Chapter 12 Linear Programming (Additional Questions)

(C) Objective function

Explanation:

Linear programming is a mathematical optimization technique used to achieve the best outcome (such as maximum profit or lowest cost) in a mathematical model whose requirements are represented by linear relationships. It involves finding the optimal value (maximum or minimum) of a linear function, known as the objective function, subject to a set of linear constraints.

The objective function is a linear expression (e.g., $Z = ax + by$) that represents the quantity to be maximized or minimized.

Let's consider the other options:

• (A) Equation: Linear equations are often part of the constraints, but linear programming optimizes a function, not just a single equation.

• (B) Inequality: Linear inequalities are typically used to define the feasible region (the set of all possible solutions) through constraints, but the goal is to optimize the objective function, not an inequality itself.

• (D) Constraint: Constraints are the limitations or restrictions (expressed as linear equations or inequalities) that must be satisfied by the solution. The objective function is optimized *subject to* these constraints, but the constraint itself is not what is being optimized.

Therefore, linear programming is fundamentally about optimizing a linear objective function.

(B) Feasible region

Explanation:

In linear programming, a system of linear inequalities defines the constraints of the problem. The set of all points that satisfy all these inequalities simultaneously forms a region in the coordinate plane (or higher-dimensional space).

This region is known as the feasible region. It represents all possible solutions to the problem that satisfy all the given constraints.

Let's consider the other options:

• (A) Solution set: While the feasible region contains solutions, "solution set" is a more general term that can apply to various mathematical problems, not specifically to the graphical region in linear programming.

• (C) Optimal region: The optimal solution (which gives the maximum or minimum value of the objective function) is typically found at one of the corner points (vertices) of the feasible region. There isn't an "optimal region" in the same sense; rather, there's an optimal point or set of points within or on the boundary of the feasible region.

• (D) Constraint region: While the region is defined by constraints, the standard term for this region in the context of linear programming is the "feasible region."

Therefore, the region represented by a system of linear inequalities in the context of linear programming is called the feasible region.

(B) Linear

Explanation:

By definition, a linear programming problem (LPP) involves optimizing (maximizing or minimizing) a linear objective function subject to a set of linear constraints.

A linear function is one where the variables are raised to the power of 1. For example, if $x$ and $y$ are decision variables, a linear objective function would be of the form $Z = ax + by + c$, where $a$, $b$, and $c$ are constants.

Let's consider why the other options are incorrect:

• (A) Quadratic: If the objective function were quadratic (e.g., $Z = ax^2 + by^2 + cxy + dx + ey + f$), the problem would be a quadratic programming problem, not a linear programming problem.

• (C) Exponential: An exponential objective function (e.g., $Z = a^x + b^y$) would make it a non-linear programming problem.

• (D) Logarithmic: A logarithmic objective function (e.g., $Z = a \log(x) + b \log(y)$) would also categorize it as a non-linear programming problem.

Therefore, the objective function in a linear programming problem is always linear.

(B) Inequalities

Explanation:

In a linear programming problem, constraints define the limitations or restrictions on the decision variables. These restrictions are typically expressed as a system of linear inequalities. Sometimes, constraints can also be linear equations, but inequalities are more common as they represent conditions like "at least", "at most", "not more than", or "not less than".

For example, a constraint might be $2x + 3y \leq 100$ (representing a resource limitation) or $x \geq 0$ (a non-negativity constraint, which is a type of linear inequality).

Let's consider the other options:

• (A) Equations: While linear equations can be part of the constraints (e.g., $x + y = 50$), inequalities are more general and frequently used to define the feasible region.

• (C) Functions: The objective function is a linear function. Constraints are relationships that involve linear expressions, but they are typically expressed as inequalities or equations, not just as "functions" in isolation of a relational operator.

• (D) Variables: Variables (e.g., $x$, $y$) are the quantities we are trying to determine. Constraints are conditions imposed on these variables.

Therefore, constraints in a linear programming problem are usually expressed as a system of linear inequalities, which may also include linear equations as a special case.

(C) Vertex (corner point) of the feasible region

Explanation:

This is a fundamental concept in linear programming known as the Corner Point Theorem or the Fundamental Theorem of Linear Programming.

The theorem states that if a linear programming problem has an optimal solution (maximum or minimum value of the objective function), and the feasible region is non-empty and bounded (or if unbounded, the optimal solution exists), then the optimal solution must occur at one or more of the vertices (corner points) of the feasible region.

Let's consider the options:

• (A) Interior point of the feasible region: If the optimal solution were at an interior point, you could always move in some direction within the feasible region to either increase or decrease the objective function's value (unless the objective function is constant, in which case the vertices would still be optimal). Thus, an interior point is generally not where the unique optimum lies.

• (B) Edge of the feasible region: It is possible for the optimal solution to occur along an entire edge of the feasible region. This happens when the objective function line is parallel to that edge. However, in such cases, the vertices at both ends of that edge are also optimal solutions. So, stating that it occurs at a vertex is more fundamental and always true if an optimal solution exists at an edge.

• (C) Vertex (corner point) of the feasible region: This is the correct answer according to the Corner Point Theorem. The optimal value is always achieved at one of the vertices. If it's achieved along an edge, it's also achieved at the vertices defining that edge.

• (D) Point outside the feasible region: A point outside the feasible region does not satisfy all the constraints of the problem, so it cannot be a solution, let alone an optimal solution.

Therefore, the optimal solution (if it exists) for a linear programming problem with a convex polygonal feasible region occurs at a vertex (corner point) of the feasible region.

(B) May or may not exist

Explanation:

If the feasible region of a linear programming problem is unbounded, the existence of an optimal solution (maximum or minimum) for the objective function depends on the nature of the objective function and the direction(s) in which the feasible region is unbounded.

Consider the following scenarios for maximizing an objective function $Z$:

Scenario 1: Maximum value may not exist.

Let the objective function be $Z = x + y$.

Constraints:

$x \geq 0$

$y \geq 0$

The feasible region is the first quadrant, which is unbounded. As $x$ and $y$ can increase indefinitely, the value of $Z = x + y$ can also increase indefinitely. In this case, there is no maximum value.

Scenario 2: Maximum value may exist.

Let the objective function be $Z = -x - y$.

Constraints:

$x \geq 0$

$y \geq 0$

The feasible region is still the first quadrant (unbounded). However, as $x$ and $y$ increase, the value of $Z = -x - y$ becomes more negative (decreases). The maximum value of $Z$ occurs at $x=0, y=0$, where $Z=0$. In this case, a maximum value exists.

Scenario 3: Maximum value may exist (different example).

Let the objective function be $Z = x$.

Constraints:

$x \leq 5$

$x \geq 0$

$y \geq 0$

The feasible region is a semi-infinite strip defined by $0 \leq x \leq 5$ and $y \geq 0$. This region is unbounded in the positive $y$-direction. The objective function $Z=x$ depends only on $x$. The maximum value of $x$ in the feasible region is 5. Thus, the maximum value of $Z$ is 5. In this case, a maximum value exists despite the unbounded feasible region.

Therefore, if the feasible region is unbounded, the maximum value of the objective function may or may not exist. It depends on whether the objective function can increase indefinitely within the feasible region.

• (A) Always exists: This is false, as shown in Scenario 1.

• (C) Never exists: This is false, as shown in Scenario 2 and Scenario 3.

• (D) Is always zero: This is false. In Scenario 2, the maximum was zero, but in Scenario 3, the maximum was 5. It could be any finite value if it exists.

(B) Non-negativity constraints

Explanation:

In linear programming problems, decision variables (like $x$ and $y$) often represent quantities that cannot be negative, such as the number of items produced, amount of resources used, or time spent on an activity.

The constraints $x \ge 0$ and $y \ge 0$ ensure that these decision variables take on non-negative values. These are universally referred to as non-negativity constraints.

Let's look at the other options:

• (A) Functional constraints: These are the main constraints of the problem that arise from the specific limitations or requirements of the situation being modeled, beyond just non-negativity. For example, $2x + 5y \leq 100$ would be a functional constraint related to resource availability.

• (C) Objective constraints: This term is not standard. The objective function is what we aim to optimize; constraints are limitations on the variables within that function.

• (D) Boundary constraints: While $x=0$ and $y=0$ do define boundaries of the feasible region (typically the axes in a 2D problem), the specific term for $x \ge 0$ and $y \ge 0$ is "non-negativity constraints." "Boundary constraints" is a more general term that could refer to any constraint defining a boundary.

Therefore, $x \ge 0$ and $y \ge 0$ are specifically known as non-negativity constraints.

(C) Empty

Explanation:

In a Linear Programming Problem (LPP), the feasible region is the set of all points that satisfy all the given constraints simultaneously.

If the constraints are inconsistent, it means that there is no combination of values for the decision variables that can satisfy all the constraints at the same time. In other words, the conditions imposed by the constraints are contradictory.

For example, consider the following inconsistent constraints:

$x \leq 2$

$x \geq 5$

There is no value of $x$ that can be both less than or equal to 2 and greater than or equal to 5 simultaneously. Therefore, the set of points satisfying both these constraints is empty.

When the feasible region contains no points, it is called an empty set (or null set). In such cases, the LPP has no solution.

Let's consider the other options:

• (A) Bounded: A bounded feasible region is one that can be enclosed within a circle of finite radius. If a region is bounded, it contains points, so the constraints are consistent.

• (B) Unbounded: An unbounded feasible region extends indefinitely in at least one direction. If a region is unbounded, it contains points, so the constraints are consistent.

• (D) A single point: If the feasible region is a single point, it means there is exactly one solution that satisfies all constraints. This implies the constraints are consistent.

Therefore, if the constraints in an LPP are inconsistent, the feasible region is empty.

(A) $(0, 0), (4, 0), (0, 4)$

Explanation:

The given constraints are:

The constraints $x \ge 0$ and $y \ge 0$ indicate that the feasible region lies in the first quadrant of the Cartesian plane.

To find the corner points of the feasible region, we consider the corresponding equations for the inequalities and find their points of intersection:

1. Boundary line for $x + y \le 4$ is $x + y = 4$.

2. Boundary line for $x \ge 0$ is $x = 0$ (the y-axis).

3. Boundary line for $y \ge 0$ is $y = 0$ (the x-axis).

Now, we find the intersection points of these lines:

a) Intersection of $x = 0$ and $y = 0$:

Substituting $x=0$ and $y=0$ gives the point $(0, 0)$.

Checking with constraint (i): $0 + 0 = 0 \le 4$. (Satisfied)

So, $(0, 0)$ is a corner point.

b) Intersection of $x = 0$ (y-axis) and $x + y = 4$:

Substitute $x = 0$ into $x + y = 4$:

$0 + y = 4 \Rightarrow y = 4$.

This gives the point $(0, 4)$.

Checking with other constraints: $0 \ge 0$ (Satisfied), $4 \ge 0$ (Satisfied).

So, $(0, 4)$ is a corner point.

c) Intersection of $y = 0$ (x-axis) and $x + y = 4$:

Substitute $y = 0$ into $x + y = 4$:

$x + 0 = 4 \Rightarrow x = 4$.

This gives the point $(4, 0)$.

Checking with other constraints: $4 \ge 0$ (Satisfied), $0 \ge 0$ (Satisfied).

So, $(4, 0)$ is a corner point.

The feasible region is a triangle formed by these three points. The corner points are $(0, 0)$, $(4, 0)$, and $(0, 4)$.

Let's check why other options are incorrect:

• Option (B) includes $(4, 4)$. For $(4, 4)$, $x+y = 4+4 = 8$, which is not $\le 4$. So $(4,4)$ is not in the feasible region.

• Option (C) includes $(2, 2)$. For $(2, 2)$, $x+y = 2+2 = 4 \le 4$; $x=2 \ge 0$; $y=2 \ge 0$. While $(2,2)$ is on the boundary line $x+y=4$ and within the feasible region, it is not a corner point (vertex) formed by the intersection of two distinct boundary lines that define the polygon.

• Option (D) includes $(1, 3)$ and $(3, 1)$. Both points satisfy all constraints (e.g., for $(1,3)$, $1+3=4 \le 4$, $1 \ge 0$, $3 \ge 0$). However, like $(2,2)$, these are points on the line segment $x+y=4$ between $(0,4)$ and $(4,0)$, not corner points.

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

(B) $12$ (at $(4,0)$)

Explanation:

To find the maximum value of the objective function $Z = 3x + 2y$, we evaluate $Z$ at each of the given corner points of the feasible region.

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

The objective function is $Z = 3x + 2y$.

We can create a table to evaluate $Z$ at each corner point:

Comparing the values of $Z$ at the corner points:

• At $(0, 0)$, $Z = 0$.

• At $(4, 0)$, $Z = 12$.

• At $(0, 4)$, $Z = 8$.

The maximum value of $Z$ is $12$, which occurs at the corner point $(4, 0)$.

Therefore, the maximum value of $Z$ is $12$ at $(4,0)$.

(A) Decision variables

(C) Linear constraints

(D) Maximizing or minimizing the objective function

Explanation:

A linear programming problem (LPP) consists of several key components:

• (A) Decision variables: These are the variables representing the unknown quantities that need to be determined to find the optimal solution. For example, in a production problem, decision variables might be the number of units of different products to manufacture. These are fundamental to an LPP.

• (B) Non-linear objective function: This is incorrect. A defining characteristic of a linear programming problem is that the objective function must be linear. If the objective function is non-linear, the problem is classified as a non-linear programming problem.

• (C) Linear constraints: These are the restrictions or limitations on the decision variables, expressed as a system of linear equations or linear inequalities. For example, constraints on resources, production capacity, or demand. These are essential for defining the feasible region of an LPP.

• (D) Maximizing or minimizing the objective function: The core purpose of a linear programming problem is to find the values of the decision variables that either maximize (e.g., profit) or minimize (e.g., cost) the linear objective function, subject to the given constraints. This optimization goal is a defining component.

Therefore, the components of a linear programming problem are decision variables, linear constraints, and the goal of maximizing or minimizing the (linear) objective function.

(A) Both A and R are true and R is the correct explanation of A.

Explanation:

Analysis of Assertion (A):

Assertion (A): "If the feasible region of an LPP is unbounded, the minimum value of the objective function may or may not exist."

This statement is true.

Consider two examples with an unbounded feasible region (e.g., $x \ge 0, y \ge 0$):

1. Objective function $Z = x + y$. The minimum value is $0$ at $(0,0)$. Here, the minimum exists.

2. Objective function $Z = -x - y$. As $x$ and $y$ can increase indefinitely in the feasible region, $Z$ can become arbitrarily small (more negative). Thus, there is no minimum value. Here, the minimum does not exist.

Since the minimum can exist in some cases and not in others for an unbounded feasible region, Assertion (A) is true.

Analysis of Reason (R):

Reason (R): "If the half-plane corresponding to $ax+by < Z_{\text{min}}$ has no point in common with the feasible region, then the minimum value exists at a corner point."

This statement is true. Here, $ax+by$ represents the objective function $Z$, and $Z_{\text{min}}$ should be interpreted as the smallest value of the objective function found among the corner points of the unbounded feasible region. Let this value be $m$. The standard theorem states: If the feasible region for an LPP is unbounded, let $m$ be the minimum value of $Z$ at the corner points. Then, $m$ is the minimum value of $Z$ if the open half-plane defined by $ax+by < m$ has no point in common with the feasible region. Otherwise, $Z$ has no minimum value.

Reason (R) accurately reflects the condition under which the minimum value (found at a corner point) is indeed the global minimum for an unbounded feasible region.

Relationship between A and R:

Assertion (A) states that the minimum value "may or may not exist". Reason (R) provides the specific condition that determines whether the minimum value exists or not.

If the condition in (R) is met (i.e., the open half-plane $ax+by < m$ has no common points with the feasible region), then the minimum value exists (and is $m$, occurring at a corner point).

If the condition in (R) is not met (i.e., the open half-plane $ax+by < m$ *does* have common points with the feasible region), then the minimum value does not exist.

Therefore, Reason (R) explains the criteria that lead to the two possibilities ("may exist" or "may not exist") mentioned in Assertion (A). It provides the decision rule for determining which case applies.

Thus, both A and R are true, and R is the correct explanation of A.

The correct matches are:

(i) Objective function $\longleftrightarrow$ (d) Linear function to be optimized

(ii) Feasible region $\longleftrightarrow$ (c) Region satisfying all constraints

(iii) Corner point $\longleftrightarrow$ (a) Intersection of boundary lines of constraints

(iv) Optimal solution $\longleftrightarrow$ (b) Value of objective function at a corner point (More accurately, the point(s) where the objective function reaches its maximum or minimum value. Option (b) describes the *value* at such a point if it's optimal, but the optimal solution refers to the point itself or the set of points.)

Explanation:

• (i) Objective function: This is the linear mathematical expression that represents the quantity to be maximized (e.g., profit) or minimized (e.g., cost) in a linear programming problem. So, it matches with (d) Linear function to be optimized.

• (ii) Feasible region: This is the set of all points (combinations of decision variables) that satisfy all the constraints of the LPP, including non-negativity constraints. So, it matches with (c) Region satisfying all constraints.

• (iii) Corner point: These are the vertices of the feasible region. They are formed by the intersection of the boundary lines of the constraints. The optimal solution, if it exists, is found at one or more of these corner points. So, it matches with (a) Intersection of boundary lines of constraints.

• (iv) Optimal solution: This refers to the specific combination of decision variables within the feasible region that yields the maximum or minimum value of the objective function. While the optimal value is found by evaluating the objective function at corner points, the "optimal solution" itself can be the set of coordinates $(x,y)$ that achieves this value. Option (b) "Value of objective function at a corner point" is somewhat ambiguous. If an optimal solution exists, it occurs at a corner point, and the value of the objective function there is the optimal value. However, not every value of the objective function at *any* corner point is the optimal solution. The optimal solution is the *specific* corner point(s) (or points on an edge if multiple optima exist) where the objective function is maximized or minimized. Given the choices, (b) is the closest, implying the value achieved at the optimal corner point(s). More precisely, an optimal solution is a point in the feasible region where the objective function attains its optimal value (maximum or minimum). If such a point is a corner point, then its value is the optimal value.

Revisiting (iv) and (b): The "optimal solution" usually refers to the point (or set of points) $(x_1, x_2, ..., x_n)$ that gives the optimal value, and the "optimal value" is the value $Z$ achieved at that point. However, in multiple choice questions, sometimes the terms are used a bit loosely. If we consider "optimal solution" to mean the best outcome, (b) is the most fitting description among the given options, representing the result achieved.

(D) D (This answer is conditional on assumptions about the unspecified graph, as explained below.)

Explanation:

The fundamental theorem of linear programming states that if an optimal solution (maximum or minimum) for a linear programming problem exists, it must occur at one of the corner points (vertices) of the feasible region.

The objective function to be maximized is $Z = 5x + 4y$.

To determine at which point (A, B, C, or D) the maximum value of $Z$ occurs, we would typically follow these steps:

1. Identify the coordinates $(x, y)$ for each corner point A, B, C, and D from the provided graph.

2. Substitute the coordinates of each corner point into the objective function $Z = 5x + 4y$.

3. Compare the calculated values of $Z$. The point that yields the highest value of $Z$ is where the maximum occurs.

Important Note: The image (graph) of the feasible region is not provided in the prompt (the `src` attribute for the `img` tag is empty). Without the visual representation of the feasible region and the specific coordinates of the points A, B, C, and D, it is impossible to definitively determine which point yields the maximum value for $Z$.

However, we can discuss the general principle for maximization. Since the coefficients of $x$ (which is 5) and $y$ (which is 4) in the objective function $Z = 5x + 4y$ are positive, the value of $Z$ increases as $x$ increases or as $y$ increases. Therefore, the maximum value of $Z$ is "likely" to occur at a corner point that is "furthest out" from the origin in the direction that increases both $x$ and $y$ values, weighted by their coefficients. This is the vertex that the line $5x + 4y = k$ (where $k$ is a constant) last touches as $k$ is increased and the line is moved parallel to itself across the feasible region.

If we are forced to choose an option without the graph, we might make an assumption based on common labeling conventions in textbook problems. Sometimes, points are labeled sequentially, and the "last" label (like D) might be intended to represent an "extreme" point of the feasible region where the maximum could occur. Assuming that point D represents the vertex that is furthest in the direction of increasing $Z$, then D would be the point where the maximum is likely to occur.

For example, if the points were A=(0,0), B=(10,0), C=(5,8), and D=(8,10):

• $Z_A = 5(0) + 4(0) = 0$

• $Z_B = 5(10) + 4(0) = 50$

• $Z_C = 5(5) + 4(8) = 25 + 32 = 57$

• $Z_D = 5(8) + 4(10) = 40 + 40 = 80$

In this hypothetical scenario, point D would yield the maximum value.

Conclusion: A definitive answer requires the graph. Assuming D represents the relevant "extreme" corner point based on a hypothetical common convention, it is chosen. In a real scenario, one must use the actual coordinates from the graph.

(C) Edge of the feasible region

Explanation:

In a linear programming problem (LPP), if an optimal solution exists, it will occur at one or more corner points (vertices) of the feasible region.

If a linear programming problem has multiple optimal solutions, it means that the objective function attains its optimal (maximum or minimum) value at more than one point. This occurs when the slope of the objective function line is the same as the slope of one of the boundary lines (edges) of the feasible region that defines the optimal solutions.

In such a scenario:

1. At least two adjacent corner points of the feasible region will be optimal solutions (i.e., they will yield the same optimal value for the objective function).

2. Every point on the line segment (which is an edge of the feasible region) connecting these two optimal corner points will also be an optimal solution. This results in infinitely many optimal solutions along that edge.

Let's consider the options:

• (A) Interior of the feasible region: An optimal solution generally does not occur in the interior. If an interior point were optimal, it would usually imply the objective function is constant over a region, which means the boundary points (edges/vertices) of that region are also optimal.

• (B) Vertices of the feasible region: While it is true that if multiple optimal solutions exist, at least two vertices will be optimal, this option doesn't fully describe the set of all multiple optimal solutions. The solutions also include all points on the edge connecting these vertices.

• (C) Edge of the feasible region: This is the correct answer. When multiple optimal solutions exist, they lie along an entire edge (or potentially a face in higher dimensions) of the feasible region. This edge connects two (or more, if the edge itself is made up of collinear vertices that are all optimal) optimal vertices.

• (D) Outside the feasible region: Points outside the feasible region do not satisfy all constraints and therefore cannot be solutions, let alone optimal solutions.

Therefore, if an LPP has multiple optimal solutions, they occur on an edge of the feasible region (which includes the two optimal corner points that form that edge).

(B) Two

Explanation:

The graphical method for solving Linear Programming Problems (LPPs) involves plotting the constraints as lines on a graph to identify the feasible region and then determining the corner points of this region. The objective function is then evaluated at these corner points to find the optimal solution.

• One decision variable: If there is only one decision variable (e.g., $x$), the constraints would be of the form like $ax \le b$ or $ax \ge b$. The feasible region would be a segment on a number line. While this can be visualized, it's a very simple case and often doesn't require the full "graphical method" as typically understood for LPPs. For instance, to maximize $Z=cx$ subject to $x \le k_1$ and $x \ge k_2$, the solution is straightforward.

• Two decision variables: If there are two decision variables (e.g., $x$ and $y$), the constraints are linear inequalities that can be plotted as lines in a 2-dimensional Cartesian plane. The feasible region is a polygon, and its corner points can be easily identified. This is the primary scenario where the graphical method is taught and is most practically applicable due to ease of visualization and manual computation.

• Three decision variables: If there are three decision variables (e.g., $x, y, z$), the constraints would represent planes in a 3-dimensional space. The feasible region would be a polyhedron. While it is theoretically possible to visualize and solve graphically, it becomes significantly more complex and difficult to draw accurately and identify all corner points by hand. It is generally not considered "suitable" for manual graphical solution.

• More than two decision variables (specifically more than three): If there are more than three decision variables, it is impossible to visualize the feasible region in standard Euclidean space (e.g., 4-dimensional space or higher). Therefore, the graphical method is not applicable.

The term "suitable" implies practicality and ease of use. The graphical method is most clearly and effectively applied when there are two decision variables, as it allows for a clear visual representation on a 2D plane.

(D) $(0, 0)$

Given:

The constraints for the Linear Programming Problem (LPP) are:

To Find:

Which of the given points $(0, 6)$, $(6, 0)$, $(2, 2)$, $(0, 0)$ is NOT a corner point of the feasible region.

Solution:

To find the corner points of the feasible region, we first convert the inequalities into equations to represent the boundary lines:

$L_1: x + 2y = 6$

$L_2: 2x + y = 6$

$L_3: x = 0$ (y-axis)

$L_4: y = 0$ (x-axis)

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

The inequalities $x+2y \ge 6$ and $2x+y \ge 6$ mean that the feasible region lies on or above these lines.

Let's find the intersection points of these lines:

1. Intersection of $L_1: x + 2y = 6$ and $L_2: 2x + y = 6$:

From $L_1$, $x = 6 - 2y$. Substitute this into $L_2$:

$2(6 - 2y) + y = 6$

$12 - 4y + y = 6$

$12 - 3y = 6$

$3y = 12 - 6$

$3y = 6 \Rightarrow y = 2$

Substitute $y=2$ back into $x = 6 - 2y$:

$x = 6 - 2(2) = 6 - 4 = 2$

So, the intersection point is $(2, 2)$. Let's check if it satisfies all constraints:

• $2 + 2(2) = 6 \ge 6$ (True)

• $2(2) + 2 = 6 \ge 6$ (True)

• $2 \ge 0$ (True)

• $2 \ge 0$ (True)

Thus, $(2, 2)$ is a corner point. (This matches option (C))

2. Intersection of boundary lines with the axes:

Consider the y-axis ($x=0$, $L_3$):

For $L_1: 0 + 2y = 6 \Rightarrow y = 3$. Point is $(0,3)$.

For $L_2: 2(0) + y = 6 \Rightarrow y = 6$. Point is $(0,6)$.

The feasible region requires $x+2y \ge 6$ (so for $x=0$, $2y \ge 6 \Rightarrow y \ge 3$) AND $2x+y \ge 6$ (so for $x=0$, $y \ge 6$). Both conditions must be met, so $y \ge 6$. Thus, the corner point on the y-axis is $(0, 6)$.

Let's check if $(0,6)$ satisfies all constraints:

• $0 + 2(6) = 12 \ge 6$ (True)

• $2(0) + 6 = 6 \ge 6$ (True)

• $0 \ge 0$ (True)

• $6 \ge 0$ (True)

Thus, $(0, 6)$ is a corner point. (This matches option (A))

Consider the x-axis ($y=0$, $L_4$):

For $L_1: x + 2(0) = 6 \Rightarrow x = 6$. Point is $(6,0)$.

For $L_2: 2x + 0 = 6 \Rightarrow x = 3$. Point is $(3,0)$.

The feasible region requires $x+2y \ge 6$ (so for $y=0$, $x \ge 6$) AND $2x+y \ge 6$ (so for $y=0$, $2x \ge 6 \Rightarrow x \ge 3$). Both conditions must be met, so $x \ge 6$. Thus, the corner point on the x-axis is $(6, 0)$.

Let's check if $(6,0)$ satisfies all constraints:

• $6 + 2(0) = 6 \ge 6$ (True)

• $2(6) + 0 = 12 \ge 6$ (True)

• $6 \ge 0$ (True)

• $0 \ge 0$ (True)

Thus, $(6, 0)$ is a corner point. (This matches option (B))

The corner points of the feasible region are $(0, 6)$, $(2, 2)$, and $(6, 0)$. The feasible region is unbounded.

Now let's check the point $(0, 0)$ (Option (D)):

Substitute $x=0, y=0$ into the constraints:

• Constraint (i): $0 + 2(0) = 0$. Is $0 \ge 6$? False.

• Constraint (ii): $2(0) + 0 = 0$. Is $0 \ge 6$? False.

• Constraint (iii): $0 \ge 0$ (True)

• Constraint (iv): $0 \ge 0$ (True)

Since $(0, 0)$ does not satisfy constraints (i) and (ii), it is not in the feasible region. Therefore, $(0, 0)$ cannot be a corner point of the feasible region.

Comparing our findings with the options:

• (A) $(0, 6)$ is a corner point.

• (B) $(6, 0)$ is a corner point.

• (C) $(2, 2)$ is a corner point.

• (D) $(0, 0)$ is NOT a corner point.

Therefore, the point that is NOT a corner point of the feasible region is $(0, 0)$.

(B) $2x + y \le 100$

Explanation:

Let $x$ be the number of Pen A produced.

Let $y$ be the number of Pen B produced.

Labour requirement for Pen A:

Each Pen A requires 2 hours of labour.

So, $x$ units of Pen A will require $2 \times x = 2x$ hours of labour.

Labour requirement for Pen B:

Each Pen B requires 1 hour of labour.

So, $y$ units of Pen B will require $1 \times y = y$ hours of labour.

Total labour required:

The total labour required to produce $x$ units of Pen A and $y$ units of Pen B is the sum of the labour for Pen A and the labour for Pen B.

Total labour = $2x + y$ hours.

Available labour constraint:

The company has a total of 100 hours of labour available per day.

This means that the total labour used cannot exceed the total labour available.

Therefore, the total labour required must be less than or equal to 100 hours.

This can be written as the inequality:

$2x + y \le 100$

Let's examine the options:

• (A) $2x + y \ge 100$: This would mean the labour used must be at least 100 hours, which contradicts the idea of an availability limit.

• (B) $2x + y \le 100$: This correctly states that the total labour used is at most 100 hours.

• (C) $x + 2y \le 100$: This incorrectly assigns 1 hour to Pen A and 2 hours to Pen B.

• (D) $x + 2y \ge 100$: This incorrectly assigns labour hours and uses the wrong inequality sign for an availability constraint.

Thus, the inequality representing the labour constraint is $2x + y \le 100$.

(B) All points on the line segment joining these two corner points.

Explanation:

This question relates to the concept of multiple optimal solutions in linear programming.

If the objective function $Z = ax + by$ attains its maximum (or minimum) value at two distinct corner points of the feasible region (which is a convex polygon), say $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$, it means that the line representing the objective function ($ax + by = Z_{\text{max}}$) is parallel to the edge of the feasible region connecting $P_1$ and $P_2$.

Let the maximum value be $Z_{\text{max}}$. So, $ax_1 + by_1 = Z_{\text{max}}$ and $ax_2 + by_2 = Z_{\text{max}}$.

Consider any point $P = (x, y)$ on the line segment joining $P_1$ and $P_2$. Such a point can be expressed as a convex combination of $P_1$ and $P_2$:

$P = (1-\lambda)P_1 + \lambda P_2$ for $0 \le \lambda \le 1$.

So, $x = (1-\lambda)x_1 + \lambda x_2$ and $y = (1-\lambda)y_1 + \lambda y_2$.

Now, let's evaluate the objective function $Z$ at point $P$:

$Z(P) = a((1-\lambda)x_1 + \lambda x_2) + b((1-\lambda)y_1 + \lambda y_2)$

$Z(P) = a(1-\lambda)x_1 + a\lambda x_2 + b(1-\lambda)y_1 + b\lambda y_2$

$Z(P) = (1-\lambda)(ax_1 + by_1) + \lambda(ax_2 + by_2)$

Since $ax_1 + by_1 = Z_{\text{max}}$ and $ax_2 + by_2 = Z_{\text{max}}$, we substitute these values:

$Z(P) = (1-\lambda)Z_{\text{max}} + \lambda Z_{\text{max}}$

$Z(P) = Z_{\text{max}} - \lambda Z_{\text{max}} + \lambda Z_{\text{max}}$

$Z(P) = Z_{\text{max}}$

This shows that every point on the line segment joining the two corner points $P_1$ and $P_2$ also yields the same maximum value $Z_{\text{max}}$. Therefore, all points on this line segment are also optimal solutions.

Let's consider the options:

• (A) All points on the boundary of the feasible region: This is generally false. The objective function will only be maximized along the specific edge connecting the two optimal corner points, not necessarily the entire boundary.

• (B) All points on the line segment joining these two corner points: This is correct, as shown by the derivation above. This line segment is an edge of the feasible region.

• (C) The midpoint of the line segment joining these two corner points only: While the midpoint is indeed an optimal solution (it's a point on the line segment), it's not the *only* other optimal solution. All points on the segment are optimal.

• (D) Only at these two corner points: This is false because if it's maximized at two corner points, it's also maximized at every point on the line segment connecting them.

Thus, if the objective function is maximized at two corner points, it is also maximized at all points on the line segment joining these two corner points.

(B) $1$

Given:

Objective function: Maximize $Z = x+y$

Subject to constraints:

To Find:

The maximum value of $Z$.

Solution:

First, we identify the corner points of the feasible region defined by the constraints.

The boundary lines are:

$L_1: x + y = 1$

$L_2: x = 0$ (y-axis)

$L_3: y = 0$ (x-axis)

Corner Points:

1. Intersection of $x=0$ and $y=0$: This gives the point $(0, 0)$.

   Check with constraint (i): $0 + 0 = 0 \le 1$. (Satisfied)

2. Intersection of $x=0$ and $x+y=1$: Substitute $x=0$ into $x+y=1 \Rightarrow 0+y=1 \Rightarrow y=1$. This gives the point $(0, 1)$.

   Check with other constraints: $0 \ge 0$, $1 \ge 0$. (Satisfied)

3. Intersection of $y=0$ and $x+y=1$: Substitute $y=0$ into $x+y=1 \Rightarrow x+0=1 \Rightarrow x=1$. This gives the point $(1, 0)$.

   Check with other constraints: $1 \ge 0$, $0 \ge 0$. (Satisfied)

The corner points of the feasible region are $(0, 0)$, $(0, 1)$, and $(1, 0)$. The feasible region is a triangle bounded by these points.

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

Comparing the values of $Z$:

• At $(0, 0)$, $Z = 0$.

• At $(0, 1)$, $Z = 1$.

• At $(1, 0)$, $Z = 1$.

The maximum value of $Z$ is $1$. This maximum value occurs at two corner points, $(0, 1)$ and $(1, 0)$, and therefore at all points on the line segment connecting $(0, 1)$ and $(1, 0)$ (which is the line $x+y=1$ for $0 \le x \le 1$).

Alternatively, observe that the objective function is $Z = x+y$. The constraint is $x+y \le 1$. Since we want to maximize $Z$, and $Z$ is identical to the left side of the constraint inequality, the maximum value $Z$ can take is limited by the right side of the constraint, which is $1$. This maximum is achieved when $x+y=1$, which is the boundary of the feasible region.

Therefore, the maximum value of $Z$ is $1$.

(B) A solution that satisfies all the constraints, including non-negativity constraints.

Explanation:

In a Linear Programming Problem (LPP), a feasible solution is a set of values for the decision variables that satisfies all the given constraints of the problem simultaneously. These constraints include both the structural constraints (e.g., resource limitations, production requirements) and the non-negativity constraints (e.g., $x \ge 0$, $y \ge 0$).

The set of all feasible solutions forms the feasible region.

Let's analyze the given options:

• (A) A solution that satisfies the objective function: The objective function is an expression to be optimized (maximized or minimized). A solution doesn't "satisfy" the objective function in the same way it satisfies a constraint. Any set of values for the decision variables will yield *some* value for the objective function. This doesn't define feasibility.

• (B) A solution that satisfies all the constraints, including non-negativity constraints: This is the precise definition of a feasible solution. Any point within or on the boundary of the feasible region is a feasible solution.

• (C) A solution that maximizes or minimizes the objective function: This describes an optimal solution, not just any feasible solution. An optimal solution is a specific type of feasible solution that gives the best possible value for the objective function.

• (D) A solution at a corner point: A solution at a corner point is a feasible solution, and if an optimal solution exists, it will be found at one or more corner points. However, not all feasible solutions are at corner points (e.g., interior points of the feasible region or points on an edge that are not vertices are also feasible solutions). This option describes a subset of feasible solutions which are candidates for optimality.

Therefore, the most accurate description of a feasible solution is one that satisfies all the constraints, including non-negativity constraints.

(A) Both A and R are true and R is the correct explanation of A.

Explanation:

Analysis of Assertion (A):

Assertion (A): "The feasible region of an LPP is always a convex set."

A set is convex if for any two points within the set, the line segment connecting these two points is entirely contained within the set.

The feasible region of a Linear Programming Problem (LPP) is defined by a system of linear inequalities (constraints). Each linear inequality $ax + by \le c$ or $ax + by \ge c$ defines a closed half-plane. If the constraint is a linear equation $ax + by = c$, it defines a line (which is also a convex set).

The feasible region is the intersection of these half-planes (and lines, if any). It is a well-known property in mathematics that the intersection of any number of convex sets is also a convex set. Since half-planes are convex sets, their intersection (the feasible region) must also be a convex set. This set can be a convex polygon, an unbounded convex region, a line segment, a single point, or an empty set (which is trivially convex).

So, Assertion (A) is true.

Analysis of Reason (R):

Reason (R): "The intersection of half-planes (which are convex sets) is always a convex set."

First, a half-plane defined by a linear inequality (e.g., $ax + by \le c$) is indeed a convex set. If you take any two points $P_1$ and $P_2$ in the half-plane, the line segment connecting $P_1$ and $P_2$ will also lie entirely within that half-plane.

Second, it is a fundamental property of convex sets that the intersection of any collection of convex sets is also a convex set. Since each constraint in an LPP defines a convex set (a half-plane or a line), and the feasible region is the intersection of all these sets, the feasible region itself must be convex.

So, Reason (R) is true.

Relationship between A and R:

Assertion (A) states that the feasible region of an LPP is always a convex set. Reason (R) explains why this is the case: the feasible region is formed by the intersection of half-planes (derived from the linear constraints), and half-planes are convex sets, and the intersection of convex sets is always convex.

Therefore, Reason (R) provides the correct mathematical justification for Assertion (A).

Thus, both A and R are true, and R is the correct explanation of A.

(D) $(3, 3)$

Given:

The constraints are:

To Find:

Which of the given points is in the feasible region.

A point is in the feasible region if it satisfies all the constraints simultaneously.

Solution:

We will check each point against all constraints.

(A) Point $(1, 1)$:

• Constraint (i): $x+y \ge 5 \Rightarrow 1+1 = 2$. Is $2 \ge 5$? False.

Since constraint (i) is not satisfied, $(1, 1)$ is not in the feasible region.

(B) Point $(4, 2)$:

• Constraint (i): $x+y \ge 5 \Rightarrow 4+2 = 6$. Is $6 \ge 5$? True.

• Constraint (ii): $x \le 3 \Rightarrow 4 \le 3$. False.

Since constraint (ii) is not satisfied, $(4, 2)$ is not in the feasible region.

(C) Point $(2, 4)$:

• Constraint (i): $x+y \ge 5 \Rightarrow 2+4 = 6$. Is $6 \ge 5$? True.

• Constraint (ii): $x \le 3 \Rightarrow 2 \le 3$. True.

• Constraint (iii): $y \le 3 \Rightarrow 4 \le 3$. False.

Since constraint (iii) is not satisfied, $(2, 4)$ is not in the feasible region.

(D) Point $(3, 3)$:

• Constraint (i): $x+y \ge 5 \Rightarrow 3+3 = 6$. Is $6 \ge 5$? True.

• Constraint (ii): $x \le 3 \Rightarrow 3 \le 3$. True.

• Constraint (iii): $y \le 3 \Rightarrow 3 \le 3$. True.

• Constraint (iv): $x \ge 0 \Rightarrow 3 \ge 0$. True.

• Constraint (v): $y \ge 0 \Rightarrow 3 \ge 0$. True.

Since the point $(3, 3)$ satisfies all the given constraints, it is in the feasible region.

Therefore, the point $(3, 3)$ is in the feasible region.

(B) Feasible region

Explanation:

The question describes the core process of the graphical method in linear programming.

1. Plotting constraints as linear equations: Each linear inequality constraint (e.g., $ax+by \le c$ or $ax+by \ge c$) has an associated linear equation ($ax+by=c$) which represents the boundary line of the half-plane defined by the inequality.

2. Identifying the region that satisfies all inequalities: For each inequality, we determine which side of its boundary line (or the line itself, if it's an equality constraint) contains the points satisfying that inequality. The region that simultaneously satisfies *all* these inequalities (including non-negativity constraints like $x \ge 0, y \ge 0$) is the set of all possible solutions to the LPP.

This region is specifically known as the feasible region.

Let's look at the options:

• (A) Objective function region: The objective function is a linear function to be optimized, not a region itself. Lines representing constant values of the objective function (iso-profit or iso-cost lines) are used in conjunction with the feasible region to find the optimal solution, but the region defined by constraints is not called the "objective function region."

• (B) Feasible region: This is the correct term. It is the set of all points that satisfy all the constraints of the LPP.

• (C) Infeasible region: This refers to the area *outside* the feasible region, where at least one constraint is violated. If there is no region that satisfies all constraints simultaneously, the problem itself is said to be infeasible, and the feasible region is empty.

• (D) Optimal solution: The optimal solution is a specific point (or set of points) *within* the feasible region (typically a corner point or an edge) where the objective function reaches its maximum or minimum value. It is not the entire region itself.

Therefore, the region that satisfies all inequalities in the graphical method of solving an LPP is called the feasible region.

(A) $0$ (at $(0,0)$)

Explanation:

To find the minimum value of the objective function $Z = 10x + 7y$, we need to evaluate $Z$ at each of the given corner points of the feasible region.

The given corner points are $(0, 0)$, $(5, 0)$, $(3, 2)$, and $(0, 4)$.

The objective function is $Z = 10x + 7y$.

We can create a table to evaluate $Z$ at each corner point:

Comparing the values of $Z$ at the corner points:

• At $(0, 0)$, $Z = 0$.

• At $(5, 0)$, $Z = 50$.

• At $(3, 2)$, $Z = 44$.

• At $(0, 4)$, $Z = 28$.

The minimum value among these is $0$, which occurs at the corner point $(0, 0)$.

Therefore, the minimum value of $Z$ is $0$ at $(0,0)$.

(A) $Z = 5000x + 4000y$

Explanation:

The objective function in a linear programming problem represents the quantity that we want to maximize or minimize.

In this problem, the farmer wants to maximize profit.

Let $x$ be the number of acres of wheat grown.

Let $y$ be the number of acres of barley grown.

Profit from wheat:

The profit from growing wheat is $\textsf{₹} 5000$ per acre.

So, the total profit from $x$ acres of wheat is $5000 \times x = 5000x$.

Profit from barley:

The profit from growing barley is $\textsf{₹} 4000$ per acre.

So, the total profit from $y$ acres of barley is $4000 \times y = 4000y$.

Total profit:

The total profit, $Z$, is the sum of the profit from wheat and the profit from barley.

$Z = \text{Profit from wheat} + \text{Profit from barley}$

$Z = 5000x + 4000y$

This is the function that the farmer wants to maximize.

Let's examine the options:

• (A) $Z = 5000x + 4000y$: This correctly represents the total profit.

• (B) $Z = 4000x + 5000y$: This incorrectly assigns the profit of $\textsf{₹} 4000$ to wheat ($x$) and $\textsf{₹} 5000$ to barley ($y$).

• (C) $x + y \le 100$: This is a constraint representing the total land availability (100 acres). It's not the objective function.

• (D) $x+y$: This represents the total number of acres used, not the total profit.

Therefore, the objective function to maximize profit is $Z = 5000x + 4000y$.

(B) Does not intersect the feasible region.

Explanation:

This question pertains to the condition for the existence of a minimum value of an objective function when the feasible region is unbounded.

Let $Z = ax+by$ be the objective function.

If the feasible region is unbounded, we first find the minimum value of $Z$ at the corner points of this unbounded region. Let this value be $m$. So, $Z_{\text{min}}$ in the question refers to this value $m$.

The theorem for an unbounded feasible region states:

• If the open half-plane defined by $ax+by < m$ (for minimization) or $ax+by > M$ (for maximization, where $M$ is the maximum value at corner points) has no point in common with the feasible region, then $m$ (or $M$) is the minimum (or maximum) value of the objective function.

• If this open half-plane has points in common with the feasible region, then the objective function does not have a finite minimum (or maximum) value; it is unbounded in that direction.

The question states that the objective function $Z = ax+by$ has a minimum value ($Z_{\text{min}}$). According to the theorem, for this to be true in an unbounded feasible region, the condition must be that the open half-plane $ax+by < Z_{\text{min}}$ (where $Z_{\text{min}}$ is the smallest value found at the corner points) must have no points in common with the feasible region.

Let's analyze the options:

• (A) Intersects the feasible region: If the half-plane $ax+by < Z_{\text{min}}$ intersects the feasible region, it means there are points $(x,y)$ in the feasible region for which $ax+by$ is even smaller than $Z_{\text{min}}$. This would imply that $Z_{\text{min}}$ (the value found at corner points) is not actually the true minimum, and the objective function can decrease further, potentially indefinitely. This contradicts the premise that $Z_{\text{min}}$ is the minimum value.

• (B) Does not intersect the feasible region: This is the correct condition. If there are no points in the feasible region where $ax+by$ is less than $Z_{\text{min}}$, then $Z_{\text{min}}$ (obtained at one or more corner points) is indeed the overall minimum value of the objective function.

• (C) Contains the origin: Whether the half-plane $ax+by < Z_{\text{min}}$ contains the origin depends on the specific values of $a, b,$ and $Z_{\text{min}}$. For example, if $Z_{\text{min}} = 5$ and the objective function is $x+y$, the half-plane is $x+y < 5$. The origin $(0,0)$ satisfies $0+0 < 5$. However, if $Z_{\text{min}} = -5$, the half-plane $x+y < -5$ does not contain the origin. This option is not universally true.

• (D) Is always above the line $ax+by = Z_{\text{min}}$: The half-plane $ax+by < Z_{\text{min}}$ represents all points "below" or "to one side" of the line $ax+by = Z_{\text{min}}$, where the value of the function $ax+by$ is strictly less than $Z_{\text{min}}$. The term "above" is ambiguous without knowing the orientation determined by $a$ and $b$. The critical aspect is that it represents values *less than* $Z_{\text{min}}$.

Therefore, if the feasible region is unbounded and the objective function $Z = ax+by$ has a minimum value $Z_{\text{min}}$, then the half-plane $ax+by < Z_{\text{min}}$ does not intersect the feasible region.

(A) $\textsf{₹} 1000$ (at $(12, 12)$) (with clarification below regarding the profit value)

Given:

Let $x$ be the number of Toy A produced.

Let $y$ be the number of Toy B produced.

Constraints:

Objective function to maximize profit: $Z = 50x + 40y$.

To Find:

The maximum profit the factory can earn.

Solution:

First, we find the corner points of the feasible region defined by the constraints.

1. Origin: $(0,0)$ is a corner point.

2. Intersection with x-axis ($y=0$):

From (i): $2x + 3(0) = 60 \Rightarrow 2x = 60 \Rightarrow x = 30$. Point $(30,0)$.

From (ii): $3x + 2(0) = 60 \Rightarrow 3x = 60 \Rightarrow x = 20$. Point $(20,0)$.

The feasible x-intercept is $(20,0)$ because $2(20)+3(0)=40 \le 60$ and $3(20)+2(0)=60 \le 60$. For $(30,0)$, constraint (ii) $3(30)+0 = 90 \not\le 60$ is violated.

So, $(20,0)$ is a corner point.

3. Intersection with y-axis ($x=0$):

From (i): $2(0) + 3y = 60 \Rightarrow 3y = 60 \Rightarrow y = 20$. Point $(0,20)$.

From (ii): $3(0) + 2y = 60 \Rightarrow 2y = 60 \Rightarrow y = 30$. Point $(0,30)$.

The feasible y-intercept is $(0,20)$ because $2(0)+3(20)=60 \le 60$ and $3(0)+2(20)=40 \le 60$. For $(0,30)$, constraint (i) $2(0)+3(30) = 90 \not\le 60$ is violated.

So, $(0,20)$ is a corner point.

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

Multiply the first equation by 2: $4x + 6y = 120$.

Multiply the second equation by 3: $9x + 6y = 180$.

Subtract the new first from the new second: $(9x-4x) + (6y-6y) = 180-120 \Rightarrow 5x = 60 \Rightarrow x=12$.

Substitute $x=12$ into $2x + 3y = 60$: $2(12) + 3y = 60 \Rightarrow 24 + 3y = 60 \Rightarrow 3y = 36 \Rightarrow y=12$.

So, $(12,12)$ is a corner point. This point satisfies both original inequalities: $2(12)+3(12)=60 \le 60$ and $3(12)+2(12)=60 \le 60$.

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

Now, evaluate the objective function $Z = 50x + 40y$ at these corner points:

The maximum value of $Z$ from the corner points is $\textsf{₹} 1080$, which occurs at $(x,y) = (12,12)$.

Analyzing the options:

• (A) $\textsf{₹} 1000$ (at $(12, 12)$): The point $(12,12)$ is where the actual maximum profit occurs. However, the profit at $(12,12)$ is $\textsf{₹} 1080$, not $\textsf{₹} 1000$. The value $\textsf{₹} 1000$ is achieved at $(20,0)$.

• (B) $\textsf{₹} 1200$ (at $(20, 0)$): The profit at $(20,0)$ is $\textsf{₹} 1000$. The stated profit $\textsf{₹} 1200$ is incorrect for this point and is higher than the actual maximum achievable profit of $\textsf{₹} 1080$.

• (C) $\textsf{₹} 1040$ (at $(0, 15)$): The point $(0,15)$ is feasible ($2(0)+3(15)=45 \le 60$; $3(0)+2(15)=30 \le 60$). The profit at $(0,15)$ is $Z = 50(0) + 40(15) = 600$. The stated profit $\textsf{₹} 1040$ is incorrect for this point. (Profit $\textsf{₹} 1040$ is achievable, e.g., at $(12,11)$, but it's not the maximum).

• (D) $\textsf{₹} 0$ (at $(0, 0)$): The profit at $(0,0)$ is indeed $\textsf{₹} 0$. This is a correct statement but $0$ is not the maximum profit.

The question asks for "What is the maximum profit the factory can earn?". The calculated maximum profit is $\textsf{₹} 1080$. This value is not explicitly listed as the profit in any option. However, option (A) refers to the point $(12,12)$ where this true maximum profit occurs. Interpreting the question to select the option associated with the highest achievable profit by evaluating $Z$ at the points specified in each option:

• Point from (A) $(12,12) \Rightarrow Z = 1080$
• Point from (B) $(20,0) \Rightarrow Z = 1000$
• Point from (C) $(0,15) \Rightarrow Z = 600$
• Point from (D) $(0,0) \Rightarrow Z = 0$

The highest profit obtained from evaluating the points in the options is $\textsf{₹}1080$, associated with point $(12,12)$ from option (A). Therefore, despite the discrepancy in the profit value stated in option (A)'s text, it is the most appropriate choice as it refers to the optimal point.

The maximum profit the factory can earn is $\textsf{₹} 1080$.

(B) Above the line $2x+3y=6$ (including the line)

Explanation:

The inequality is $2x + 3y \ge 6$.

The boundary line for this inequality is $2x + 3y = 6$.

To determine which half-plane represents the inequality, we can use a test point that is not on the line. A common test point is the origin $(0, 0)$.

Substitute $(0, 0)$ into the inequality $2x + 3y \ge 6$:

$2(0) + 3(0) \ge 6$

$0 + 0 \ge 6$

$0 \ge 6$

This statement is false. This means that the half-plane containing the origin $(0, 0)$ does not satisfy the inequality.

Therefore, the region defined by $2x + 3y \ge 6$ is the half-plane on the opposite side of the line $2x + 3y = 6$ from the origin.

To understand "above" or "below", let's consider the y-intercept of the line. If $x=0$, $3y=6 \Rightarrow y=2$. So the line passes through $(0,2)$. If $y=0$, $2x=6 \Rightarrow x=3$. So the line passes through $(3,0)$.

Consider the y-axis (where $x=0$). The inequality becomes $3y \ge 6$, or $y \ge 2$. This means for points on the y-axis, the y-values must be 2 or greater. This corresponds to the region above (and including) the point $(0,2)$ on the y-axis.

Generally, for an inequality of the form $Ax + By \ge C$ (where $B > 0$), the region is "above" the line $Ax+By=C$. If $B < 0$, it would be "below". If $B=0$, it's a vertical line, and the region is to the right or left.

In our case, $2x + 3y \ge 6$, the coefficient of $y$ (which is $3$) is positive. Since the origin (which is "below" the line in a general sense if we consider the y-intercept) did not satisfy the inequality, the region satisfying the inequality must be "above" the line.

The "$\ge$" sign means that the line $2x + 3y = 6$ itself is included in the region.

Therefore, the region defined by the inequality $2x + 3y \ge 6$ is the half-plane above the line $2x+3y=6$ (including the line).

(A) Always exist and occur at corner points.

Explanation:

This question refers to a fundamental theorem in linear programming regarding bounded feasible regions.

Theorem: If the feasible region of a linear programming problem is non-empty and bounded, then the objective function will have both a maximum and a minimum value within that region. Furthermore, these optimal values (maximum and minimum) will occur at one or more of the corner points (vertices) of the feasible region.

Let's analyze the options based on this theorem:

• (A) Always exist and occur at corner points: This statement is directly supported by the theorem. Since the feasible region is bounded (and presumably non-empty, as is standard in such questions unless specified otherwise), both a maximum and a minimum value for the linear objective function are guaranteed to exist. These optimal values are achieved at the vertices of the feasible region.

• (B) May exist at any point within the feasible region: While it's true that if the optimal value occurs along an edge, any point on that edge (which is within the feasible region) is optimal, the theorem specifically guarantees existence at corner points. The phrase "any point" is too broad and doesn't highlight the special role of corner points. The primary locations for optimal values are the corner points. If an interior point were optimal, it would imply the objective function is constant over a region, and thus also optimal at the boundary (including corner points) of that region.

• (C) May not exist: This is false for a bounded, non-empty feasible region. The boundedness ensures that the objective function cannot increase or decrease indefinitely within the feasible region, so finite optimal values must exist.

• (D) Only exist if the objective function is constant: This is false. Optimal values exist even if the objective function is not constant. If the objective function is constant, then every point in the feasible region (including all corner points) is an optimal solution, and the maximum and minimum values are the same.

Therefore, if the feasible region is bounded, the maximum and minimum values of the objective function always exist and occur at corner points.

(D) $(0, 3)$ (Note: Point (A) $(0,0)$ is also not a corner point of the feasible region.)

Given:

The constraints for the Linear Programming Problem (LPP) are:

The boundary lines are $L_1: x+y=3$, $L_2: x-y=1$, $L_3: x=0$, $L_4: y=0$.

To Find:

Which of the given points is NOT a corner point of the feasible region.

Solution:

A corner point of the feasible region must satisfy all the given constraints and be an intersection of two boundary lines.

Let's find the intersection points of the boundary lines and check their feasibility:

1. Intersection of $L_1: x+y=3$ and $L_2: x-y=1$:

As given in the problem statement, adding the two equations: $2x = 4 \implies x=2$.

Substituting $x=2$ into $x+y=3 \implies 2+y=3 \implies y=1$.

The point is $(2,1)$.

Check feasibility:

• $2+1=3 \le 3$ (True)

• $2-1=1 \ge 1$ (True)

• $2 \ge 0$ (True)

• $1 \ge 0$ (True)

So, $(2,1)$ is a corner point of the feasible region.

2. Intersection of $L_2: x-y=1$ and $L_4: y=0$ (x-axis):

Substitute $y=0$ into $x-y=1 \implies x-0=1 \implies x=1$.

The point is $(1,0)$.

Check feasibility:

• $1+0=1 \le 3$ (True)

• $1-0=1 \ge 1$ (True)

• $1 \ge 0$ (True)

• $0 \ge 0$ (True)

So, $(1,0)$ is a corner point of the feasible region. This corresponds to option (B).

3. Intersection of $L_1: x+y=3$ and $L_4: y=0$ (x-axis):

Substitute $y=0$ into $x+y=3 \implies x+0=3 \implies x=3$.

The point is $(3,0)$.

Check feasibility:

• $3+0=3 \le 3$ (True)

• $3-0=3 \ge 1$ (True)

• $3 \ge 0$ (True)

• $0 \ge 0$ (True)

So, $(3,0)$ is a corner point of the feasible region. This corresponds to option (C).

The corner points of the feasible region are $(1,0)$, $(3,0)$, and $(2,1)$.

Now, let's check the given options:

(A) $(0, 0)$:

Check feasibility:

• $0+0=0 \le 3$ (True)

• $0-0=0$. Is $0 \ge 1$? False.

Since $(0,0)$ is not feasible, it is NOT a corner point of the feasible region.

(B) $(1, 0)$:

As determined above, $(1,0)$ is a corner point of the feasible region.

(C) $(3, 0)$:

As determined above, $(3,0)$ is a corner point of the feasible region.

(D) $(0, 3)$:

This point is the intersection of $L_1: x+y=3$ and $L_3: x=0$.

Check feasibility:

• $0+3=3 \le 3$ (True)

• $0-3=-3$. Is $-3 \ge 1$? False.

Since $(0,3)$ is not feasible, it is NOT a corner point of the feasible region.

Both option (A) $(0,0)$ and option (D) $(0,3)$ are not corner points of the feasible region. In a single-choice question context, this indicates a potential issue with the question or options. However, if forced to choose one, both are technically correct answers to "Which of the following is NOT a corner point". The typical interpretation is to identify points that are vertices of the valid feasible area. Points $(1,0), (3,0), (2,1)$ form these vertices.

(C) Convex polygon

Explanation:

The feasible region of a Linear Programming Problem (LPP) is defined by a set of linear inequalities. Each linear inequality defines a half-plane, which is a convex set. The feasible region is the intersection of these half-planes.

The intersection of a finite number of closed half-planes is known as a convex polygonal region (if bounded) or an unbounded convex region. In the context of typical LPPs where a solution space is formed, this region is a convex set.

Let's analyze the options:

• (A) Triangle: A triangle is a specific type of convex polygon. The feasible region can be a triangle (e.g., if there are three effective constraints forming a bounded region), but it can also be a quadrilateral, pentagon, or an unbounded convex region. So, "triangle" is too specific.

• (B) Circle: A circle is defined by a non-linear equation (e.g., $x^2 + y^2 = r^2$) and its interior by a non-linear inequality. Linear constraints do not form circular regions.

• (C) Convex polygon: This is the most accurate general description. If the feasible region is bounded, it forms a convex polygon. If it is unbounded, it still forms a convex region (an unbounded convex polygon). The term "polygon" here can be understood to include unbounded regions with straight edges. The key property is convexity.

• (D) Straight line: The feasible region could be a straight line segment if, for example, two constraints are equalities that define a line and other constraints restrict it to a segment. However, this is a very specific case. More generally, it's a region, not just a line.

Given the options, "Convex polygon" is the best fit. It encompasses bounded feasible regions which are polygons and maintains the essential property of convexity that always holds for feasible regions of LPPs. If the region is unbounded, it's an unbounded convex polygonal region.

It's important to note that the feasible region can also be a single point or an empty set, both of which are considered convex. However, among the choices provided that describe a non-degenerate region, "convex polygon" is the most appropriate.

(B) The optimal value occurs at all points on the line segment joining these two points.

Explanation:

This scenario describes the case of multiple optimal solutions in a linear programming problem.

If an objective function $Z = ax + by$ achieves the same optimal value (say $Z_{\text{opt}}$) at two distinct corner points, $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$, of the feasible region, it means that the line representing the objective function ($ax + by = Z_{\text{opt}}$) coincides with the edge of the feasible region connecting $P_1$ and $P_2$.

Any point $P(x,y)$ on the line segment joining $P_1$ and $P_2$ can be written as a convex combination: $P = (1-\lambda)P_1 + \lambda P_2$ for $0 \le \lambda \le 1$.

The value of the objective function at point $P$ is:

$Z(P) = a((1-\lambda)x_1 + \lambda x_2) + b((1-\lambda)y_1 + \lambda y_2)$

$Z(P) = (1-\lambda)(ax_1 + by_1) + \lambda(ax_2 + by_2)$

Since $ax_1 + by_1 = Z_{\text{opt}}$ and $ax_2 + by_2 = Z_{\text{opt}}$,

$Z(P) = (1-\lambda)Z_{\text{opt}} + \lambda Z_{\text{opt}} = Z_{\text{opt}}$

This shows that every point on the line segment connecting $P_1$ and $P_2$ also yields the same optimal value $Z_{\text{opt}}$.

Let's analyze the options:

• (A) The optimal value occurs at exactly two points: This is false. While it occurs at the two corner points, it also occurs at infinitely many other points on the segment connecting them.

• (B) The optimal value occurs at all points on the line segment joining these two points: This is correct. This line segment is an edge of the feasible region.

• (C) The optimal value occurs only at one of the two points: This contradicts the premise that it has the same optimal value at *two distinct* corner points.

• (D) The problem has no optimal solution: This is false. The problem does have an optimal solution; in fact, it has infinitely many optimal solutions.

Therefore, if the objective function has the same optimal value at two distinct corner points of a feasible region, then the optimal value occurs at all points on the line segment joining these two points.

(D) Both (B) and (C) are true.

Explanation:

The given constraints are:

We are also given that $a > 0$ and $b > 0$.

Let's analyze these constraints:

• Constraints (iii) $x \ge 0$ and (iv) $y \ge 0$ restrict the feasible region to the first quadrant (including the axes).

• Constraint (i) $x \le a$ means that all points in the feasible region must have an x-coordinate less than or equal to $a$. This defines the region to the left of or on the vertical line $x=a$.

• Constraint (ii) $y \le b$ means that all points in the feasible region must have a y-coordinate less than or equal to $b$. This defines the region below or on the horizontal line $y=b$.

Combining these, the feasible region is the set of points $(x,y)$ such that $0 \le x \le a$ and $0 \le y \le b$. This describes a rectangle.

The vertices of this rectangle are formed by the intersections of the boundary lines $x=0, x=a, y=0, y=b$:

• Intersection of $x=0$ and $y=0$: $(0, 0)$

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

• Intersection of $x=a$ and $y=b$: $(a, b)$

• Intersection of $x=0$ and $y=b$: $(0, b)$

Now let's evaluate the options:

(A) The feasible region is unbounded.

This is false. Since $0 \le x \le a$ and $0 \le y \le b$, the region is bounded (it can be enclosed in a circle of finite radius).

(B) The feasible region is a rectangle with vertices at $(0, 0), (a, 0), (a, b), (0, b)$.

This is true, as derived above. The sides of the rectangle are of length $a$ (along the x-direction) and $b$ (along the y-direction).

(C) The feasible region is a square if $a=b$.

A square is a special type of rectangle where all sides are of equal length. If $a=b$, then the length of the side along the x-direction ($a$) is equal to the length of the side along the y-direction ($b$). In this case, the rectangle becomes a square with side length $a$ (or $b$). This statement is true.

(D) Both (B) and (C) are true.

Since statement (B) is true (the region is always a rectangle with the given vertices) and statement (C) is also true (it becomes a square under the specific condition $a=b$), this option is correct.

Therefore, both (B) and (C) are true.

(B) Graphical method (Corner Point Method)

Explanation:

The question describes a key procedure in solving Linear Programming Problems (LPPs), especially when the number of decision variables is small (typically two).

The steps involved are:

1. Formulate the LPP (define objective function and constraints).

2. Graph the constraints to identify the feasible region.

3. Determine the coordinates of all corner points (vertices) of the feasible region.

4. Evaluate the objective function at each of these corner points.

5. The point that yields the optimal (maximum or minimum, as required) value for the objective function is the optimal solution.

This entire process, particularly steps 2 through 5, is characteristic of the Graphical Method. Because a crucial part of this method is identifying and testing the corner points, it is also often referred to as the Corner Point Method.

Let's look at the other options:

• (A) Simplex method: The Simplex method is an algebraic, iterative algorithm used to solve LPPs, especially those with many variables where graphical methods are not feasible. While it also explores corner points in a systematic way, the description directly matches the graphical approach's core idea.

• (C) Substitution method: This is a general algebraic technique used to solve systems of equations (e.g., finding the intersection of two lines, which is a sub-step in finding corner points), but it is not the name of the overall LPP solving method described.

• (D) Elimination method: Similar to the substitution method, this is another algebraic technique for solving systems of linear equations, not the name of the LPP solving method itself.

Therefore, the method described is best known as the Graphical method, and the "Corner

(C) No feasible solution, hence no optimal solution.

Explanation:

The feasible region of a Linear Programming Problem (LPP) is the set of all points that satisfy all the given constraints simultaneously.

An optimal solution is a feasible solution that results in the maximum (or minimum) value of the objective function.

If the feasible region is empty, it means that there are no points $(x,y)$ that satisfy all the constraints of the LPP. This situation arises when the constraints are inconsistent or contradictory.

If there are no feasible solutions, then there cannot be an optimal solution, because an optimal solution must, by definition, be a feasible solution.

Let's analyze the options:

• (A) A unique optimal solution: This would require a non-empty feasible region where the objective function reaches its optimum at a single point.

• (B) Multiple optimal solutions: This would also require a non-empty feasible region where the objective function reaches its optimum along an edge or at multiple corner points.

• (C) No feasible solution, hence no optimal solution: This is correct. If there's no point that satisfies all constraints, then there's no solution to choose from, and thus no "best" (optimal) solution among them.

• (D) An unbounded solution: An unbounded solution (more accurately, an unbounded objective function value) occurs when the feasible region is unbounded, and the objective function can increase or decrease indefinitely within that region. This scenario implies the existence of a feasible region, albeit an unbounded one. An empty feasible region is different.

Therefore, if the feasible region of an LPP is empty, it means there are no feasible solutions, and consequently, there can be no optimal solution.

In a Linear Programming Problem (LPP), the feasible region is the set of all possible solutions that satisfy all the constraints of the problem. These constraints are typically in the form of linear inequalities or equalities.

Yes, the feasible region in a Linear Programming Problem is always convex. A region is convex if for any two points within the region, the line segment joining those two points is also entirely contained within the region.

This property arises from the fact that the constraints are linear. The intersection of linear inequalities (or half-spaces) always results in a convex region. Therefore, the feasible region, which is formed by the intersection of these half-spaces defined by the constraints, will always be convex.

Let $x$ be the number of Type A cameras and $y$ be the number of Type B cameras.

Given:

• Maximum storage space: 50 cameras
• Cost of Type A camera: $\textsf{₹} 3000$
• Cost of Type B camera: $\textsf{₹} 5000$
• Total budget: $\textsf{₹} 2,00,000$

Constraints:

Based on the information provided, we can formulate the following linear inequalities:

1. Storage constraint: $x + y \leq 50$
2. Budget constraint: $3000x + 5000y \leq 200000$ which simplifies to $3x + 5y \leq 200$
3. Non-negativity constraints: $x \geq 0$ and $y \geq 0$ (since the number of cameras cannot be negative)

Therefore, the system of linear inequalities is:

$x + y \leq 50$

$3x + 5y \leq 200$

$x \geq 0$

$y \geq 0$

Given:

• Objective function: $Z = 3x + 4y$
• Constraints:
  • $x \geq 0$
  • $y \geq 0$
  • $x + y \leq 4$

To find the corner points of the feasible region, we need to identify the intersection points of the constraint lines.

1. $x = 0$ and $y = 0$: Intersection point is $(0, 0)$.
2. $x = 0$ and $x + y = 4$: Substituting $x = 0$ into $x + y = 4$, we get $y = 4$. Intersection point is $(0, 4)$.
3. $y = 0$ and $x + y = 4$: Substituting $y = 0$ into $x + y = 4$, we get $x = 4$. Intersection point is $(4, 0)$.

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

In a Linear Programming Problem (LPP):

Objective Function: The objective function is a linear function that we want to either maximize or minimize. It represents the goal or objective of the problem. The objective function depends on the decision variables.

Constraints: Constraints are a set of linear inequalities or equalities that define the limitations or restrictions on the decision variables. These constraints represent the feasible region within which the optimal solution must lie. They represent real-world limitations, such as resource availability, production capacity, or demand requirements.

Example of an Objective Function:

Maximize $Z = 5x + 3y$, where $x$ and $y$ are decision variables representing quantities of two different products, and the coefficients 5 and 3 represent the profit per unit of each product, respectively. The goal is to maximize the total profit $Z$.

Given:

• Objective Function: Minimize $Z = 2x + 3y$
• Constraints:
  • $x \geq 0$
  • $y \geq 0$
  • $x + y \geq 4$

First, we need to graph the feasible region defined by the constraints.

1. $x \geq 0$ and $y \geq 0$ imply that the feasible region is in the first quadrant.
2. $x + y \geq 4$ can be graphed by first graphing the line $x + y = 4$. The feasible region is the area above and to the right of this line since we have $x + y \geq 4$.

The corner points of the feasible region are the intersection points of the constraint lines:

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

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

• At $(0, 4)$: $Z = 2(0) + 3(4) = 12$
• At $(4, 0)$: $Z = 2(4) + 3(0) = 8$

Since we are minimizing $Z$, the minimum value occurs at the point $(4, 0)$, where $Z = 8$.

Therefore, the solution to the LPP is $x = 4$, $y = 0$, and the minimum value of $Z$ is 8.

Optimal Solution: In the context of a Linear Programming Problem (LPP), an optimal solution is a feasible solution (a solution that satisfies all constraints) that yields the best possible value of the objective function. "Best possible" means either the maximum value (if the objective is to maximize) or the minimum value (if the objective is to minimize).

Relationship to Corner Points: A fundamental theorem in linear programming states that if an optimal solution exists, it will always occur at one of the corner points (also known as vertices or extreme points) of the feasible region. This is a crucial property that simplifies the process of finding the optimal solution.

This theorem is based on the fact that the objective function is linear, and the feasible region is convex. As we move along the edge of the feasible region, the value of the objective function changes linearly. Therefore, the optimal value must occur at one of the vertices where the edges intersect.

To find the optimal solution, we typically evaluate the objective function at each corner point of the feasible region and select the point that gives the best value (maximum or minimum, depending on the objective).

In summary, the optimal solution to an LPP, if it exists, is located at a corner point of the feasible region. This allows us to solve the LPP by examining only the corner points rather than infinitely many points within the feasible region.

Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively.

$\angle ACP = \angle QCD$.

For the first circle (passing through A, B, C, P), consider the arc AP. The angles subtended by the arc AP at the points C and B on the remaining part of the circle are $\angle ACP$ and $\angle ABP$.

We know that angles in the same segment of a circle are equal.

For the second circle (passing through D, B, C, Q), consider the arc DQ. The angles subtended by the arc DQ at the points C and B on the remaining part of the circle are $\angle QCD$ and $\angle QBD$.

Again, we know that angles in the same segment of a circle are equal.

We are given that ABD and PBQ are line segments. Since lines AD and PQ intersect at point B, the vertically opposite angles are equal.

From equations (i), (ii), and (iii), we can conclude that:

$\angle ACP = \angle ABP$ (from equation (i))

$\angle ABP = \angle QBD$ (from equation (iii))

$\angle QBD = \angle QCD$ (from equation (ii))

By combining these equalities, we get:

$\angle ACP = \angle QCD$

Hence, Proved.

The given Linear Programming Problem (LPP) is:

Maximize $Z = x + y$

Subject to the constraints:

$x + y \leq 5$

$x \geq 0$

$y \geq 0$

First, we will plot the constraints on a graph. The constraints $x \geq 0$ and $y \geq 0$ restrict the feasible region to the first quadrant of the Cartesian plane.

Next, we consider the inequality $x + y \leq 5$. To plot this, we first draw the line corresponding to the equation $x + y = 5$.

We can find two points on this line to plot it:

To determine the region represented by the inequality $x + y \leq 5$, we use a test point, typically the origin (0, 0).

Substituting (0, 0) into the inequality:

$0 + 0 \leq 5$

$0 \leq 5$, which is a true statement.

This means the region satisfying $x + y \leq 5$ is the half-plane containing the origin.

The intersection of all these regions forms the feasible region. In this case, the feasible region is a triangle with vertices at the origin O(0,0), point A(5,0) on the x-axis, and point B(0,5) on the y-axis.

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

According to the corner point method, the maximum or minimum value of the objective function occurs at one of the corner points of the feasible region. We will now evaluate the objective function $Z = x + y$ at each of these corner points.

From the table above, the maximum value of Z is 5. This maximum value is achieved at two corner points, A(5, 0) and B(0, 5). When the optimal value occurs at more than one corner point, it also occurs at every point on the line segment connecting these points.

Therefore, the maximum value of Z is 5, which occurs at all points on the line segment joining (5, 0) and (0, 5).

To formulate the constraints for this production problem, let's first define the variables.

Let $x$ be the number of units of Type A pens produced.

Let $y$ be the number of units of Type B pens produced.

The information given in the problem can be summarized in the following table:

Formulation of Constraints:

1. Machine Time Constraint:

The total machine time required is the sum of the time required for Type A pens and Type B pens.

Time for $x$ units of Type A = $2x$ hours.

Time for $y$ units of Type B = $3y$ hours.

The total machine time used is $2x + 3y$. Since the maximum available machine time is 100 hours, the constraint is:

2. Labor Time Constraint:

The total labor time required is the sum of the labor time required for Type A pens and Type B pens.

Time for $x$ units of Type A = $3x$ hours.

Time for $y$ units of Type B = $2y$ hours.

The total labor time used is $3x + 2y$. Since the maximum available labor time is 120 hours, the constraint is:

3. Non-Negativity Constraints:

The number of pens produced cannot be negative. Therefore, we have the following constraints:

Thus, the required constraints for the production problem are:

$2x + 3y \leq 100$

$3x + 2y \leq 120$

$x \geq 0$

$y \geq 0$

Concept of an Unbounded Feasible Region

In a Linear Programming Problem (LPP), the feasible region is the set of all points (x, y) that satisfy all the given constraints simultaneously. A feasible region is said to be unbounded if it extends infinitely in at least one direction. This means you cannot enclose the entire region within a circle of any finite radius.

An unbounded region typically occurs in problems where the constraints are of the 'greater than or equal to' ($\ge$) type, which define half-planes that do not form a closed, bounded polygon.

For example, the set of constraints below results in an unbounded feasible region:

The region defined by these inequalities is the area above the line $x+y=5$ in the first quadrant, which extends infinitely upwards and to the right.

Existence of an Optimal Solution in an Unbounded Region

Yes, an LPP with an unbounded feasible region can have an optimal (maximum or minimum) solution, but it is not guaranteed. The existence of an optimal solution depends entirely on the objective function, $Z$.

For an LPP with an unbounded feasible region, one of two outcomes is possible:

1. An optimal solution exists at a corner point (or along a line segment connecting two corner points).
2. The solution is unbounded, meaning the objective function can be increased (for maximization) or decreased (for minimization) indefinitely. In this case, no optimal solution exists.

Detailed Analysis of Optimal Solutions

The key is to compare the direction of optimization of the objective function with the direction in which the feasible region is unbounded.

Case 1: An Optimal Solution Exists

An optimal solution exists if the objective function's direction of optimization is opposite to the direction of unboundedness of the feasible region.

Example (Minimization):

Let's consider the objective function Minimize $Z = 4x + 2y$ with an unbounded feasible region defined by:

The feasible region is unbounded away from the origin. To minimize $Z$, we want to find the smallest possible values of $x$ and $y$. This means we are moving the line $4x+2y=k$ towards the origin. The line will eventually touch one of the corner points of the feasible region, and this will be the point of minimum value. Thus, an optimal solution exists.

Case 2: The Solution is Unbounded (No Optimal Solution)

No optimal solution exists if the objective function's value can be improved indefinitely in the direction of unboundedness.

Example (Maximization):

Let's consider the objective function Maximize $Z = 4x + 2y$ with the same unbounded feasible region as above.

The feasible region extends infinitely in the positive x and y directions. Since the objective function $Z = 4x + 2y$ increases as $x$ and $y$ increase, we can make $Z$ arbitrarily large by choosing points in the feasible region that are farther and farther from the origin. There is no upper limit to the value of $Z$. Therefore, no maximum value exists, and the solution is described as unbounded.

The given Linear Programming Problem (LPP) is:

Minimize $Z = 5x + 7y$

Subject to the constraints:

$x \geq 0$

$y \geq 0$

$x \geq 3$

$y \leq 5$

Solution:

First, we will determine the feasible region by plotting the given constraints on a graph.

• The constraints $x \geq 0$ and $y \geq 0$ restrict the feasible region to the first quadrant.
• The constraint $x \geq 3$ represents the region to the right of the vertical line $x = 3$.
• The constraint $y \leq 5$ represents the region below the horizontal line $y = 5$.

Combining all the constraints, the feasible region is the area in the first quadrant that is to the right of the line $x=3$ and between the lines $y=0$ (x-axis) and $y=5$. This region is unbounded in the positive x-direction.

The corner points of this unbounded feasible region are the points of intersection of the boundary lines. Let's find them:

• Point A: Intersection of $x = 3$ and $y = 0$. The coordinates are A(3, 0).
• Point B: Intersection of $x = 3$ and $y = 5$. The coordinates are B(3, 5).

Now, we evaluate the objective function $Z = 5x + 7y$ at these corner points.

From the table, the minimum value of Z at the corner points is 15, which occurs at point A(3, 0).

Since the feasible region is unbounded, we need to check whether this is the true minimum value. To do this, we graph the inequality corresponding to the objective function being less than the current minimum value:

$5x + 7y < 15$

We need to determine if there are any common points between the feasible region and the open half-plane defined by $5x + 7y < 15$.

For any point $(x, y)$ in the feasible region, we have $x \geq 3$ and $y \geq 0$.

Let's evaluate the objective function with these minimum values:

$Z = 5x + 7y \ge 5(3) + 7(0)$

$Z \ge 15 + 0$

$Z \ge 15$

This shows that for any point in the feasible region, the value of Z will always be greater than or equal to 15. The open half-plane $5x + 7y < 15$ has no points in common with the feasible region.

Therefore, the minimum value of Z is indeed 15.

The minimum value of Z is 15, which occurs at the point (3, 0).

To formulate this problem as a Linear Programming Problem (LPP), we need to define the decision variables, the objective function, and the constraints.

Step 1: Define Decision Variables

Let $x$ be the number of units of Product P1 to be manufactured.

Let $y$ be the number of units of Product P2 to be manufactured.

Step 2: Formulate the Objective Function

The goal is to maximize the profit. The profit is given by:

Profit from Product P1 = $\textsf{₹} 50$ per unit.

Profit from Product P2 = $\textsf{₹} 70$ per unit.

The total profit, $Z$, can be expressed as:

The objective is to Maximize Z.

Step 3: Formulate the Constraints

The constraints are based on the limited resources, which are assembly time and finishing time. The information can be summarized as follows:

Based on the table, we can formulate the constraints:

1. Assembly Time Constraint:

The total assembly time used for $x$ units of P1 and $y$ units of P2 must be less than or equal to the 10 hours available.

2. Finishing Time Constraint:

The total finishing time used for $x$ units of P1 and $y$ units of P2 must be less than or equal to the 12 hours available.

3. Non-Negativity Constraints:

The number of units produced cannot be negative.

The Complete LPP Formulation:

The complete Linear Programming Problem is formulated as follows:

Maximize $Z = 50x + 70y$

Subject to the constraints:

$x + 2y \leq 10$

$2x + y \leq 12$

$x \geq 0$

$y \geq 0$

The given Linear Programming Problem (LPP) is:

Maximize $Z = 3x + 2y$

Subject to the constraints:

$x + y \leq 2$

$x \geq 0$

$y \geq 0$

Solution:

First, we will graph the constraints to determine the feasible region.

The constraints $x \geq 0$ and $y \geq 0$ indicate that the feasible region is located in the first quadrant of the Cartesian plane.

To plot the inequality $x + y \leq 2$, we first draw the line for the equation $x + y = 2$.

We find the intercepts of this line:

• If $x = 0$, then $y = 2$. This gives the point (0, 2).
• If $y = 0$, then $x = 2$. This gives the point (2, 0).

To determine the region for the inequality $x + y \leq 2$, we can use the origin (0, 0) as a test point.

Substituting (0, 0) into the inequality: $0 + 0 \leq 2$, which simplifies to $0 \leq 2$. This is a true statement, so the feasible region lies on the side of the line that includes the origin.

The feasible region is the triangle formed by the intersection of these three constraints. Let's call the vertices O, A, and B.

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

• O(0, 0) - The origin.
• A(2, 0) - The x-intercept of the line $x + y = 2$.
• B(0, 2) - The y-intercept of the line $x + y = 2$.

According to the corner point theorem, the maximum or minimum value of the objective function will occur at one of these corner points. We now evaluate the objective function $Z = 3x + 2y$ at each vertex.

Comparing the values of Z at the corner points, we find that the maximum value is 6.

Therefore, the maximum value of Z is 6, which is achieved at the point (2, 0).

In the context of a Linear Programming Problem (LPP), the terms "feasible solution" and "feasible region" are fundamental concepts.

Feasible Solution

A feasible solution is a set of values for the decision variables (for example, a point $(x, y)$) that satisfies all the given constraints of the LPP simultaneously. This includes both the structural constraints (like $2x + y \leq 10$) and the non-negativity constraints (like $x \geq 0$ and $y \geq 0$).

Any point that lies within the feasible region or on its boundary is considered a feasible solution.

For instance, if the constraints are $x + y \leq 5$, $x \geq 0$, and $y \geq 0$, then the point $(1, 3)$ is a feasible solution because it satisfies all three conditions:

• $1 + 3 = 4$, which is $\leq 5$.
• $1 \geq 0$.
• $3 \geq 0$.

Conversely, the point $(4, 2)$ is an infeasible solution because it violates the first constraint ($4 + 2 = 6$, which is not $\leq 5$).

Feasible Region

The feasible region is the set of all possible feasible solutions of an LPP. When represented graphically, it is the common area or region formed by the intersection of the graphs of all the linear constraints.

Each constraint in an LPP defines a half-plane. The feasible region is the intersection of all these half-planes.

Key characteristics of a feasible region:

• It is always a convex set (a line segment connecting any two points in the region lies entirely within the region).
• It can be bounded, meaning it is a closed polygon that can be enclosed within a circle.
• It can be unbounded, meaning it extends infinitely in at least one direction.
• If there is no region that satisfies all constraints simultaneously, the LPP has no feasible solution, and the feasible region is empty.

The optimal solution (maximum or minimum value of the objective function) to an LPP, if it exists, is always found at one of the corner points (vertices) of the feasible region.

The given Linear Programming Problem (LPP) is:

Minimize $Z = x + 2y$

Subject to: $2x + y \geq 3$, $x \geq 0$, $y \geq 0$.

Solution:

The feasible region is determined by the constraints. It is an unbounded region in the first quadrant, bounded below by the line $2x + y = 3$.

The corner points of the feasible region are the intercepts of the line $2x + y = 3$ with the axes.

• Point A (x-intercept): Setting $y=0$ in $2x+y=3$ gives $x=1.5$. So, A = (1.5, 0).
• Point B (y-intercept): Setting $x=0$ in $2x+y=3$ gives $y=3$. So, B = (0, 3).

We evaluate the objective function $Z = x + 2y$ at these corner points.

The minimum value of Z from the table is 1.5. Since the feasible region is unbounded, we must check if a smaller value is possible by graphing the inequality $x + 2y < 1.5$.

The open half-plane defined by $x + 2y < 1.5$ has no points in common with the feasible region.

Thus, the minimum value found is the true minimum.

The minimum value of Z is 1.5, which occurs at the point (1.5, 0).

To formulate the constraints for this problem, we first need to define the decision variables.

Let x be the area of land (in hectares) used for growing Crop A.

Let y be the area of land (in hectares) used for growing Crop B.

The given information can be summarized in the following table:

Formulation of Constraints:

1. Land Area Constraint:

The total area of land used for both crops cannot exceed the total available land, which is 10 hectares.

2. Fertilizer Constraint:

The total amount of fertilizer used for both crops cannot exceed the maximum available amount of 24 units.

Fertilizer for $x$ hectares of Crop A = $2x$ units.

Fertilizer for $y$ hectares of Crop B = $3y$ units.

The total fertilizer constraint is:

3. Non-Negativity Constraints:

The area of land allocated to each crop cannot be negative, as it is a physical quantity.

Therefore, the complete set of constraints for this problem is:

$x + y \leq 10$

$2x + 3y \leq 24$

$x \geq 0$

$y \geq 0$

The given Linear Programming Problem (LPP) is:

Maximize $Z = 4x + y$

Subject to the constraints:

$x + y \leq 5$

$x \geq 1$

$x \geq 0$

$y \geq 0$

Solution:

First, we determine the feasible region by plotting the constraints on a graph.

• The constraints $x \geq 0$ and $y \geq 0$ restrict the solution to the first quadrant.
• The constraint $x \geq 1$ represents the region to the right of the vertical line $x = 1$.
• The constraint $x + y \leq 5$ represents the region on and below the line $x + y = 5$.

The feasible region is the polygon formed by the intersection of these regions. We find the corner points (vertices) of this region by finding the points of intersection of the boundary lines.

The corner points are:

• Point A: Intersection of $x = 1$ and $y = 0$. Coordinates are A(1, 0).
• Point B: Intersection of $x + y = 5$ and $y = 0$. This gives $x=5$. Coordinates are B(5, 0).
• Point C: Intersection of $x = 1$ and $x + y = 5$. Substituting $x=1$ gives $1+y=5$, so $y=4$. Coordinates are C(1, 4).

According to the corner point method, the maximum value of the objective function Z will occur at one of these vertices. We now evaluate $Z = 4x + y$ at each corner point.

By comparing the values of Z at the corner points, we find that the maximum value is 20.

The maximum value of Z is 20, which is achieved at the point (5, 0).

In a Linear Programming Problem (LPP), the "constraint set" and its "boundary" are fundamental concepts that define the space of possible solutions.

Constraint Set

The constraint set is the collection of all points that satisfy all the given constraints of the LPP simultaneously. This includes both the main constraints and the non-negativity constraints (e.g., $x \geq 0, y \geq 0$).

This set is more commonly referred to as the feasible region. Any point that lies within the constraint set or on its boundary is known as a feasible solution.

For example, if the constraints are:

The constraint set is the triangular area in the first quadrant formed by these three inequalities.

Boundary of the Constraint Set

The boundary of the constraint set (or feasible region) is the perimeter or edge of this region. It is composed of the lines or line segments that are formed by converting the constraint inequalities into equations.

Using the same example, the boundary is formed by the lines:

• $x + y = 6$
• $x = 0$ (the y-axis)
• $y = 0$ (the x-axis)

The boundary consists of the line segments of these lines that enclose the feasible region. The boundary is crucial because the corner points (vertices) of the feasible region lie on it. According to the fundamental theorem of linear programming, the optimal solution (maximum or minimum value of the objective function), if it exists, will always be found at one of these corner points on the boundary.

The given Linear Programming Problem (LPP) is:

Minimize $Z = x + y$

Subject to the constraints:

$x + y \leq 8$

$2x + y \geq 6$

$x \geq 0$

$y \geq 0$

Solution:

First, we will graph the constraints to identify the feasible region.

• The constraints $x \geq 0$ and $y \geq 0$ confine the feasible region to the first quadrant.
• Constraint 1: $x + y \leq 8$. We draw the line $x + y = 8$. The intercepts are (8, 0) and (0, 8). The region is on and below this line (towards the origin).
• Constraint 2: $2x + y \geq 6$. We draw the line $2x + y = 6$. The intercepts are (3, 0) and (0, 6). The region is on and above this line (away from the origin).

The feasible region is the polygon formed by the intersection of these regions. This is a bounded region (a quadrilateral). We find the corner points (vertices) of this feasible region by finding the points of intersection of the boundary lines.

The corner points are:

• Point A: Intersection of $2x + y = 6$ and $y = 0$ (x-axis). This gives $2x=6$, so $x=3$. The point is A(3, 0).
• Point B: Intersection of $x + y = 8$ and $y = 0$ (x-axis). This gives $x=8$. The point is B(8, 0).
• Point C: Intersection of $x + y = 8$ and $x = 0$ (y-axis). This gives $y=8$. The point is C(0, 8).
• Point D: Intersection of $2x + y = 6$ and $x = 0$ (y-axis). This gives $y=6$. The point is D(0, 6).

According to the corner point method, the minimum value of the objective function Z will occur at one of these vertices. We now evaluate $Z = x + y$ at each corner point.

By comparing the values of Z at the corner points, we find that the minimum value is 3.

Since the feasible region is bounded, this minimum value is the optimal solution.

The minimum value of Z is 3, which is achieved at the point (3, 0).

To formulate the constraints for this profit maximization problem, we first define the decision variables.

Let x be the number of units of Model A produced.

Let y be the number of units of Model B produced.

The information given in the problem can be organized into the following table:

Formulation of Constraints:

1. Machine M1 Constraint:

The total time used on machine M1 for producing $x$ units of Model A and $y$ units of Model B cannot exceed the maximum available time of 12 hours.

2. Machine M2 Constraint:

The total time used on machine M2 for producing $x$ units of Model A and $y$ units of Model B cannot exceed the maximum available time of 14 hours.

This constraint can be simplified by dividing the entire inequality by 2:

3. Non-Negativity Constraints:

The number of units produced for each model cannot be negative.

The complete set of constraints for this LPP is:

$x + 3y \leq 12$

$x + y \leq 7$

$x \geq 0$

$y \geq 0$

The given Linear Programming Problem (LPP) is:

Maximize $Z = 2x + 5y$

Subject to the constraints:

$x + 2y \leq 10$

$3x + y \leq 12$

$x \geq 0$

$y \geq 0$

Solution:

First, we will graph the constraints to find the feasible region.

• The constraints $x \geq 0$ and $y \geq 0$ restrict the feasible region to the first quadrant.
• Constraint 1: $x + 2y \leq 10$. The boundary line is $x + 2y = 10$. Its intercepts are (10, 0) and (0, 5). The region is on and below this line.
• Constraint 2: $3x + y \leq 12$. The boundary line is $3x + y = 12$. Its intercepts are (4, 0) and (0, 12). The region is on and below this line.

The feasible region is the polygon formed by the intersection of these areas. We now find the corner points (vertices) of this feasible region.

The corner points are:

• Point O: The origin, O(0, 0).
• Point A: The x-intercept of $3x + y = 12$, which is A(4, 0).
• Point C: The y-intercept of $x + 2y = 10$, which is C(0, 5).
• Point B: The intersection of the lines $x + 2y = 10$ and $3x + y = 12$.

To find point B, we solve the system of equations:

From equation (ii), $y = 12 - 3x$. Substituting this into equation (i):

$x + 2(12 - 3x) = 10$

$x + 24 - 6x = 10$

$-5x = -14$

$x = \frac{14}{5}$

Now, we find y:

$y = 12 - 3(\frac{14}{5}) = 12 - \frac{42}{5} = \frac{60 - 42}{5} = \frac{18}{5}$

So, the point of intersection is B($\frac{14}{5}, \frac{18}{5}$).

The corner points of the feasible region are O(0, 0), A(4, 0), B($\frac{14}{5}, \frac{18}{5}$), and C(0, 5).

Now, we evaluate the objective function $Z = 2x + 5y$ at each corner point.

By comparing the values of Z at the corner points, we find that the maximum value is 25.

The maximum value of Z is 25, which is achieved at the point (0, 5).

Slack and surplus variables are non-negative variables that are introduced into a Linear Programming Problem (LPP) to convert inequality constraints into equality constraints. This conversion is a crucial step for putting the LPP into its standard form, which is required for solution methods like the Simplex algorithm.

Slack Variable

A slack variable is used to convert a "less than or equal to" ($\leq$) constraint into an equation. It represents the unused amount or "slack" of a resource.

• It is added to the left-hand side of a $\leq$ inequality.
• It is always non-negative ($s \geq 0$).

Example:

Consider the resource constraint:

To convert this into an equation, we introduce a slack variable, let's say $s_1$. The equation becomes:

Here, $s_1$ represents the amount of the resource that is left unused. If $s_1=0$, the resource is fully utilized. If $s_1 > 0$, there is some resource to spare.

Surplus Variable

A surplus variable is used to convert a "greater than or equal to" ($\geq$) constraint into an equation. It represents the amount by which the left-hand side exceeds the minimum requirement.

• It is subtracted from the left-hand side of a $\geq$ inequality.
• It is always non-negative ($s \geq 0$).

Example:

Consider the requirement constraint:

To convert this into an equation, we introduce a surplus variable, let's say $s_2$. The equation becomes:

Here, $s_2$ represents the surplus amount over the minimum requirement of 50. If $s_2=0$, the requirement is exactly met. If $s_2 > 0$, the requirement is exceeded.

Use in the Standard Form of an LPP

The standard form of an LPP requires that:

1. The objective function is to be maximized (or minimized).
2. All constraints are expressed as equations (with non-negative right-hand sides).
3. All decision variables are non-negative.

Slack and surplus variables are essential for meeting the second requirement. For every $\leq$ constraint, a slack variable is added, and for every $\geq$ constraint, a surplus variable is subtracted. These new variables are also included in the objective function, but with a coefficient of zero, so they do not affect the optimal value of the original problem.

The problem is to determine the number of slotted and Phillips screwdrivers to manufacture per week to maximize profit, subject to constraints on manufacturing time and raw material costs. We can formulate this as a Linear Programming Problem (LPP).

Step 1: Define Decision Variables

Let $x$ be the number of slotted screwdrivers manufactured per week.

Let $y$ be the number of Phillips screwdrivers manufactured per week.

Step 2: Formulate the Objective Function

The profit on a slotted screwdriver is $\textsf{₹} 15$ and on a Phillips screwdriver is $\textsf{₹} 20$. The total profit, $Z$, is to be maximized.

Maximize $Z = 15x + 20y$

Step 3: Formulate the Constraints

The problem data can be summarized as follows:

Based on this table, we derive the constraints:

1. Manufacturing Time Constraint:

   $2x + 3y \leq 60$

2. Raw Material Cost Constraint:

   $10x + 15y \leq 300$

   This inequality can be simplified by dividing by 5:

   $2x + 3y \leq 60$

3. Non-Negativity Constraints:

   The number of screwdrivers cannot be negative.

   $x \geq 0, y \geq 0$

The complete LPP is:

Maximize $Z = 15x + 20y$

Subject to:

$2x + 3y \leq 60$

$x \geq 0$

$y \geq 0$

Step 4: Graphical Solution

We will graph the constraints to find the feasible region. The constraints $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

We plot the line corresponding to the main constraint: $2x + 3y = 60$.

• When $x = 0$, $3y = 60 \implies y = 20$. This gives the point (0, 20).
• When $y = 0$, $2x = 60 \implies x = 30$. This gives the point (30, 0).

The inequality $2x + 3y \leq 60$ represents the region on and below this line, including the origin.

The feasible region is a bounded triangle with vertices at the origin and the intercepts of the line.

Step 5: Identify the Corner Points

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

• O(0, 0)
• A(30, 0)
• B(0, 20)

Step 6: Evaluate the Objective Function

We evaluate the profit function $Z = 15x + 20y$ at each corner point to find the maximum value.

Step 7: Conclusion

From the table, the maximum value of the objective function Z is $\textsf{₹} 450$, which occurs at the point (30, 0).

Therefore, to maximize profit, the factory should manufacture 30 slotted screwdrivers and 0 Phillips screwdrivers per week.

The maximum profit is $\textsf{₹} 450$.

This problem can be solved by formulating it as a Linear Programming Problem (LPP).

Step 1: Define Decision Variables

Let $x$ be the number of bags of Type A fertilizer to be purchased.

Let $y$ be the number of bags of Type B fertilizer to be purchased.

Step 2: Formulate the Objective Function

The cost of a bag of Type A is $\textsf{₹} 100$ and Type B is $\textsf{₹} 120$. The farmer wants to minimize the total cost, $Z$.

Minimize $Z = 100x + 120y$

Step 3: Formulate the Constraints

The information about the nutrient content per bag (100 kg) is as follows:

Based on this table, we derive the constraints:

1. Nitrogen Requirement Constraint:

   $10x + 5y \geq 14$

2. Phosphoric Acid Requirement Constraint:

   $6x + 10y \geq 14$

   This can be simplified by dividing by 2:

   $3x + 5y \geq 7$

3. Non-Negativity Constraints:

   $x \geq 0, y \geq 0$

Step 4: Graphical Solution

We graph the inequalities to find the feasible region. The region is in the first quadrant and is unbounded.

• Line 1: $10x + 5y = 14$. Intercepts are (1.4, 0) and (0, 2.8). The region is on and above this line.
• Line 2: $3x + 5y = 7$. Intercepts are ($\frac{7}{3}$, 0) or approx. (2.33, 0) and (0, 1.4). The region is on and above this line.

Step 5: Identify the Corner Points

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

• Point A: The x-intercept of $3x+5y=7$, which is A($\frac{7}{3}$, 0).
• Point C: The y-intercept of $10x+5y=14$, which is C(0, 2.8).
• Point B: The intersection of the lines $10x + 5y = 14$ and $3x + 5y = 7$.

  Subtracting the second equation from the first:

  $(10x + 5y) - (3x + 5y) = 14 - 7$

  $7x = 7 \implies x = 1$

  Substituting $x=1$ into $3x + 5y = 7$:

  $3(1) + 5y = 7 \implies 5y = 4 \implies y = \frac{4}{5} = 0.8$

  So, the point of intersection is B(1, 0.8).

Step 6: Evaluate the Objective Function

We evaluate the cost function $Z = 100x + 120y$ at each corner point.

Step 7: Conclusion

The minimum cost from the corner points is $\textsf{₹} 196$ at point B(1, 0.8). Since the feasible region is unbounded, we must verify this result. We check if the open half-plane $100x + 120y < 196$ has any points in common with the feasible region. Graphically, it can be seen that there is no common region.

Therefore, the minimum cost is achieved at point B.

To minimize the cost, the farmer should purchase 1 bag of Type A fertilizer and 0.8 bags of Type B fertilizer.

The minimum cost will be $\textsf{₹} 196$.

This problem can be formulated and solved as a Linear Programming Problem (LPP).

Step 1: Define Decision Variables

Let $x$ be the number of units of food item F1.

Let $y$ be the number of units of food item F2.

Step 2: Formulate the Objective Function

The cost of food F1 is $\textsf{₹} 4$ per unit and F2 is $\textsf{₹} 6$ per unit. The objective is to minimize the total cost, $Z$.

Minimize $Z = 4x + 6y$

Step 3: Formulate the Constraints

The nutrient information is summarized below:

Based on this table, the constraints are:

1. Vitamin A Requirement Constraint:

   $3x + 6y \geq 80$

2. Minerals Requirement Constraint:

   $4x + 3y \geq 100$

3. Non-Negativity Constraints:

   $x \geq 0, y \geq 0$

Step 4: Graphical Solution

We will graph the constraints to find the feasible region. The constraints $x \geq 0$ and $y \geq 0$ place the region in the first quadrant. The feasible region is unbounded.

• Line 1: $3x + 6y = 80$. The intercepts are ($\frac{80}{3}$, 0) or approx. (26.67, 0) and (0, $\frac{80}{6}$) or approx. (0, 13.33). The region is on and above this line.
• Line 2: $4x + 3y = 100$. The intercepts are (25, 0) and (0, $\frac{100}{3}$) or approx. (0, 33.33). The region is on and above this line.

Step 5: Identify the Corner Points

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

• Point A: The x-intercept of $3x + 6y = 80$, which is A($\frac{80}{3}$, 0).
• Point C: The y-intercept of $4x + 3y = 100$, which is C(0, $\frac{100}{3}$).
• Point B: The intersection of the lines $3x + 6y = 80$ and $4x + 3y = 100$.

  To solve for the intersection, we multiply the second equation by 2 and subtract the first equation:

  $2(4x + 3y) - (3x + 6y) = 2(100) - 80$

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

  $5x = 120 \implies x = 24$

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

  $4(24) + 3y = 100 \implies 96 + 3y = 100 \implies 3y = 4 \implies y = \frac{4}{3}$

  The point of intersection is B(24, $\frac{4}{3}$).

Step 6: Evaluate the Objective Function

We evaluate the cost function $Z = 4x + 6y$ at each corner point.

Step 7: Conclusion

The minimum cost from the corner points is $\textsf{₹} 104$ at point B(24, $\frac{4}{3}$). Since the feasible region is unbounded, we must verify this by checking if the open half-plane $4x + 6y < 104$ has any points in common with the feasible region. Graphically, this region does not overlap with the feasible region, confirming our result.

To minimize the cost, one should purchase 24 units of food F1 and $\frac{4}{3}$ units of food F2.

The minimum cost to satisfy the dietary requirements is $\textsf{₹} 104$.

This problem can be formulated and solved as a Linear Programming Problem (LPP) to maximize profit.

Step 1: Define Decision Variables

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

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

Step 2: Formulate the Objective Function

The profit on toy A is $\textsf{₹} 30$ and on toy B is $\textsf{₹} 40$. The objective is to maximize the total profit, $Z$.

Maximize $Z = 30x + 40y$

Step 3: Formulate the Constraints

The resource information is summarized below:

Based on this table, the constraints are:

1. Machine Time Constraint:

   $2x + 3y \leq 200$

2. Craftsman Time Constraint:

   $3x + 2y \leq 250$

3. Non-Negativity Constraints:

   $x \geq 0, y \geq 0$

Step 4: Graphical Solution

We will graph the constraints to find the feasible region. The constraints $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant. The feasible region is bounded.

• Line 1: $2x + 3y = 200$. Intercepts are (100, 0) and (0, $\frac{200}{3} \approx 66.7$).
• Line 2: $3x + 2y = 250$. Intercepts are ($\frac{250}{3} \approx 83.3$, 0) and (0, 125).

Step 5: Identify the Corner Points

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

• Point O: The origin, O(0, 0).
• Point A: The x-intercept of $3x+2y=250$, which is A($\frac{250}{3}$, 0).
• Point C: The y-intercept of $2x+3y=200$, which is C(0, $\frac{200}{3}$).
• Point B: The intersection of the lines $2x + 3y = 200$ and $3x + 2y = 250$.

  Multiply the first equation by 3 and the second by 2:

  $6x + 9y = 600$

  $6x + 4y = 500$

  Subtracting the second new equation from the first:

  $5y = 100 \implies y = 20$

  Substitute $y=20$ into $2x + 3y = 200$:

  $2x + 3(20) = 200 \implies 2x + 60 = 200 \implies 2x = 140 \implies x = 70$

  The point of intersection is B(70, 20).

Step 6: Evaluate the Objective Function

We evaluate the profit function $Z = 30x + 40y$ at each corner point.

Step 7: Conclusion

The maximum value of the objective function is $\textsf{₹} 2900$, which occurs at the point (70, 20).

Therefore, to maximize profit, the company should produce 70 units of toy A and 20 units of toy B.

The maximum profit will be $\textsf{₹} 2900$.

The given Linear Programming Problem (LPP) is:

Maximize $Z = 5x + 3y$

Subject to the constraints:

$3x + 5y \leq 15$

$5x + 2y \leq 10$

$x \geq 0$

$y \geq 0$

Solution:

First, we will graph the constraints to determine the feasible region. The constraints $x \geq 0$ and $y \geq 0$ restrict the feasible region to the first quadrant.

• Constraint 1: $3x + 5y \leq 15$. We draw the line $3x + 5y = 15$.

  The intercepts are: when $x=0$, $y=3$ (Point (0,3)); when $y=0$, $x=5$ (Point (5,0)). The region is on and below this line.

• Constraint 2: $5x + 2y \leq 10$. We draw the line $5x + 2y = 10$.

  The intercepts are: when $x=0$, $y=5$ (Point (0,5)); when $y=0$, $x=2$ (Point (2,0)). The region is on and below this line.

The feasible region is the bounded polygon formed by the intersection of these regions.

Identify the Corner Points:

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

• Point O: The origin, O(0, 0).
• Point A: The x-intercept of the line $5x + 2y = 10$, which is A(2, 0).
• Point C: The y-intercept of the line $3x + 5y = 15$, which is C(0, 3).
• Point B: The intersection of the lines $3x + 5y = 15$ and $5x + 2y = 10$.

To find point B, we solve the system of equations:

Multiply equation (i) by 2 and equation (ii) by 5:

$6x + 10y = 30$

$25x + 10y = 50$

Subtracting the first new equation from the second:

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

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

Substitute $x = \frac{20}{19}$ into equation (ii):

$5(\frac{20}{19}) + 2y = 10 \implies \frac{100}{19} + 2y = 10$

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

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

So, the point of intersection is B($\frac{20}{19}, \frac{45}{19}$).

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

Evaluate the Objective Function:

We evaluate the objective function $Z = 5x + 3y$ at each corner point.

Conclusion:

Comparing the values of Z, the maximum value is $\frac{235}{19}$ (approximately 12.37).

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

The given Linear Programming Problem (LPP) is:

Minimize $Z = 200x + 500y$

Subject to the constraints:

$x + 2y \geq 10$

$3x + 4y \leq 24$

$x \geq 0$

$y \geq 0$

Solution:

First, we will graph the constraints to identify the feasible region. The constraints $x \geq 0$ and $y \geq 0$ restrict the feasible region to the first quadrant.

• Constraint 1: $x + 2y \geq 10$. The boundary line is $x + 2y = 10$. Its intercepts are (10, 0) and (0, 5). The region is on and above this line (away from the origin).
• Constraint 2: $3x + 4y \leq 24$. The boundary line is $3x + 4y = 24$. Its intercepts are (8, 0) and (0, 6). The region is on and below this line (towards the origin).

The feasible region is the bounded polygon (a triangle) formed by the intersection of these regions.

Identify the Corner Points:

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

• Point A: Intersection of $x + 2y = 10$ and the y-axis ($x=0$). This gives $2y=10$, so $y=5$. The point is A(0, 5).
• Point C: Intersection of $3x + 4y = 24$ and the y-axis ($x=0$). This gives $4y=24$, so $y=6$. The point is C(0, 6).
• Point B: Intersection of the lines $x + 2y = 10$ and $3x + 4y = 24$.

  From the first equation, $x = 10 - 2y$. Substituting this into the second equation:

  $3(10 - 2y) + 4y = 24$

  $30 - 6y + 4y = 24$

  $30 - 2y = 24$

  $6 = 2y \implies y = 3$

  Now, find x: $x = 10 - 2(3) = 10 - 6 = 4$.

  The point of intersection is B(4, 3).

The corner points of the feasible region are A(0, 5), B(4, 3), and C(0, 6).

Evaluate the Objective Function:

We evaluate the objective function $Z = 200x + 500y$ at each corner point.

Conclusion:

By comparing the values of Z at the corner points, we find that the minimum value is 2300.

The minimum value of Z is 2300, which is achieved at the point (4, 3).

This problem can be formulated and solved as a Linear Programming Problem (LPP) to maximize the total profit for the cooperative society.

Step 1: Formulate the LPP

Decision Variables:

Let $x$ be the area of land (in hectares) used for crop X.

Let $y$ be the area of land (in hectares) used for crop Y.

Objective Function:

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

Maximize $Z = 10500x + 9000y$

Constraints:

1. Land Constraint: The total land available is 50 hectares.

   $x + y \leq 50$

2. Herbicide Constraint: The total herbicide available is 800 litres. Crop X requires 20 litres/hectare and Crop Y requires 10 litres/hectare.

   $20x + 10y \leq 800$

   This can be simplified by dividing by 10:

   $2x + y \leq 80$

3. Weeding Cost Constraint: The total budget for weeding is $\textsf{₹} 5000$. Cost for crop X is $\textsf{₹} 100$/hectare and for crop Y is $\textsf{₹} 200$/hectare.

   $100x + 200y \leq 5000$

   This can be simplified by dividing by 100:

   $x + 2y \leq 50$

4. Non-Negativity Constraints: The area of land cannot be negative.

   $x \geq 0, y \geq 0$

The constraint $x+y \leq 50$ is redundant as any point satisfying $2x+y \leq 80$ and $x+2y \leq 50$ will also satisfy $x+y \leq 50$. (Adding the two main constraints gives $3x+3y \leq 130$, which implies $x+y \leq 43.33$, so $x+y \leq 50$ is always true for the feasible region). The feasible region is bounded by $2x+y=80$ and $x+2y=50$.

Step 2: Graphical Solution

We will graph the constraints to find the feasible region. The region is bounded and lies in the first quadrant.

• Line 1: $2x + y = 80$. Intercepts are (40, 0) and (0, 80).
• Line 2: $x + 2y = 50$. Intercepts are (50, 0) and (0, 25).

Step 3: Identify the Corner Points

The corner points of the feasible region are:

• Point O: The origin, O(0, 0).
• Point A: The x-intercept of $2x + y = 80$, which is A(40, 0).
• Point C: The y-intercept of $x + 2y = 50$, which is C(0, 25).
• Point B: The intersection of the lines $2x + y = 80$ and $x + 2y = 50$.

  From $2x + y = 80$, we have $y = 80 - 2x$. Substituting this into the second equation:

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

  $x + 160 - 4x = 50$

  $-3x = -110 \implies x = \frac{110}{3}$

  Now, find y: $y = 80 - 2(\frac{110}{3}) = \frac{240 - 220}{3} = \frac{20}{3}$

  The point of intersection is B($\frac{110}{3}, \frac{20}{3}$).

The vertices of the feasible region are O(0,0), A(40,0), B($\frac{110}{3}, \frac{20}{3}$), and C(0,25).

Step 4: Evaluate the Objective Function

We evaluate the profit function $Z = 10500x + 9000y$ at each corner point.

Step 5: Conclusion

The maximum value of the objective function Z is $\textsf{₹} 445,000$, which occurs at point B($\frac{110}{3}, \frac{20}{3}$).

Therefore, to maximize profit, the society should grow $\frac{110}{3}$ hectares of crop X and $\frac{20}{3}$ hectares of crop Y.

The maximum profit will be $\textsf{₹} 445,000$.

This problem can be formulated and solved as a Linear Programming Problem (LPP) to find the production mix that maximizes profit.

Step 1: Formulate the LPP

Decision Variables:

Let $x$ be the number of Model A bicycles produced per week.

Let $y$ be the number of Model B bicycles produced per week.

Objective Function:

The profit from Model A is $\textsf{₹} 25$ per unit and from Model B is $\textsf{₹} 35$ per unit. The objective is to maximize the total profit, $Z$.

Maximize $Z = 25x + 35y$

Constraints:

The resource limitations are summarized below:

1. Assembly Time Constraint:

   $2x + y \leq 100$

2. Painting Time Constraint:

   $x + 2y \leq 120$

3. Non-Negativity Constraints:

   $x \geq 0, y \geq 0$

Step 2: Graphical Solution

We graph the constraints to find the feasible region. The region is bounded and lies in the first quadrant.

• Line 1: $2x + y = 100$. Intercepts are (50, 0) and (0, 100).
• Line 2: $x + 2y = 120$. Intercepts are (120, 0) and (0, 60).

Step 3: Identify the Corner Points

The vertices of the feasible region are:

• Point O: The origin, O(0, 0).
• Point A: The x-intercept of $2x + y = 100$, which is A(50, 0).
• Point C: The y-intercept of $x + 2y = 120$, which is C(0, 60).
• Point B: The intersection of the lines $2x + y = 100$ and $x + 2y = 120$.

  From $2x + y = 100$, we get $y = 100 - 2x$. Substitute this into the second equation:

  $x + 2(100 - 2x) = 120$

  $x + 200 - 4x = 120$

  $-3x = -80 \implies x = \frac{80}{3}$

  Now, find y: $y = 100 - 2(\frac{80}{3}) = \frac{300 - 160}{3} = \frac{140}{3}$

  The point of intersection is B($\frac{80}{3}, \frac{140}{3}$).

The corner points are O(0, 0), A(50, 0), B($\frac{80}{3}, \frac{140}{3}$), and C(0, 60).

Step 4: Evaluate the Objective Function

We evaluate the profit function $Z = 25x + 35y$ at each corner point.

Step 5: Conclusion

The maximum value of the objective function Z is $\textsf{₹} 2300$, which occurs at the point B($\frac{80}{3}, \frac{140}{3}$).

Therefore, to maximize profit, the manufacturer should produce $\frac{80}{3}$ units of Model A and $\frac{140}{3}$ units of Model B per week. Since the number of bicycles must be an integer, the optimal integer solution would be found near this vertex. However, based on the graphical LPP solution method, this fractional value gives the theoretical maximum.

The maximum profit is $\textsf{₹} 2300$.

The given Linear Programming Problem (LPP) is:

Maximize $Z = 3x + 4y$

Subject to the constraints:

$x + y \leq 4$

$x \geq 0$

$y \geq 0$

Solution:

First, we will graph the constraints to determine the feasible region. The constraints $x \geq 0$ and $y \geq 0$ restrict the solution to the first quadrant of the Cartesian plane.

We then plot the line for the constraint $x + y = 4$.

• The x-intercept is found by setting $y=0$, which gives $x=4$. The point is (4, 0).
• The y-intercept is found by setting $x=0$, which gives $y=4$. The point is (0, 4).

The inequality $x + y \leq 4$ represents the region on and below this line. The feasible region is a bounded triangle formed by the vertices at the origin and the intercepts of the line.

Identify the Corner Points:

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

• O(0, 0) - The origin.
• A(4, 0) - The x-intercept.
• B(0, 4) - The y-intercept.

Evaluate the Objective Function:

According to the corner point method, the maximum value of the objective function Z will occur at one of these vertices. We evaluate $Z = 3x + 4y$ at each corner point.

Conclusion:

By comparing the values of Z at the corner points, we find that the maximum value is 16.

The maximum value of Z is 16, which is achieved at the point (0, 4).

The given Linear Programming Problem (LPP) is:

Minimize $Z = 3x + 2y$

Subject to the constraints:

$x + y \geq 8$

$3x + 5y \leq 15$

$x \geq 0$

$y \geq 0$

Graphical Analysis of Constraints:

To solve this LPP, we first need to determine the feasible region by plotting the constraints on a graph. The constraints $x \geq 0$ and $y \geq 0$ restrict the solution to the first quadrant.

1. Constraint: $x + y \geq 8$

We first draw the line $x + y = 8$.

• The x-intercept is (8, 0).
• The y-intercept is (0, 8).

Testing the origin (0, 0), we get $0 + 0 \geq 8$, which is false. Therefore, the region for this constraint is the half-plane on the side of the line that does not contain the origin (i.e., the region above the line).

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

We draw the line $3x + 5y = 15$.

• The x-intercept is found by setting $y=0$, so $3x=15 \implies x=5$. The point is (5, 0).
• The y-intercept is found by setting $x=0$, so $5y=15 \implies y=3$. The point is (0, 3).

Testing the origin (0, 0), we get $3(0) + 5(0) \leq 15$, which is true. Therefore, the region for this constraint is the half-plane on the side of the line that does contain the origin (i.e., the region below the line).

Existence of the Feasible Region:

The feasible region is the common area that satisfies all constraints simultaneously. When we plot both regions on the graph:

• The region for $x + y \geq 8$ is above the line connecting (8,0) and (0,8).
• The region for $3x + 5y \leq 15$ is below the line connecting (5,0) and (0,3).

Graphically, it is clear that these two regions do not overlap. There are no points $(x, y)$ in the first quadrant that are simultaneously above the line $x+y=8$ and below the line $3x+5y=15$.

Justification and Conclusion:

No, the feasible region does not exist.

This can be justified as follows: The set of points satisfying $3x + 5y \leq 15$ along with $x \geq 0, y \geq 0$ is a triangle with vertices (0,0), (5,0), and (0,3). For any point $(x, y)$ in this triangle, the maximum value of $x$ is 5 and the maximum value of $y$ is 3. Therefore, the maximum possible value for the sum $x+y$ for any point in this region is $5+3=8$. However, even the point (5,3) itself does not satisfy $3x+5y \leq 15$ since $3(5)+5(3) = 30 > 15$. In fact, for any point in this region, $x+y < 8$.

The first constraint requires that $x+y \geq 8$. Since there is no point that can satisfy both $x+y \geq 8$ and $3x+5y \leq 15$ at the same time, there is no common region.

Because there is no feasible region, there are no feasible solutions. Consequently, the LPP has no solution, and the minimum value of Z cannot be found.

This problem can be formulated and solved as a Linear Programming Problem (LPP) to find the production mix that maximizes profit.

Step 1: Formulate the LPP

Decision Variables:

Let $x$ be the number of units of product A manufactured per week.

Let $y$ be the number of units of product B manufactured per week.

Objective Function:

The profit on unit A is $\textsf{₹} 20$ and on unit B is $\textsf{₹} 30$. The objective is to maximize the total profit, $Z$.

Maximize $Z = 20x + 30y$

Constraints:

The constraints are based on the available machine hours.

1. Grinding Time Constraint: There are 2 grinding machines, each working for 40 hours.

   Total available grinding time = $2 \times 40 = 80$ hours per week.

   The constraint is $2x + 4y \leq 80$, which can be simplified to:

   $x + 2y \leq 40$

2. Polishing Time Constraint: There are 3 polishing machines, each working for 40 hours.

   Total available polishing time = $3 \times 40 = 120$ hours per week.

   $3x + 2y \leq 120$

3. Non-Negativity Constraints: The number of units produced cannot be negative.

   $x \geq 0, y \geq 0$

Step 2: Graphical Solution

We will graph the constraints to find the feasible region. The region is bounded and lies in the first quadrant.

• Line 1: $x + 2y = 40$. Intercepts are (40, 0) and (0, 20).
• Line 2: $3x + 2y = 120$. Intercepts are (40, 0) and (0, 60).

Step 3: Identify the Corner Points

The feasible region is the area in the first quadrant that is below both lines. The corner points (vertices) of this region are:

• Point O: The origin, O(0, 0).
• Point A: The intersection of the lines $x + 2y = 40$ and $3x + 2y = 120$. Subtracting the first equation from the second gives $2x = 80$, so $x=40$. Substituting back gives $40+2y=40$, so $y=0$. The point is A(40, 0). This is also the x-intercept for both lines.
• Point B: The y-intercept of the feasible region. This is the y-intercept of the line $x + 2y = 40$ because it is the inner boundary. The point is B(0, 20).

The vertices of the feasible region are O(0, 0), A(40, 0), and B(0, 20).

Step 4: Evaluate the Objective Function

We evaluate the profit function $Z = 20x + 30y$ at each corner point.

Step 5: Conclusion

The maximum value of the objective function Z is $\textsf{₹} 800$, which occurs at the point A(40, 0).

Therefore, to maximize profit, the firm should manufacture 40 units of product A and 0 units of product B per week.

The maximum profit will be $\textsf{₹} 800$.

The given Linear Programming Problem (LPP) is:

Minimize and Maximize $Z = x + 2y$

Subject to the constraints:

$x + 2y \geq 100$

$2x - y \leq 0$ (or $y \geq 2x$)

$2x + y \leq 200$

$x \geq 0$

$y \geq 0$

Solution:

First, we will graph the given constraints to determine the feasible region. The constraints $x \geq 0$ and $y \geq 0$ restrict the solution to the first quadrant.

• Line 1: $x + 2y = 100$. The intercepts are (100, 0) and (0, 50). The region for $x+2y \geq 100$ is on and above this line.
• Line 2: $y = 2x$. This line passes through the origin (0, 0). The region for $y \geq 2x$ is on and above this line.
• Line 3: $2x + y = 200$. The intercepts are (100, 0) and (0, 200). The region for $2x + y \leq 200$ is on and below this line.

The feasible region is the bounded polygon formed by the intersection of these regions.

Identify the Corner Points:

The corner points (vertices) of the feasible region are found by finding the points of intersection of the boundary lines.

• Point A: Intersection of $x + 2y = 100$ and $x=0$ (y-axis). This gives $2y = 100$, so $y=50$. The point is A(0, 50).
• Point B: Intersection of $2x + y = 200$ and $x=0$ (y-axis). This gives $y=200$. The point is B(0, 200).
• Point C: Intersection of $2x + y = 200$ and $y = 2x$.

  Substituting $y = 2x$ gives $2x + (2x) = 200 \implies 4x = 200 \implies x = 50$.

  Then, $y = 2(50) = 100$. The point is C(50, 100).

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

  Substituting $y = 2x$ gives $x + 2(2x) = 100 \implies 5x = 100 \implies x = 20$.

  Then, $y = 2(20) = 40$. The point is D(20, 40).

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

Evaluate the Objective Function:

We evaluate the objective function $Z = x + 2y$ at each corner point.

Conclusion:

From the table, we can determine the minimum and maximum values of Z.

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

The minimum value of Z is 100. This value occurs at two corner points, A(0, 50) and D(20, 40). When an optimal value occurs at more than one corner point, it occurs at every point on the line segment connecting these points. In this case, the line segment connecting A and D is a part of the line $x + 2y = 100$. Since the objective function is $Z = x + 2y$, any point on this segment will yield a Z value of 100. Therefore, the minimum value occurs at all points on the line segment joining (0, 50) and (20, 40).