and $3x + 5y = 225$: Setting $x=0$, $5y=225 \implies y=45$. Point (0, 45). Check $2x+y \leq 80$: $2(0)+45=45 \leq 80$ (True). Feasible.

d) Intersection of $3x + 5y = 225$ and $2x + y = 80$. From the second equation, $y = 80 - 2x$. Substitute into the first equation:

$3x + 5(80 - 2x) = 225$

$3x + 400 - 10x = 225$

$-7x = 225 - 400$

$-7x = -175 \implies x = \frac{175}{7} = 25$

Substitute $x=25$ into $y = 80 - 2x$:

$y = 80 - 2(25) = 80 - 50 = 30$

Point (25, 30).

The vertices of the feasible region are (0, 0), (40, 0), (25, 30), and (0, 45).

Evaluate the objective function $Z = 1000x + 500y$ at each vertex:

Comparing the values of Z at the vertices (0, 40000, 40000, 22500), the maximum value is 40000.

The maximum profit is $\textsf{₹}40000$, which occurs at two vertices: (40, 0) and (25, 30). This means any point on the line segment connecting (40, 0) and (25, 30) that is within the feasible region also yields the maximum profit. However, since the number of bikes must be integers, we look at the vertices.

Conclusion:

The maximum profit is $\textsf{₹}40000$.

This maximum profit can be achieved by producing either:

40 units of Model X and 0 units of Model Y, OR

25 units of Model X and 30 units of Model Y.

Given:

Cost per tablet:

Tablet X: $\textsf{₹}2$

Tablet Y: $\textsf{₹}1$

To Find:

The number of tablets of type X and type Y to take to minimize the total cost while satisfying the minimum daily requirement of nutrients.

Formulation as LPP:

Let $x$ be the number of tablets of type X taken.

Let $y$ be the number of tablets of type Y taken.

These are our decision variables. Since the number of tablets must be whole, this is ideally an Integer Programming problem, but we will solve it as a standard LPP and check the vertex solution.

The objective is to minimize the total cost.

The objective function to minimize is:

$Z = 2x + y$

The constraints are based on the minimum required amounts of iron, calcium, and vitamins.

Constraints:

1. Iron constraint:

$6x + 2y \geq 18$

Simplifying by dividing by 2:

$3x + y \geq 9$

2. Calcium constraint:

$3x + 3y \geq 21$

Simplifying by dividing by 3:

$x + y \geq 7$

3. Vitamin constraint:

$2x + 4y \geq 16$

Simplifying by dividing by 2:

$x + 2y \geq 8$

Also, the number of tablets cannot be negative.

Non-negativity constraints:

$x \geq 0$

$y \geq 0$

Solving the LPP:

Minimize $Z = 2x + y$

Subject to:

$3x + y \geq 9$

$x + y \geq 7$

$x + 2y \geq 8$

$x \geq 0, y \geq 0$

The feasible region is in the first quadrant ($x \geq 0, y \geq 0$) and is the area above or on the lines $3x+y=9$, $x+y=7$, and $x+2y=8$.

We find the vertices (corner points) of this feasible region:

a) Intersection of $x=0$ and $3x+y=9$: $(0, 9)$. Check $0+9 \geq 7$ (True), $0+2(9)=18 \geq 8$ (True). Vertex: (0, 9).

b) Intersection of $y=0$ and $x+2y=8$: $(8, 0)$. Check $3(8)+0=24 \geq 9$ (True), $8+0 \geq 7$ (True). Vertex: (8, 0).

c) Intersection of $3x+y=9$ and $x+y=7$:

Subtract (ii) from (i): $(3x+y) - (x+y) = 9-7 \implies 2x = 2 \implies x=1$.

Substitute $x=1$ into $x+y=7$: $1+y=7 \implies y=6$. Point (1, 6).

Check $x+2y \geq 8$: $1+2(6) = 1+12 = 13 \geq 8$ (True). Feasible. Vertex: (1, 6).

d) Intersection of $x+y=7$ and $x+2y=8$:

Subtract (iii) from (iv): $(x+2y) - (x+y) = 8-7 \implies y = 1$.

Substitute $y=1$ into $x+y=7$: $x+1=7 \implies x=6$. Point (6, 1).

Check $3x+y \geq 9$: $3(6)+1 = 18+1 = 19 \geq 9$ (True). Feasible. Vertex: (6, 1).

e) Intersection of $3x+y=9$ and $x+2y=8$. (As determined in thought process, this point (2, 3) is not feasible with $x+y \geq 7$).

The vertices of the feasible region are (0, 9), (1, 6), (6, 1), and (8, 0).

We evaluate the objective function $Z = 2x + y$ at each vertex:

Comparing the values of Z at the vertices (9, 8, 13, 16), the minimum value is 8.

This minimum cost occurs at the vertex (1, 6).

Since the coordinates of the optimal vertex are integers, the integer solution is the same as the real solution.

Conclusion:

To satisfy the requirements at the minimum cost, the person should take 1 tablet of type X and 6 tablets of type Y.

The minimum cost will be $\textsf{₹}8$.

Given:

Minimum required calculators:

Model A: 6400

Model B: 4000

Model C: 4800

Production capacity per day:

Factory I: 50 (A), 50 (B), 30 (C)

Factory II: 40 (A), 20 (B), 40 (C)

Operating cost per day:

Factory I: $\textsf{₹}12000$

Factory II: $\textsf{₹}15000$

Objective: Minimise the total operating cost.

To Find:

The number of days each factory should operate to meet the demand at the minimum operating cost, and the minimum cost.

Formulation as LPP:

Let $x$ be the number of days Factory I operates.

Let $y$ be the number of days Factory II operates.

These are our decision variables.

The objective is to minimize the total operating cost.

The objective function to minimize is:

$Z = 12000x + 15000y$

The constraints are based on meeting the minimum demand for each model.

Constraints:

1. Constraint for Model A demand:

Total Model A produced = (Production from Factory I) + (Production from Factory II)

Total Model A produced = $50x + 40y$

Minimum required = 6400

$50x + 40y \geq 6400$

Dividing by 10:

$5x + 4y \geq 640$

2. Constraint for Model B demand:

Total Model B produced = $50x + 20y$

Minimum required = 4000

$50x + 20y \geq 4000$

Dividing by 10:

$5x + 2y \geq 400$

3. Constraint for Model C demand:

Total Model C produced = $30x + 40y$

Minimum required = 4800

$30x + 40y \geq 4800$

Dividing by 10:

$3x + 4y \geq 480$

Also, the number of days operated cannot be negative.

Non-negativity constraints:

$x \geq 0$

$y \geq 0$

Since the number of days must be integers, $x$ and $y$ are also restricted to be integers. We will solve as a standard LPP and check the optimal vertex.

Solving the LPP:

Minimize $Z = 12000x + 15000y$

Subject to:

$5x + 4y \geq 640$

$5x + 2y \geq 400$

$3x + 4y \geq 480$

$x \geq 0, y \geq 0$

We find the vertices of the feasible region, which is the area in the first quadrant satisfying all '$\geq$' inequalities. The vertices are the intersection points of the boundary lines that define the boundaries of this feasible region.

Boundary lines are $5x + 4y = 640$, $5x + 2y = 400$, $3x + 4y = 480$, $x=0$, and $y=0$.

The vertices of the feasible region are:

a) Intersection of $x=0$ and $5x + 4y = 640$ is (0, 160). Check other constraints: $5(0)+2(160)=320 \geq 400$ (False). Not a vertex of the feasible region.

b) Intersection of $x=0$ and $3x + 4y = 480$ is (0, 120). Check other constraints: $5(0)+2(120)=240 \geq 400$ (False). Not a vertex of the feasible region.

c) Intersection of $x=0$ and $5x + 2y = 400$ is (0, 200). Check other constraints: $5(0)+4(200)=800 \geq 640$ (True), $3(0)+4(200)=800 \geq 480$ (True). Vertex: (0, 200).

d) Intersection of $y=0$ and $5x + 2y = 400$ is (80, 0). Check other constraints: $5(80)+4(0)=400 \geq 640$ (False). Not a vertex of the feasible region.

e) Intersection of $y=0$ and $3x + 4y = 480$ is (160, 0). Check other constraints: $5(160)+4(0)=800 \geq 640$ (True), $5(160)+2(0)=800 \geq 400$ (True). Vertex: (160, 0).

f) Intersection of $5x + 4y = 640$ and $5x + 2y = 400$. Subtracting the second from the first: $2y = 240 \implies y = 120$. Substituting into $5x+2y=400$: $5x + 2(120) = 400 \implies 5x + 240 = 400 \implies 5x = 160 \implies x = 32$. Point (32, 120). Check third constraint: $3(32)+4(120) = 96 + 480 = 576 \geq 480$ (True). Vertex: (32, 120).

g) Intersection of $5x + 4y = 640$ and $3x + 4y = 480$. Subtracting the second from the first: $2x = 160 \implies x = 80$. Substituting into $3x+4y=480$: $3(80)+4y=480 \implies 240+4y=480 \implies 4y=240 \implies y=60$. Point (80, 60). Check second constraint: $5(80)+2(60)=400+120=520 \geq 400$ (True). Vertex: (80, 60).

h) Intersection of $5x + 2y = 400$ and $3x + 4y = 480$. (As determined in thought process, this point is not in the feasible region).

The vertices of the feasible region are (0, 200), (32, 120), (80, 60), and (160, 0).

We evaluate the objective function $Z = 12000x + 15000y$ at each vertex:

Comparing the values of Z at the vertices (3,000,000, 2,184,000, 1,860,000, 1,920,000), the minimum value is 1,860,000.

The feasible region is unbounded, but since the objective function involves minimizing with positive coefficients and the constraints are all '$\geq$' type, the minimum value exists and occurs at a corner point.

The optimal vertex is (80, 60), which consists of integer values, satisfying the implicit requirement that the number of days must be a whole number.

Conclusion:

To minimize the operating costs and still meet the demand, Factory I should operate for 80 days and Factory II should operate for 60 days.

The minimum operating cost will be $\textsf{₹}1,860,000$.

Given:

Objective function: Maximize and Minimize $Z = 3x – 4y$

Constraints:

$x – 2y \leq 0$

$- 3x + y \leq 4$

$x – y \leq 6$

$x \geq 0$

$y \geq 0$

To Find:

The maximum and minimum values of Z subject to the given constraints.

Solution:

We first identify the feasible region defined by the given constraints. The constraints $x \geq 0$ and $y \geq 0$ mean the feasible region lies in the first quadrant.

The boundary lines for the other constraints are:

1. $x – 2y = 0 \implies y = \frac{1}{2}x$

2. $- 3x + y = 4 \implies y = 3x + 4$

3. $x – y = 6 \implies y = x – 6$

We find the vertices (corner points) of the feasible region by finding the intersection points of these boundary lines within the first quadrant that satisfy all the inequalities.

The vertices of the feasible region are found by intersecting the boundary lines:

a) Intersection of $x=0$ and $y=0$: (0, 0). This point satisfies all constraints ($0 \leq 0, 0 \leq 4, 0 \leq 6, 0 \geq 0, 0 \geq 0$). Vertex: (0, 0).

b) Intersection of $x=0$ and $y=3x+4$: Setting $x=0$ in $y=3x+4$ gives $y=4$. Point (0, 4). Check other constraints: $0-2(4)=-8 \leq 0$ (T), $-3(0)+4=4 \leq 4$ (T), $0-4=-4 \leq 6$ (T), $0 \geq 0, 4 \geq 0$ (T). Vertex: (0, 4).

c) Intersection of $y=x/2$ and $y=x-6$: Substitute $y=x/2$ into $y=x-6$: $\frac{x}{2} = x - 6 \implies x = 2x - 12 \implies x = 12$. Then $y = 12/2 = 6$. Point (12, 6). Check other constraints: $-3(12)+6 = -36+6 = -30 \leq 4$ (T), $12-6 = 6 \leq 6$ (T), $12 \geq 0, 6 \geq 0$ (T). Vertex: (12, 6).

d) Intersection of $y=3x+4$ and $x=12$: Setting $x=12$ in $y=3x+4$ gives $y=3(12)+4 = 36+4 = 40$. Point (12, 40). Check other constraints: $12-2(40)=12-80=-68 \leq 0$ (T), $-3(12)+40=-36+40=4 \leq 4$ (T), $12-40=-28 \leq 6$ (T), $12 \geq 0, 40 \geq 0$ (T). Vertex: (12, 40).

Other intersections (e.g., $y=x/2$ and $y=3x+4$, $y=3x+4$ and $y=x-6$, $y=x-6$ and $y=0$) do not result in vertices of the feasible region in the first quadrant satisfying all inequalities.

The feasible region is a bounded polygon with vertices (0, 0), (0, 4), (12, 40), and (12, 6).

We evaluate the objective function $Z = 3x - 4y$ at each vertex:

Comparing the values of Z at the vertices (0, -16, 12, -124), we find the minimum and maximum values.

The minimum value is -124.

The maximum value is 12.

Conclusion:

The minimum value of Z is -124, which occurs at the vertex (12, 40).

The maximum value of Z is 12, which occurs at the vertex (12, 6).

Given:

The corner points of the feasible region are (0, 0), (0, 40), (20, 40), (60, 20), and (60, 0).

Objective Function:

$Z = 4x + 3y$

To Find:

The maximum value of Z and compare it with 325.

Solution:

We evaluate the objective function $Z = 4x + 3y$ at each of the given corner points.

The values of Z at the corner points are 0, 120, 200, 300, and 240.

The maximum value of Z is the largest among these values, which is 300.

So, the quantity in Column A (Maximum of Z) is 300.

Comparison:

Column A = Maximum of Z = 300

Column B = 325

Comparing Column A and Column B: $300 < 325$

The quantity in Column B is greater than the quantity in Column A.

Conclusion:

The quantity in Column B is greater.

Therefore, the correct option is (B) The quantity in column B is greater.

Given:

The feasible region is shown in the figure.

The corner points of the feasible region, as visible from the figure, are (0, 0), (5, 0), (4, 10), and (0, 8).

Objective Function:

$Z = 3x - 4y$

To Find:

The corner point where the minimum value of Z occurs.

Solution:

We evaluate the objective function $Z = 3x - 4y$ at each of the corner points of the feasible region.

The values of Z at the corner points are 0, 15, -28, and -32.

The minimum value of Z is the smallest among these values, which is -32.

This minimum value of Z occurs at the corner point (0, 8).

Conclusion:

The minimum of Z occurs at the point (0, 8).

Therefore, the correct option is (B) (0, 8).

Given:

This question refers to Exercise 27. From the figure in Exercise 27, the corner points of the feasible region are (0, 0), (5, 0), (4, 10), and (0, 8).

Objective Function:

From Exercise 27, the objective function is $Z = 3x - 4y$.

To Find:

The corner point where the maximum value of Z occurs.

Solution:

We evaluate the objective function $Z = 3x - 4y$ at each of the corner points of the feasible region, as determined in Exercise 27.

The values of Z at the corner points are 0, 15, -28, and -32.

The maximum value of Z is the largest among these values, which is 15.

This maximum value of Z occurs at the corner point (5, 0).

Conclusion:

The maximum of Z occurs at the point (5, 0).

Therefore, the correct option is (A) (5, 0).

Given:

This question refers to Exercise 27.

From the solution of Exercise 27, the minimum value of the objective function $Z = 3x - 4y$ in the given feasible region is -32.

From the solution of Exercise 28, the maximum value of the objective function $Z = 3x - 4y$ in the given feasible region is 15.

To Find:

The sum of the maximum and minimum values of Z.

Solution:

Minimum value of Z = -32

Maximum value of Z = 15

We need to calculate (Maximum value of Z + Minimum value of Z).

Sum = Maximum value of Z + Minimum value of Z

Sum = $15 + (-32)$

Sum = $15 - 32$

Sum = $-17$

Conclusion:

The sum of the maximum and minimum values of Z is -17.

Therefore, the correct option is (D) – 17.

Given:

The feasible region is shown in the figure. The corner points of the feasible region, as identified from the figure, are (8, 0), (12, 0), (6, 5), and (0, 8).

Objective Function:

$F = 3x - 4y$

To Find:

The maximum value of the objective function F in the given feasible region.

Solution:

We evaluate the objective function $F = 3x - 4y$ at each of the corner points of the feasible region.

The values of F at the corner points are 24, 36, -2, and -32.

The maximum value of F is the largest among these values, which is 36.

Conclusion:

The maximum value of F is 36.

(Note: The calculated maximum value 36 is not among the given options. Based on the provided problem statement and figure, the maximum value is 36).

Given:

This question refers to Exercise 30. From the figure in Exercise 30, the corner points of the feasible region are (8, 0), (12, 0), (6, 5), and (0, 8).

Objective Function:

From Exercise 30, the objective function is $F = 3x - 4y$.

To Find:

The minimum value of the objective function F in the given feasible region.

Solution:

We evaluate the objective function $F = 3x - 4y$ at each of the corner points of the feasible region.

The values of F at the corner points are 24, 36, -2, and -32.

The minimum value of F is the smallest among these values, which is -32.

Conclusion:

The minimum value of F is -32.

Upon comparing this value with the given options, it is observed that -32 is not among the options (A) 0, (B) – 16, (C) 12, (D) does not exist. Since the feasible region is bounded, a minimum value must exist.

Based on the provided figure and objective function, the minimum value calculated is -32.

Given:

The corner points of the feasible region are (0, 2), (3, 0), (6, 0), (6, 8), and (0, 5).

Objective Function:

$F = 4x + 6y$

To Find:

The point(s) where the minimum value of F occurs.

Solution:

We evaluate the objective function $F = 4x + 6y$ at each of the given corner points of the feasible region.

The values of F at the corner points are 12, 12, 24, 72, and 30.

The minimum value of F is the smallest among these values, which is 12.

This minimum value occurs at two distinct corner points: (0, 2) and (3, 0).

According to the fundamental theorem of linear programming, if the optimal value (maximum or minimum) of the objective function occurs at two corner points of the feasible region, then it occurs at every point on the line segment joining these two points.

Conclusion:

The minimum value of F occurs at any point on the line segment joining the points (0, 2) and (3, 0).

Therefore, the correct option is (D) any point on the line segment joining the points (0, 2) and (3, 0).

Given:

This question refers to Exercise 32. From the solution to Exercise 32, we evaluated the objective function $F = 4x + 6y$ at the corner points of the feasible region.

Values of F at Corner Points (from Exercise 32):

F(0, 2) = 12

F(3, 0) = 12

F(6, 0) = 24

F(6, 8) = 72

F(0, 5) = 30

Maximum and Minimum Values of F:

From the values calculated above:

The minimum value of F is 12.

The maximum value of F is 72.

To Find:

Maximum of F - Minimum of F.

Solution:

Maximum of F - Minimum of F = $72 - 12$

Maximum of F - Minimum of F = $60$

Conclusion:

The difference between the maximum and minimum values of F is 60.

Therefore, the correct option is (A) 60.

Given:

The corner points of the feasible region are (0, 3), (1, 1), and (3, 0).

The objective function is $Z = px + qy$, where $p > 0$ and $q > 0$.

The minimum value of Z occurs at the points (3, 0) and (1, 1).

To Find:

The condition on p and q for the minimum of Z to occur at both (3, 0) and (1, 1).

Solution:

If the minimum value of the objective function occurs at two corner points of the feasible region, then the minimum value of the objective function is the same at these two points. This also implies that the minimum value occurs at every point on the line segment joining these two points.

Given that the minimum of Z occurs at (3, 0) and (1, 1), the value of Z at (3, 0) must be equal to the value of Z at (1, 1).

Let's evaluate the objective function $Z = px + qy$ at these two points:

Value of Z at (3, 0):

$Z(3, 0) = p(3) + q(0) = 3p$

Value of Z at (1, 1):

$Z(1, 1) = p(1) + q(1) = p + q$

Since the minimum occurs at both points, we have:

Now, we solve for the relationship between p and q:

$3p - p = q$

$2p = q$

This relationship can also be written as $p = \frac{q}{2}$.

Let's verify if this value is indeed the minimum by evaluating Z at the remaining corner point (0, 3).

Value of Z at (0, 3):

$Z(0, 3) = p(0) + q(3) = 3q$

Substitute $q = 2p$ (since $p, q > 0$):

$Z(0, 3) = 3(2p) = 6p$

Now compare the values:

$Z(3, 0) = 3p$

$Z(1, 1) = p + q = p + 2p = 3p$

$Z(0, 3) = 6p$

Since $p > 0$, we have $3p < 6p$. Thus, $3p$ is indeed the minimum value among the corner points when $q = 2p$.

Conclusion:

The condition on p and q for the minimum of Z to occur at (3, 0) and (1, 1) is $2p = q$ or $p = \frac{q}{2}$.

Comparing with the given options, the correct option is (B) $p = \frac{q}{2}$.

Answer: constraints

Explanation:

In a Linear Programming Problem (LPP), the goal is to optimize (maximize or minimize) a linear objective function subject to certain limitations or restrictions.

These restrictions are expressed in the form of linear inequalities or sometimes linear equations involving the decision variables.

These linear inequalities or equations that limit the values of the variables are formally called constraints.

For example, if you are producing two types of goods, X and Y, and have limited raw materials and labor hours, the inequalities representing these limitations (e.g., amount of raw material used $\le$ total raw material available) are the constraints.

The set of all points satisfying all the constraints simultaneously forms the feasible region, which represents all possible valid solutions to the problem.

Common examples of constraints include resource constraints (like material, labor, machine time), demand constraints, and non-negativity constraints (variables like quantity produced cannot be negative, i.e., $x \ge 0, y \ge 0$).

Answer: linear

Explanation:

A Linear Programming Problem (LPP) is a mathematical method for determining a way to achieve the best outcome (such as maximum profit or lowest cost) in a mathematical model whose requirements are represented by linear relationships.

The term "Linear" in Linear Programming Problem specifically refers to the fact that both the objective function and the constraints are expressed as linear functions of the decision variables.

The objective function is the function that you want to maximize or minimize. In an LPP, this function is always a linear combination of the variables.

For example, if the variables are $x$ and $y$, a linear objective function would look like $Z = ax + by$, where $a$ and $b$ are constants.

Functions like $Z = x^2 + y$ or $Z = xy$ are not linear, and therefore, problems involving such functions as the objective function are not considered LPPs in the standard sense.

Answer: unbounded

Explanation:

In a Linear Programming Problem (LPP), the feasible region is the set of all points that satisfy all the constraints. The optimal solution (maximum or minimum value of the objective function) is usually found at one of the corner points of the feasible region.

If the feasible region is bounded, it means it can be enclosed within a finite region. In this case, the objective function $Z = ax + by$ will always attain both a maximum and a minimum value at one or more of the corner points.

However, if the feasible region is unbounded, it extends infinitely in one or more directions. In such a case, the objective function $Z = ax + by$:

1. May have a minimum value but no maximum value.

2. May have a maximum value but no minimum value.

3. May have neither a maximum nor a minimum value.

For example, if the feasible region is the first quadrant ($x \ge 0, y \ge 0$) and the objective function is $Z = x + y$, then Z can increase indefinitely as x and y increase, so there is no maximum. The minimum value occurs at (0, 0), which is 0.

The question states that the optimal value (maximum or minimum) "may or may not exist", which is characteristic of an unbounded feasible region.

Answer: maximum

Explanation:

In a Linear Programming Problem (LPP), the feasible region is a convex set, and the optimal value (maximum or minimum) of the linear objective function $Z = ax + by$ occurs at one or more corner points of the feasible region.

A fundamental theorem of linear programming states that if the optimal value of the objective function is attained at two distinct corner points, it is also attained at every point on the line segment joining these two points.

This means if the maximum value of Z is the same at corner point A and corner point B, then any point $(x, y)$ lying on the line segment AB will also result in the same maximum value for the objective function $Z = ax + by$.

Therefore, every point on the line segment joining the two points gives the same maximum value.

Answer: bounded

Explanation:

In the context of Linear Programming Problems (LPPs), the feasible region is the set of all points that satisfy all the given linear constraints.

A feasible region is classified as either bounded or unbounded.

A feasible region is said to be bounded if it is possible to draw a circle (of sufficiently large radius) such that the entire feasible region lies within that circle. This means the feasible region does not extend infinitely in any direction.

Conversely, a feasible region is said to be unbounded if it cannot be enclosed within any finite circle. This occurs when the feasible region extends infinitely in one or more directions.

The description "can be enclosed within a circle" is the defining characteristic of a bounded feasible region.

Answer: intersection

Explanation:

In a Linear Programming Problem (LPP), the feasible region is defined by a system of linear inequalities (constraints).

Each inequality corresponds to a half-plane, and the feasible region is the intersection of all these half-planes.

The boundary of each half-plane is a straight line, which is obtained by replacing the inequality sign with an equality sign.

A corner point (also known as a vertex) of the feasible region is a point where two or more of these boundary lines intersect.

These intersection points lie within or on the boundary of the feasible region.

According to the corner point theorem, the optimal solution (maximum or minimum value) of the objective function, if it exists, will occur at one of these corner points.

Thus, a corner point is specifically the intersection of two boundary lines of the feasible region.

Answer: convex

Explanation:

In a Linear Programming Problem (LPP), the feasible region is defined by a system of linear inequalities.

Each linear inequality in two variables, say $ax + by \le c$ or $ax + by \ge c$, represents a half-plane in the plane. A half-plane is a region on one side of a straight line $ax + by = c$.

A key property of a half-plane is that it is a convex set. A set is convex if, for any two points within the set, the entire line segment connecting those two points also lies within the set.

The feasible region of an LPP is the set of points that satisfy all the linear constraints simultaneously. In other words, it is the intersection of all the half-planes defined by the constraints.

A fundamental property of convex sets is that the intersection of any number of convex sets is also a convex set.

Since the feasible region is the intersection of half-planes (which are convex sets), the feasible region itself is always a convex set.

If the feasible region is bounded and non-empty, it forms a polygon. Because it is a convex set, this polygon will always be a convex polygon.

A convex polygon has the property that for any two points inside or on the boundary of the polygon, the line segment connecting them lies entirely within or on the boundary of the polygon.

Answer: True

Explanation:

This statement is correct. When the feasible region of a Linear Programming Problem (LPP) is unbounded, the objective function does not necessarily attain an optimal value (maximum or minimum).

While an optimal value might exist (e.g., a minimum value may exist but no maximum, or vice versa), it is not guaranteed that both a maximum and a minimum value will exist, or even that one of them will exist, depending on the direction in which the objective function increases or decreases relative to the unbounded part of the feasible region.

For example, if the feasible region is $x \ge 0, y \ge 0$ (unbounded) and the objective function is $Z = x + y$, there is no maximum value as Z can increase indefinitely. However, the minimum value is 0 at (0, 0).

If the objective function was $Z = -x - y$ over the same feasible region, there would be no minimum value, but the maximum value would be 0 at (0, 0).

Since the optimal value (maximum or minimum) is not always guaranteed to exist in an unbounded feasible region, the statement that it "may or may not exist" is true.

Answer: False

Explanation:

This statement is incorrect.

While it is true that if the maximum value of the objective function exists in a Linear Programming Problem (LPP) with a bounded feasible region, it will occur at a corner point, it does not necessarily occur at only one corner point.

According to the fundamental theorem of linear programming, if the optimal value (maximum or minimum) of the objective function occurs at two distinct corner points of the feasible region, then it occurs at every point on the line segment joining these two points.

This means the maximum value can occur at:

1. Exactly one corner point.

2. At two (or more) corner points and every point on the line segment(s) connecting them.

Therefore, the statement that the maximum value always occurs at only one corner point is false, as it can occur at infinitely many points if it occurs at two distinct corner points.

Answer: False

Explanation:

This statement is incorrect.

If the origin (0, 0) is a corner point of the feasible region, the value of the objective function $Z = ax + by$ at the origin is:

While the objective function has a value of 0 at the origin, this value is not necessarily the minimum value of Z over the entire feasible region.

The minimum value of the objective function occurs at one or more corner points of the feasible region. If there are other corner points where the objective function takes a negative value, then that negative value would be the minimum, which is less than 0.

For example, consider the objective function $Z = x - y$ and a feasible region with corner points (0, 0), (5, 0), and (0, 3). The origin (0, 0) is a corner point.

Evaluate Z at the corner points:

$Z(0, 0) = 0 - 0 = 0$

$Z(5, 0) = 5 - 0 = 5$

$Z(0, 3) = 0 - 3 = -3$

In this example, the minimum value of Z is -3, which occurs at the point (0, 3), even though the origin (0, 0) is a corner point and Z(0, 0) = 0.

Therefore, the minimum value of the objective function is not always 0 just because the origin is a corner point. It depends on the coefficients $a$ and $b$ and the location of the other corner points of the feasible region.

Answer: False

Explanation:

This statement is incorrect.

In a Linear Programming Problem (LPP), the maximum value of the objective function $Z = ax + by$ is not always finite.

The existence and finiteness of the maximum value depend on the nature of the feasible region:

1. If the feasible region is bounded (can be enclosed within a circle), then the objective function will always attain both a maximum and a minimum value, and these values will be finite.

2. If the feasible region is unbounded (extends infinitely in at least one direction), the objective function may or may not have a maximum value. If the objective function increases indefinitely within the feasible region, then its maximum value does not exist (or is infinite).

For example, consider the LPP:

Maximize $Z = x + y$

Subject to the constraints:

$x \ge 0$

$y \ge 0$

The feasible region is the entire first quadrant, which is an unbounded region. As $x$ and $y$ increase, the value of $Z = x + y$ also increases without limit. Therefore, the maximum value of Z does not exist (it is not finite).

Since there exist LPPs with unbounded feasible regions where the maximum value of the objective function is not finite, the statement "the maximum value of the objective function Z = ax + by is always finite" is false.