Chapter 10 Linear Programming (Q & A)

Explanation:

In Linear Programming:

• (A) The objective function is always linear in the decision variables (e.g., maximize \( Z = 3x + 2y \)).
• (B) The constraints are always expressed as linear equations or inequalities.
• (C) The non-negativity conditions (\( x \geq 0, y \geq 0 \)) are linear inequalities.

Thus, all the given components involve linear relationships.

Correct Option: (D) All of the above

Explanation:

In Linear Programming, the objective function is designed either to maximize profit or minimize cost. The coefficients of the decision variables in this function indicate how much each unit of the variable contributes to the objective.

For example, in the objective function:

$Z = 4x + 3y$

The coefficients 4 and 3 represent the profit (or cost) contributed per unit of variables $x$ and $y$, respectively.

Correct Option: (B) Profit or cost contribution per unit of each variable

Explanation:

In a Linear Programming Problem (LPP) that focuses on minimizing cost, the objective function is structured to reflect the total cost of producing units. Each coefficient in the objective function shows the cost incurred per unit of a decision variable.

For example, the objective function:

$Z = 5x + 8y$

represents a scenario where producing one unit of $x$ costs $\textsf{₹}$5 and one unit of $y$ costs $\textsf{₹}$8. The goal is to minimize $Z$, the total cost.

Correct Option: (B) Cost per unit

Explanation:

In a graphical representation of a Linear Programming Problem (LPP), the feasible region is the area on the graph where all the constraints (including non-negativity) are satisfied simultaneously. It represents the set of all possible feasible solutions to the LPP, not just the optimal one.

Only the points lying within this region (including its boundaries) are considered for evaluation by the objective function.

Correct Option: (C) Solutions that satisfy all constraints and non-negativity restrictions

Explanation:

In Linear Programming, a point is considered an infeasible solution if it does not satisfy all the constraints of the problem. This means it lies outside the feasible region, which includes all points that fulfill the constraints and non-negativity conditions.

Only points within or on the boundary of the feasible region are valid (feasible) solutions. Any point lying outside is considered infeasible.

Correct Option: (A) Lies outside the feasible region

Explanation:

The graphical method of solving Linear Programming Problems (LPPs) involves plotting the feasible region and identifying the optimal solution using graphs. This method works effectively only when there are two decision variables because we can easily represent the feasible region and objective function in a 2D graph.

For more than two variables (e.g., three or more), it becomes impractical to draw and interpret the feasible region visually in higher dimensions.

Correct Option: (A) It is difficult to draw a feasible region in more than two dimensions.

Explanation:

In Linear Programming, a constraint of the form $\ge$ (greater than or equal to) is used when there is a minimum requirement that must be satisfied. For example, if a diet plan must provide at least 50 grams of protein daily, the constraint would be represented as:

Protein ≥ 50

This ensures that the amount is not less than the required minimum.

Correct Option: (B) $\ge$ constraint

Explanation:

In a production planning Linear Programming Problem (LPP), constraints are the conditions that must be met when planning production. These typically relate to the availability of limited resources such as:

• Labour hours
• Raw materials
• Machine time
• Storage capacity

These limitations restrict how much of each product can be produced and form the inequalities in the mathematical model.

Correct Option: (C) Limitations on resources like labour, raw materials, machine hours

Explanation:

An LPP (Linear Programming Problem) has an unbounded solution when the objective function (such as maximizing profit or minimizing cost) can be improved indefinitely without violating any constraints.

This is possible when the feasible region is unbounded and the direction in which the objective function improves lies entirely within the feasible region. For example, in a maximization problem, if profit can be increased infinitely without any restriction from the constraints, the solution is said to be unbounded.

Correct Option: (C) The feasible region is unbounded and the objective function can be increased (for maximization) or decreased (for minimization) indefinitely in the feasible region.

Explanation:

The inequality \( x \ge 3 \) means that all values of \( x \) are greater than or equal to 3. This forms a vertical boundary line at \( x = 3 \), and the solution region includes all points on and to the right of this vertical line.

Hence, the shaded region is the area to the right of the vertical line \( x = 3 \).

Correct Option: (B) To the right of the vertical line \( x = 3 \)

Explanation:

The inequality \( y \le 5 \) represents all points where the value of \( y \) is less than or equal to 5. This corresponds to the region that lies on and below the horizontal line \( y = 5 \).

Correct Option: (D) Below the horizontal line \( y = 5 \)

Explanation:

A feasible solution in linear programming is any point that satisfies all the constraints of the problem, including the non-negativity constraints. This point can lie either inside the feasible region or exactly on its boundary.

Correct Option: (C) Either on the boundary or inside the feasible region.

Explanation:

All the statements provided are correct regarding the optimal solution in Linear Programming Problems (LPP):

• (A) The optimal solution must be a feasible solution since it must satisfy all constraints.
• (B) If an optimal solution exists, it is found at a corner point (or an edge between corner points in case of multiple optima).
• (C) By definition, the optimal solution provides the best (maximum or minimum) value of the objective function among all feasible solutions.

Correct Option: (D) All of the above.

Explanation:

The feasible region in a Linear Programming Problem (LPP) is indeed a polygon (or more specifically, a convex polygon) because it results from the intersection of a finite number of half-planes. Each linear inequality defines a half-plane, and the intersection of such half-planes defines the region where all constraints are satisfied.

Therefore, both the assertion and the reason are correct, and the reason appropriately explains why the feasible region is a polygon.

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

Explanation:

In the graphical method for solving a Linear Programming Problem (LPP), the optimal value (maximum or minimum) of the objective function occurs at one of the **corner points** (also called vertices) of the feasible region. This is based on the **Corner Point Theorem**, which states that if an optimal solution exists, it must be at a vertex (or along an edge connecting two vertices in the case of multiple optimal solutions).

Note: The origin is not necessarily a corner point unless it lies within the feasible region and satisfies all constraints.

Correct Option: (B) A corner point of the feasible region

Step 1: Analyze the Constraints

• \( x \ge 0 \): This means the solution lies to the right of the y-axis.
• \( y \ge 0 \): This means the solution lies above the x-axis.
• \( x + y \ge 5 \): This means the solution lies on or above the line \( x + y = 5 \).
• \( x + y \le 3 \): This means the solution lies on or below the line \( x + y = 3 \).

Step 2: Check for Feasibility

The region defined by \( x + y \ge 5 \) is entirely above the line \( x + y = 5 \), while the region defined by \( x + y \le 3 \) is entirely below the line \( x + y = 3 \).

There is no overlap between these two regions, hence no point can satisfy both conditions at the same time.

Conclusion: The constraints are contradictory, so there is no point that satisfies all the given conditions. Therefore, the Linear Programming Problem (LPP) has no feasible solution.

Correct Option: (D) No feasible solution

Matching Explanation:

(i) Constraint inequality → (c) A half-plane

This represents a region defined by a linear inequality such as \( x + y \leq 5 \).

(ii) Feasible Region → (d) The shaded polygonal area

This is the area that satisfies all constraints.

(iii) Corner Point → (b) Intersection of boundary lines of the feasible region

These are the vertices of the feasible region polygon, where optimal solutions often lie.

(iv) Objective Function line → (a) A line or curve representing constant objective value

This line is used to identify the optimal solution by shifting parallel to itself.

Correct Matching:

• (i)-(c)
• (ii)-(d)
• (iii)-(b)
• (iv)-(a)

Correct Option: (A)

Given: Each kg of food X contains 2 units of Vitamin A and each kg of food Y contains 3 units of Vitamin A.

We are using \( x \) kg of X and \( y \) kg of Y. The total Vitamin A obtained will be \( 2x + 3y \).

Since the minimum required is 10 units, the constraint becomes:

\( 2x + 3y \ge 10 \)

Correct Option: (B)

Step 1: Understand the feasible region.

A point is feasible if it satisfies all constraints:

$x \ge 0$

$y \ge 0$

$x + y \le 5$

$2x + y \le 8$

Step 2: Check each option one by one.

Option (A): $(4, 1)$

$x = 4 \ge 0$, $y = 1 \ge 0$ → Satisfies non-negativity

$x + y = 4 + 1 = 5 \le 5$ → Satisfies

$2x + y = 2(4) + 1 = 8 + 1 = 9 \nleq 8$ → Fails

Option (B): $(3, 3)$

$x + y = 3 + 3 = 6 \nleq 5$ → Fails

Option (C): $(2, 2)$

$x + y = 2 + 2 = 4 \le 5$ → Satisfies

$2x + y = 4 + 2 = 6 \le 8$ → Satisfies

$x \ge 0$, $y \ge 0$ → Satisfies

This point satisfies all constraints.

Option (D): $(5, 0)$

$x + y = 5 + 0 = 5 \le 5$ → Satisfies

$2x + y = 10 + 0 = 10 \nleq 8$ → Fails

Final Answer: (C) $(2, 2)$

Step 1: Understand the concept of a corner point.

A corner point (or vertex) of the feasible region is the point of intersection of two or more boundary lines of the constraints, and lies within or on the edge of the feasible region.

Step 2: Analyze the given point.

The point $(4,2)$ is the intersection of the lines $x+y=6$ and $x-y=2$.

This point satisfies:

$x + y = 4 + 2 = 6$ ✅

$x - y = 4 - 2 = 2$ ✅

Also, $x = 4 \ge 0$, $y = 2 \ge 0$ → satisfies non-negativity ✅

Conclusion:

The point $(4, 2)$ is an intersection of two constraint lines and satisfies non-negativity. It can be a corner point only if it also satisfies all other constraints in the problem (if any).

Final Answer: (A) Yes, if it satisfies all other constraints as well.

Step 1: Understand the types of feasible regions in Linear Programming Problems (LPPs).

Bounded Feasible Region: A region that is enclosed within limits. It can produce a maximum and minimum value for the objective function.

Unbounded Feasible Region: A region that is not enclosed on all sides. It can still have a solution depending on the direction of optimization.

Empty Feasible Region: When the constraints are contradictory, no point satisfies all constraints, and hence, the feasible region is empty.

Curved Feasible Region: This is not possible in LPP because all constraints in an LPP are linear, and hence feasible regions are always polygonal or half-plane shaped, formed by straight lines.

Final Answer: (D) LPP with curved feasible region

Step 1: Use the objective function $Z = 5x + 4y$ to evaluate the value of $Z$ at each corner point.

Corner Point 1: $(0, 0)$

$Z = 5(0) + 4(0) = 0$

Corner Point 2: $(10, 0)$

$Z = 5(10) + 4(0) = 50$

Corner Point 3: $(5, 5)$

$Z = 5(5) + 4(5) = 25 + 20 = 45$

Corner Point 4: $(0, 10)$

$Z = 5(0) + 4(10) = 40$

Step 2: Find the minimum among these values:

$\min(0,\ 50,\ 45,\ 40) = 0$

Final Answer: (A) $5(0) + 4(0) = 0$

Step 1: In the standard form of an LPP for maximization, all constraints must be of the form $\le$.

Step 2: Given inequality is:

To convert this into a $\le$ form, multiply both sides of the inequality by $-1$:

Final Answer: (A) $-x - y \le -5$

Step 1: In linear programming problems, if the objective function has the same optimal value at two corner points of the feasible region, it means the function is constant along the line segment connecting them.

Step 2: Since the objective function is linear, and it takes the same value at both endpoints, it must take that same value at every point on the line segment between them.

This happens because linear functions change at a constant rate, and if the change from point A to B is zero, it means the value is constant along the line.

Conclusion: The optimal value occurs at every point on the line segment joining those two corner points.

Final Answer: (B) Every point on the line segment joining these two points.

Step 1: In Linear Programming Problems (LPP), the feasible region is defined by the set of points that satisfy all the linear constraints and non-negativity conditions.

Step 2: The intersection of linear inequalities forms a region known as the feasible region.

This region is always a convex set, because:

• If $P$ and $Q$ are two feasible solutions (i.e., lie in the region),
• Then any point on the line segment joining $P$ and $Q$, given by $\lambda P + (1 - \lambda) Q$ for $0 \le \lambda \le 1$, is also a feasible solution.

Conclusion: The feasible region of an LPP is always a convex set.

Final Answer: (D) convex

Step 1: In linear programming, the feasible region is the set of all points that satisfy the system of linear constraints.

Step 2: Let's analyze each option:

• (A) The feasible region is a polygon in 2D when bounded. ✅ True
• (B) The feasible region can be empty if no solution satisfies all constraints. ✅ True
• (C) The feasible region can be a single point (e.g., when all constraints intersect at one point). ✅ True
• (D) The feasible region cannot be non-convex. It is always a convex set due to the nature of linear inequalities. ❌ False

Conclusion: The FALSE statement is that the feasible region can be non-convex. This contradicts a fundamental property of feasible regions in LPPs.

Final Answer: (D) It can be non-convex.

Step 1: In a Linear Programming Problem (LPP), a solution is said to be feasible only if it satisfies:

• All the given constraints, and
• All the non-negativity conditions, i.e., $x \ge 0, y \ge 0$ etc.

Step 2: If a point violates any constraint or non-negativity condition, it is called an infeasible solution.

Even if it satisfies all the linear constraints, violating $x \ge 0$ or $y \ge 0$ (or similar conditions) makes the point infeasible.

Conclusion: A point not satisfying the non-negativity conditions is not a feasible solution.

Final Answer: (C) Infeasible

Step 1: The objective function in a Linear Programming Problem is generally written as:

$Z = ax + by$

where $a$ and $b$ are constants. The level lines (also called iso-profit or iso-cost lines) of this objective function are represented by:

$ax + by = k$

Step 2: All lines of the form $ax + by = k$ are parallel for different values of $k$. The direction of increase of $Z$ is perpendicular to these lines and depends on the sign of $a$ and $b$.

Since $a > 0$ and $b > 0$, moving away from the origin in the direction of increasing $x$ and $y$ leads to higher values of $Z$.

Conclusion: The given statement is generally true under the assumption that $a > 0$ and $b > 0$, and the direction of maximization is perpendicular to the objective function line and towards increasing values of $x$ and $y$.

Final Answer: (A) Yes

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

The constraint $x + y \ge 2$ means that the region is above the line $x + y = 2$. Since $x$ and $y$ can increase without bound while still satisfying the inequality $x + y \ge 2$, the feasible region extends infinitely.

Therefore, the feasible region is unbounded.

The correct option is (B).

In a Linear Programming Problem (LPP) aiming to maximize profit, the objective function is typically of the form: $Z = c_1x_1 + c_2x_2 + ... + c_nx_n$, where $x_i$ are the decision variables (e.g., quantity of each product to produce).

To maximize the overall profit $Z$, the coefficients $c_i$ must represent the profit contribution per unit of each product.

Therefore, the objective function should have coefficients that represent profit per unit of each product.

The correct option is (C).

Assertion (A): "Every feasible solution is a corner point feasible solution" is false. A feasible solution is any point within the feasible region, while a corner point feasible solution is specifically a vertex of the feasible region.

Reason (R): "Corner point feasible solutions are feasible solutions that lie at the vertices of the feasible region" is true. This is the definition of a corner point feasible solution.

Since A is false and R is true, the correct option is (D).

Therefore, the answer is (D).

The feasible region is a rectangle bounded by the lines $x = 0$, $y = 0$, $x = 5$, and $y = 4$.

The corner points of this rectangle are the intersections of these lines:

• $(0, 0)$
• $(5, 0)$
• $(0, 4)$
• $(5, 4)$

The point $(3, 2)$ lies within the rectangle but is not at a vertex.

Therefore, the point that is NOT a corner point is $(3, 2)$.

The correct option is (D).

If an objective function $Z = ax + by$ has the same maximum value at two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ within the feasible region, it implies that the line $ax + by = Z_{max}$ (where $Z_{max}$ is the maximum value) coincides with an edge of the feasible region.

In this case, every point on the line segment connecting $(x_1, y_1)$ and $(x_2, y_2)$ will also yield the same maximum value $Z_{max}$.

Therefore, the maximum value of Z is achieved at every point on the line segment connecting $(x_1, y_1)$ and $(x_2, y_2)$.

The correct option is (C).

In a balanced transportation problem, the total supply is equal to the total demand. This means that all available supply will be used to satisfy the demand, and all demand will be met by the available supply.

Therefore, the constraints in a balanced transportation problem are typically of the $=$ type, ensuring that the supply from each source equals the demand at each destination.

The correct option is (C).

The inequalities $x \ge 0$ and $y \ge 0$ restrict the feasible region to the first quadrant.

The inequality $x + y \ge 10$ means that the region is above the line $x + y = 10$.

Since $x$ and $y$ can increase without bound while still satisfying the inequality $x + y \ge 10$, the feasible region extends infinitely in the first quadrant, above the line $x+y=10$.

Therefore, the feasible region is unbounded.

The correct option is (B).

We need to check each point against all the constraints:

(A) $(5, 0)$:

• $x \ge 0$: $5 \ge 0$ (True)
• $y \ge 0$: $0 \ge 0$ (True)
• $2x + y \le 10$: $2(5) + 0 \le 10 \Rightarrow 10 \le 10$ (True)
• $x + 3y \le 12$: $5 + 3(0) \le 12 \Rightarrow 5 \le 12$ (True)

Therefore, $(5, 0)$ is a feasible solution.

(B) $(0, 4)$:

• $x \ge 0$: $0 \ge 0$ (True)
• $y \ge 0$: $4 \ge 0$ (True)
• $2x + y \le 10$: $2(0) + 4 \le 10 \Rightarrow 4 \le 10$ (True)
• $x + 3y \le 12$: $0 + 3(4) \le 12 \Rightarrow 12 \le 12$ (True)

Therefore, $(0, 4)$ is a feasible solution.

(C) $(3, 3)$:

• $x \ge 0$: $3 \ge 0$ (True)
• $y \ge 0$: $3 \ge 0$ (True)
• $2x + y \le 10$: $2(3) + 3 \le 10 \Rightarrow 9 \le 10$ (True)
• $x + 3y \le 12$: $3 + 3(3) \le 12 \Rightarrow 12 \le 12$ (True)

Therefore, $(3, 3)$ is a feasible solution.

(D) $(6, 0)$:

• $x \ge 0$: $6 \ge 0$ (True)
• $y \ge 0$: $0 \ge 0$ (True)
• $2x + y \le 10$: $2(6) + 0 \le 10 \Rightarrow 12 \le 10$ (False)
• $x + 3y \le 12$: $6 + 3(0) \le 12 \Rightarrow 6 \le 12$ (True)

Since $2x + y \le 10$ is false for $(6, 0)$, it is not a feasible solution.

Since the question asks for a feasible solution, and options A, B, and C are all feasible, there might be an error in the question. However, since option D is definitively NOT a feasible solution, we can assume it's looking for the one that *isn't* feasible. If we're looking for which one is NOT feasible, the answer is D.

The correct option is (D).

The primary purpose of formulating a Linear Programming Problem (LPP) mathematically is to take a real-world problem, which may be described in words and involve complex relationships, and transform it into a structured mathematical model.

This mathematical model consists of an objective function (to be maximized or minimized) and a set of constraints, all expressed as linear equations or inequalities. This allows us to use standard solution techniques, such as the Simplex method, to find the optimal solution.

Therefore, the primary purpose is to convert a real-world problem into a standard format that can be solved using known techniques.

The correct option is (B).

According to the fundamental theorem of linear programming, if an LPP has a bounded feasible region, the optimal solution (whether minimizing or maximizing the objective function) will always occur at a corner point (vertex) of the feasible region.

In this case, we want to minimize Z. Therefore, the optimal solution will occur at the corner point that gives the smallest value of Z.

Therefore, the optimal solution will occur at the corner point that gives the smallest value of Z.

The correct option is (B).

The equation $ax + by = c$, where $a$, $b$, and $c$ are constants, is the standard form of a linear equation in two variables, $x$ and $y$.

Graphically, a linear equation in two variables represents a straight line.

Therefore, a constraint of the form $ax + by = c$ graphically represents a straight line.

The correct option is (C).

Let's analyze each option:

• (A) Unique optimal solution: This is a common outcome where there's one specific point that maximizes or minimizes the objective function.
• (B) Multiple optimal solutions: This occurs when the objective function is parallel to an edge of the feasible region, leading to multiple points with the same optimal value.
• (C) Feasible solution exists, but no optimal solution (unbounded): This happens when the feasible region is unbounded, and the objective function can increase or decrease indefinitely without violating the constraints.
• (D) A non-linear optimal solution: Linear Programming Problems (LPPs) are defined by linear objective functions and linear constraints. Therefore, the solution *must* lie on the corner points formed by these linear functions. A non-linear solution is not possible within the framework of LPP.

Therefore, a non-linear optimal solution is NOT a possible outcome when solving an LPP.

The correct option is (D).

Assertion (A): "If a constraint is an equality ($ax + by = c$), the feasible region is restricted to the points lying exactly on that line segment within the intersection of other constraints" is true. The equality constraint forces all feasible solutions to lie on the line defined by the equation. The feasible region will then be the portion of that line that also satisfies all other constraints.

Reason (R): "An equality constraint defines a line, not a half-plane" is true. Equality constraints define lines, while inequality constraints define half-planes.

The reason R correctly explains why assertion A is true. Because an equality constraint defines a line, the feasible region is restricted to that line, making the assertion true.

Therefore, the answer is (A).

The correct option is (A).

Product P1 requires 1 hour on Machine A, and Product P2 requires 1 hour on Machine A.

Let $x$ be the number of units of P1 and $y$ be the number of units of P2.

The total time spent on Machine A is $1x + 1y = x + y$.

Machine A is available for a maximum of 10 hours, so the constraint is $x + y \le 10$.

Therefore, the constraint representing the time available on Machine A is $x + y \le 10>.

The correct option is (A).

Product P1 requires 2 hours on Machine B, and Product P2 requires 1 hour on Machine B.

Let $x$ be the number of units of P1 and $y$ be the number of units of P2.

The total time spent on Machine B is $2x + 1y = 2x + y$.

Machine B is available for a maximum of 15 hours, so the constraint is $2x + y \le 15$.

Therefore, the constraint representing the time available on Machine B is $2x + y \le 15$.

The correct option is (B).

The profit per unit of P1 is $\textsf{₹}300$, and the profit per unit of P2 is $\textsf{₹}400$.

Let $x$ be the number of units of P1 and $y$ be the number of units of P2.

The total profit is $300x + 400y$. We want to maximize this profit, so the objective function is Maximize $Z = 300x + 400y$.

Therefore, the objective function to maximize profit is Maximize $Z = 300x + 400y$.

The correct option is (B).

Since $x$ represents the number of units of product P1 and $y$ represents the number of units of product P2, these quantities cannot be negative.

Therefore, the number of units produced must be greater than or equal to zero. This is represented by the inequalities $x \ge 0$ and $y \ge 0$.

The non-negativity constraints are $x \ge 0, y \ge 0$.

The correct option is (B).

We need to check each point to see if it satisfies the constraints and is a corner point.

(A) $(10, 0)$:

• $x + y \le 10$: $10 + 0 \le 10 \Rightarrow 10 \le 10$ (True)
• $2x + y \le 15$: $2(10) + 0 \le 15 \Rightarrow 20 \le 15$ (False)
• $x \ge 0$: $10 \ge 0$ (True)
• $y \ge 0$: $0 \ge 0$ (True)

Since $(10, 0)$ does not satisfy all constraints, it's not a feasible solution, let alone a corner point.

(B) $(7.5, 0)$:

• $x + y \le 10$: $7.5 + 0 \le 10 \Rightarrow 7.5 \le 10$ (True)
• $2x + y \le 15$: $2(7.5) + 0 \le 15 \Rightarrow 15 \le 15$ (True)
• $x \ge 0$: $7.5 \ge 0$ (True)
• $y \ge 0$: $0 \ge 0$ (True)

$(7.5, 0)$ satisfies all constraints. This point is the intersection of $2x + y = 15$ and $y = 0$, thus it is a corner point.

(C) $(0, 15)$:

• $x + y \le 10$: $0 + 15 \le 10 \Rightarrow 15 \le 10$ (False)
• $2x + y \le 15$: $2(0) + 15 \le 15 \Rightarrow 15 \le 15$ (True)
• $x \ge 0$: $0 \ge 0$ (True)
• $y \ge 0$: $15 \ge 0$ (True)

Since $(0, 15)$ does not satisfy all constraints, it's not a feasible solution, let alone a corner point.

(D) $(5, 10)$:

• $x + y \le 10$: $5 + 10 \le 10 \Rightarrow 15 \le 10$ (False)
• $2x + y \le 15$: $2(5) + 10 \le 15 \Rightarrow 20 \le 15$ (False)
• $x \ge 0$: $5 \ge 0$ (True)
• $y \ge 0$: $10 \ge 0$ (True)

Since $(5, 10)$ does not satisfy all constraints, it's not a feasible solution, let alone a corner point.

Therefore, the correct option is (B).

We need to solve the system of equations:

Subtract equation (1) from equation (2):

Substitute $x = 5$ into equation (1):

The intersection point is $(5, 5)$.

Therefore, the correct option is (C).

We are given the objective function $Z = 300x + 400y$ and the corner point $(0, 0)$.

To evaluate the objective function at the corner point $(0, 0)$, we substitute $x = 0$ and $y = 0$ into the equation:

Therefore, the value of the objective function at $(0, 0)$ is 0.

The correct option is (C).

We are given the objective function $Z = 300x + 400y$ and the corner point $(7.5, 0)$.

To evaluate the objective function at the corner point $(7.5, 0)$, we substitute $x = 7.5$ and $y = 0$ into the equation:

Therefore, the value of the objective function at $(7.5, 0)$ is 2250.

The correct option is (A).

We are given the objective function $Z = 300x + 400y$ and the corner point $(0, 10)$.

To evaluate the objective function at the corner point $(0, 10)$, we substitute $x = 0$ and $y = 10$ into the equation:

Therefore, the value of the objective function at $(0, 10)$ is 4000.

The correct option is (C).

We are given the objective function $Z = 300x + 400y$ and the corner point $(5, 5)$.

To evaluate the objective function at the corner point $(5, 5)$, we substitute $x = 5$ and $y = 5$ into the equation:

Therefore, the value of the objective function at $(5, 5)$ is 3500.

The correct option is (B).

We have already evaluated the objective function $Z = 300x + 400y$ at the relevant corner points of the feasible region:

• $(0, 0)$: $Z = 0$
• $(7.5, 0)$: $Z = 2250$
• $(0, 10)$: $Z = 4000$
• $(5, 5)$: This point is inside the feasible region, so let's find the corner points instead.

However, $(5,5)$ is not a corner point. The corner points of the feasible region are (0,0), (7.5,0), (0,10), and (5,5).

To clarify, we need to find the corner points correctly. They are indeed:

• (0,0)
• (7.5,0) which gives a profit of 2250
• (0,10) which gives a profit of 4000

To find the 4th corner point, we must check the intersection of x+y=10 and 2x+y=15. Subtracting the first equation from the second yields x=5. Substituting x=5 into x+y=10 gives y=5. So the corner point is (5,5) which gives a profit of 3500.

Comparing these values, the maximum profit is $\textsf{₹}4000$.

Therefore, the correct option is (D).

From the previous question, we found that the maximum profit of $\textsf{₹}4000$ is achieved at the corner point $(0, 10)$.

This means that to maximize profit, the factory should produce 0 units of product P1 ($x = 0$) and 10 units of product P2 ($y = 10$).

Therefore, the optimal production plan is $x=0, y=10$.

The correct option is (D).

A feasible solution is any point that satisfies all the constraints of the LPP.

The question itself verifies that $(3, 4)$ satisfies all the constraints: $x+y \le 10$, $2x+y \le 15$, $x \ge 0$, and $y \ge 0$.

Therefore, $(3, 4)$ is a feasible solution.

The correct option is (A).

A feasible solution must satisfy *all* the constraints.

The question itself shows that the point $(8, 1)$ violates the constraint $2x + y \le 15$ since $2(8) + 1 = 17$, which is not less than or equal to 15.

Therefore, $(8, 1)$ is not a feasible solution.

The correct option is (B).

The constraint representing Machine A's availability is $x + y \le 10$.

At the point $(5, 5)$, the amount of time used on Machine A is $5 + 5 = 10$ hours.

The maximum time available on Machine A is 10 hours.

Therefore, Used: 10, Available: 10.

The correct option is (A).

The constraint representing Machine B's availability is $2x + y \le 15$.

At the point $(5, 5)$, the amount of time used on Machine B is $2(5) + 5 = 15$ hours.

The maximum time available on Machine B is 15 hours.

Therefore, Used: 15, Available: 15.

The correct option is (B).

When a constraint is satisfied exactly, it means that the solution lies directly on the line (for equality constraints) or on the boundary of the region defined by the inequality (for inequality constraints).

This type of constraint actively influences the feasible region and the optimal solution. It's called a binding constraint.

Therefore, when a constraint is satisfied exactly (the LHS equals the RHS), it is called a Binding Constraint.

The correct option is (A).

If a constraint doesn't affect the feasible region, it means that removing the constraint would not change the set of possible solutions. This implies that the constraint is unnecessary and doesn't play a role in determining the optimal solution.

Such a constraint is called a redundant constraint or a non-binding constraint.

The term redundant constraint is more appropriate here.

Therefore, a constraint that does not affect the feasible region is called a Redundant Constraint.

The correct option is (C).

Let's consider each option:

• (A) Production Mix Problem: Aims to determine the optimal combination of products to produce, usually to maximize profit or revenue, but sometimes also to minimize cost.
• (B) Diet Problem: Aims to determine the most cost-effective combination of foods to meet certain nutritional requirements, with the objective of minimizing the overall cost of the diet.
• (C) Resource Allocation Problem (to maximize output): Aims to allocate limited resources to different activities to maximize the total output, not necessarily minimizing cost.
• (D) Investment Problem (to maximize return): Aims to allocate capital to different investment opportunities to maximize the overall return, not minimizing cost.

Therefore, the LPP type that typically involves minimizing cost is the Diet Problem.

The correct option is (B).

Let's consider each option:

• (A) Empty: This occurs when the constraints are contradictory and there is no solution that satisfies all of them simultaneously.
• (B) Bounded: This occurs when the feasible region is limited in all directions, forming a closed shape.
• (C) Unbounded: This occurs when the feasible region extends infinitely in at least one direction.

All three of these scenarios are possible depending on the constraints of the LPP.

Therefore, the correct option is (D).

The correct option is (D).

When minimizing an objective function with an unbounded feasible region, there isn't always a guaranteed minimum value. If the objective function decreases indefinitely within the feasible region, the problem has an unbounded minimum.

However, if the constraints prevent the objective function from decreasing indefinitely (even though the feasible region is unbounded), there can still be a minimum value, which will occur at a corner point.

Therefore, the answer is: Yes, if the objective function doesn't decrease indefinitely in the feasible region.

The correct option is (B).

Assertion (A): "If an LPP has multiple optimal solutions, the objective function value is the same for all of them" is true. If there are multiple optimal solutions, they all must have the same objective function value since they are all *optimal*.

Reason (R): "Multiple optimal solutions mean that the objective function line coincides with one of the boundary lines of the feasible region" is true. This is the geometric interpretation of multiple optimal solutions.

Reason (R) is the correct explanation for assertion (A). The objective function line coinciding with a boundary line means every point on that boundary line segment has the same objective function value, and that value is optimal.

Therefore, the correct option is (A).

The correct option is (A).

The graphical method for solving LPPs involves the following steps:

1. Plot the constraints on a graph.
2. Identify the feasible region (the area satisfying all constraints).
3. Identify the corner points (vertices) of the feasible region.
4. Evaluate the objective function at each corner point.
5. Determine the optimal solution (the corner point that maximizes or minimizes the objective function).

Therefore, after plotting the constraints and identifying the feasible region, the next step is to identify the corner points of the feasible region.

The correct option is (C).

Non-negativity conditions $x \ge 0$ and $y \ge 0$ restrict the feasible region to where both $x$ and $y$ are non-negative.

This corresponds to the first quadrant of the Cartesian coordinate system.

Therefore, the feasible solution space is always in the first quadrant.

The correct option is (A).

For a point to be a corner point of the feasible region, it must satisfy two conditions:

1. It must be the intersection of two boundary lines (constraints).
2. It must satisfy all the constraints, meaning it lies within the feasible region.

Therefore, a point is a corner point if it is the intersection of two boundary lines and both (A) and (B) are true.

The correct option is (D).

Let's examine the requirements for a Linear Programming Problem (LPP):

• (A) The objective function must be linear. (True)
• (B) The constraints must be linear inequalities or equalities. (True)
• (C) The decision variables must be integers. (False) - This is a requirement for Integer Programming Problems (IPP), but not necessarily for LPPs. LPP decision variables can be real numbers.
• (D) Non-negativity constraints must apply to decision variables. (True - usually)

Therefore, the statement that is NOT a requirement for an LPP is "The decision variables must be integers".

The correct option is (C).

In a bounded feasible region, the optimal solution (maximum or minimum) will occur at a corner point (or along an edge connecting corner points).

If the maximum value of $Z=2x+2y$ is 10, then the line $2x+2y=10$ represents the objective function at its maximum value. This line must pass through at least one corner point to achieve the maximum value within the feasible region.

Therefore, the correct answer is (C).

The correct option is (C).

When minimizing with an unbounded feasible region, the key is whether the objective function can decrease indefinitely.

• (A) Objective function coefficients all positive: doesn't guarantee a minimum, depends on constraints.
• (B) Objective function coefficients all negative: doesn't guarantee a minimum, depends on constraints.
• (C) The objective function does not decrease indefinitely in the feasible direction: this *does* guarantee a minimum. If the function doesn't keep decreasing, there must be a lowest point.
• (D) The feasible region in the first quadrant: doesn't guarantee a minimum, depends on the objective function.

Therefore, the minimum value will exist if the objective function does not decrease indefinitely in the feasible direction.

The correct option is (C).

Let's analyze the graphical method and its advantages:

• The graphical method is primarily useful for problems with two decision variables because it allows us to visualize the constraints, the feasible region, and the movement of the objective function.
• It provides a clear, intuitive understanding of how the solution is obtained.
• The graphical method is not suitable for large problems with many variables because it becomes difficult to visualize the feasible region in higher dimensions.

Therefore, the main advantage of the graphical method is indeed the visual understanding of the problem and its solution space.

The correct option is (A).

The problem involves minimizing cost, so it's a minimization problem.

The requirement to produce "at least 100 units" translates to a constraint of the form "quantity produced $\ge$ 100". This is a "greater than or equal to" constraint.

Therefore, this is a minimization problem with a $\ge$ constraint.

The correct option is (D).

The feasible region represents the set of all points that satisfy all the constraints of the LPP simultaneously.

If the feasible region is empty, it means that there are no points that can satisfy all the constraints at the same time.

Therefore, the correct answer is: There is no point that satisfies all the given constraints simultaneously.

The correct option is (C).

The corner point theorem states that if an LPP has an optimal solution and a bounded feasible region, that solution will occur at a corner point of the feasible region.

The theorem *requires* the feasible region to be bounded. If the region is unbounded, there may not be an optimal solution, even if there are corner points.

Therefore, the corner point theorem is valid for only bounded feasible regions.

The correct option is (B).

The 'correct' side of the line is determined by the inequality sign in the constraint. If the constraint is $ax + by \le c$, then the feasible region lies on or below the line $ax + by = c$. If the constraint is $ax + by \ge c$, then the feasible region lies on or above the line $ax + by = c$.

The portion of the existing feasible region that satisfies the new constraint will remain part of the updated feasible region.

Therefore, the statement is correct.

The correct option is (A).

Let's analyze each option:

• (A) Diet Problem: Involves finding the most cost-effective combination of ingredients to meet nutritional requirements.
• (B) Transportation Problem: Involves finding the optimal way to transport goods from various sources to various destinations to minimize transportation costs.
• (C) Resource Allocation Problem: Involves allocating limited resources (like advertising budget) among different activities (print and online media) to maximize some objective (reach).
• (D) Assignment Problem: Involves assigning tasks to individuals to minimize cost or maximize efficiency.

Since the company is allocating its advertising budget (a limited resource) between different media (print and online) to maximize reach, this is an example of a Resource Allocation Problem.

The correct option is (C).

If the feasible region consists of only a single point, that point itself is considered a corner point (vertex) of the feasible region.

Therefore, the LPP has one corner point.

The correct option is (B).

The feasible region is a triangle bounded by the lines $x = 0$, $y = 0$, and $x + y = 5$. The corner points are where these lines intersect: (0,0), (5,0), and (0,5).

We are looking for a feasible solution that is *not* a corner point.

• (A) $(5,0)$ - is a corner point.
• (B) $(0,5)$ - is a corner point.
• (C) $(0,0)$ - is a corner point.
• (D) $(1,1)$ - Let's test it. $1 \ge 0$, $1 \ge 0$, and $1+1 = 2 \le 5$. So it is a feasible solution. Since it doesn't lie on the vertices, then it's not a corner point.

Therefore, the correct option is (D).

The correct option is (D).

The graphical method, used for solving linear programming problems with two variables, focuses on visually representing the problem.

It aims to:

1. Plot the constraints.
2. Identify the feasible region.
3. Determine the corner points of that region.

While the objective function is used to find the *optimal* solution *among* those points, the main focus of the graphical method itself is to understand the feasible region and its corner points.

Therefore, the primary goal is to identify the feasible region and its corner points.

The correct option is (B).

Assertion (A): "The Simplex method is used for solving LPPs with more than two variables" is true. The Simplex method is an algebraic procedure suitable for solving LPPs with any number of variables, especially when graphical methods become impractical or impossible.

Reason (R): "The graphical method is limited to problems with two variables" is true. The graphical method relies on visualizing the feasible region in a two-dimensional space, which is not possible for problems with more than two variables. We could do 3 variables by making it a 3d plot, but after that, it would be nearly impossible to visualise.

Reason (R) explains Assertion (A): since the graphical method is limited to 2 variables, another method is required to solve LPPs with more variables, which is the Simplex method.

Therefore, the correct option is (A).

Let's consider each option:

• (A) Portfolio selection: LPP can be used to determine the optimal mix of investments to maximize return while minimizing risk.
• (B) Production scheduling: LPP can be used to optimize production schedules to minimize cost or maximize profit, subject to resource constraints.
• (C) Determining customer preferences: This is more related to market research and statistical analysis, not directly solvable by LPP. While LPP can *use* information about customer demand, it doesn't *determine* preferences.
• (D) Optimal resource utilization: LPP is fundamentally about allocating limited resources to achieve the best possible outcome, such as maximizing profit or minimizing cost.

Therefore, the option that is NOT a common application of LPP is determining customer preferences.

The correct option is (C).

If the feasible region is unbounded and we are maximizing the objective function $Z = 2x + 3y$, where both coefficients are positive, the objective function can increase indefinitely as $x$ and $y$ increase within the feasible region. There is no upper bound.

While it *might* be at a corner point, that's only if the constraints happen to stop Z from increasing. But given the information that the region is unbounded, the function can increase without limit.

Therefore, the maximum value of Z is likely unbounded (infinite).

The correct option is (C).

A feasible solution is one that satisfies all the constraints of the LPP.

An optimal solution is a feasible solution that also maximizes (or minimizes) the objective function.

If a solution satisfies the constraints but not the objective function, it means it's a feasible solution, but there exists another feasible solution that yields a better value for the objective function. Thus it is not the *optimal* solution

Therefore, such a solution is feasible but not optimal.

The correct option is (A).

A redundant constraint is a constraint that does not affect the feasible region.

Therefore, if a constraint is redundant, its removal does not change the feasible region. This is the definition.

The correct option is (C).

Since $x+y \le 8$ is more restrictive than $x+y \le 10$, any solution that satisfies $x+y \le 8$ will also automatically satisfy $x+y \le 10$. Thus, $x+y \le 10$ doesn't actually restrict the feasible region any further than the other constraints already do.

Therefore, $x+y \le 10$ is a redundant constraint.

The correct option is (C).

The statement that the optimal solution of an LPP lies at a corner point of the feasible region (assuming the feasible region is bounded and an optimal solution exists) is a fundamental concept in the field of Linear Programming.

This is often referred to as the Corner Point Theorem or the Fundamental Theorem of Linear Programming.

Therefore, the correct option is (C).

The correct steps in solving an LPP using the graphical method are:

1. Formulate LPP: Define the objective function and constraints mathematically.
2. Plot constraints: Draw the lines or regions representing each constraint on a graph.
3. Identify feasible region: Determine the area that satisfies all the constraints simultaneously.
4. Identify corner points: Find the vertices of the feasible region.
5. Evaluate objective function at corner points: Calculate the value of the objective function at each corner point.

Therefore, the correct order is (A).

The correct option is (A).

Definition: In the context of a Linear Programming Problem (LPP), the feasible region is the set of all possible points that satisfy all the constraints of the problem simultaneously.

Each constraint in an LPP is typically represented by a linear inequality. When plotted on a graph (usually for problems with two variables), each inequality defines a half-plane.

The intersection of these half-planes represents the feasible region, which is often a polygonal area or an unbounded region depending on the nature of the constraints.

Key Characteristics:

• The feasible region contains all the feasible solutions of the LPP.
• Any point inside or on the boundary of the feasible region satisfies all the constraints.
• Optimal solutions (maximum or minimum values of the objective function) lie within or on the boundary of this region.

Note: If no point satisfies all the constraints simultaneously, then the feasible region is empty, and the LPP has no feasible solution.

Definition: In a Linear Programming Problem (LPP), the objective function is a linear function that is to be maximized or minimized under the given constraints.

It represents the goal of the problem — for example, maximizing profit, minimizing cost, or optimizing resources.

General form: An objective function is usually written as:

where $Z$ is the quantity to be optimized, $x$ and $y$ are decision variables, and $a$ and $b$ are constants representing coefficients.

Example:

Suppose a company makes two products A and B. The profit on each unit of A is $\textsf{₹}$50 and on B is $\textsf{₹}$30. If $x$ units of A and $y$ units of B are produced, then the objective function to maximize profit is:

Here, $Z$ is the total profit to be maximized.

Definition: In a Linear Programming Problem (LPP), constraints are the limitations or restrictions on the decision variables, typically expressed as linear inequalities or equations.

They define the conditions under which the objective function must be optimized (either maximized or minimized).

Mathematical Representation:

Constraints are usually written in the form of linear inequalities such as:

These inequalities limit the values that $x$ and $y$ can take.

Importance of Constraints:

• They define the feasible region, which contains all possible solutions to the problem.
• They ensure that the solution is practical and realistic within real-world limitations (e.g., availability of resources, time, cost, etc.).
• Without constraints, the optimization problem would be unbounded and not meaningful.

Definition: A feasible solution to a Linear Programming Problem (LPP) is a set of values of the decision variables that satisfy all the given constraints (including non-negativity conditions) simultaneously.

In other words, it is any point within or on the boundary of the feasible region.

Mathematical Example: Suppose the constraints of an LPP are:

The point $(2, 3)$ satisfies all the above inequalities, so it is a feasible solution.

Note: A feasible solution may or may not give the maximum or minimum value of the objective function. Only the best among all feasible solutions is called the optimal solution.

Definition: An optimal feasible solution to a Linear Programming Problem (LPP) is a solution that maximizes or minimizes the objective function, while still satisfying all the constraints of the problem.

This solution lies within the feasible region and gives the best possible value (either maximum or minimum) of the objective function.

Mathematical Representation:

If the objective function is:

Then the optimal feasible solution is the point $(x, y)$ in the feasible region that gives the highest or lowest value of $Z$, depending on whether we are maximizing or minimizing.

Note: The optimal solution always lies at a corner point (or on a line segment joining corner points) of the feasible region in a graphical method.

Definition: The non-negativity restrictions in a Linear Programming Problem (LPP) are conditions that require the decision variables to be greater than or equal to zero.

Mathematically, if $x$ and $y$ are decision variables, then the non-negativity constraints are:

Reason for Inclusion: These are included because, in real-world problems, quantities like production units, time, distance, profit, cost, etc., cannot be negative. Hence, non-negativity ensures the solution remains practical and meaningful.

Conclusion: Without these restrictions, the solution to the LPP might include negative values, which are not feasible in real-life scenarios.

Two essential characteristics that make a problem suitable for formulation as a Linear Programming Problem (LPP) are:

1. Linearity: The objective function and all the constraints must be linear functions of the decision variables. This means each term in the equations should be of the form $ax$, $by$, etc., and should not include powers, roots, or products of variables.

2. Objective and Constraints: There must be a clear objective — either to maximize or minimize a linear function — subject to a set of linear inequalities or equations which define the constraints.

Given Constraints:

• $x + y \leq 5$
• $x \geq 0$
• $y \geq 0$

Given Point: $(x, y) = (2, 3)$

Substituting in the first inequality:

Checking non-negativity:

• $x = 2 \geq 0$
• $y = 3 \geq 0$

Conclusion: Since the point $(2, 3)$ satisfies all three constraints, it is a feasible solution.

Given Constraints:

• $2x + y \geq 8$
• $x \geq 0$
• $y \geq 0$

Given Point: $(x, y) = (1, 5)$

Substituting into the first constraint:

Checking non-negativity:

• $x = 1 \geq 0$
• $y = 5 \geq 0$

Conclusion: Although the point satisfies the non-negativity constraints, it does not satisfy $2x + y \geq 8$. Hence, the point $(1, 5)$ is not a feasible solution.

Given Constraints:

• $x \leq 4$
• $y \leq 3$
• $x \geq 0$
• $y \geq 0$

Step-by-step construction of the feasible region:

These inequalities represent a region in the first quadrant bounded by vertical and horizontal lines:

• $x \leq 4$ represents a vertical line passing through $x = 4$; region is to the left.
• $y \leq 3$ represents a horizontal line passing through $y = 3$; region is below.
• $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

Graphical Representation:

The feasible region is the rectangular area bounded by the lines $x = 0$, $x = 4$, $y = 0$, and $y = 3$ in the first quadrant.

Corner Points of the Feasible Region:

• $(0, 0)$
• $(4, 0)$
• $(4, 3)$
• $(0, 3)$

Note: To actually draw this feasible region, plot the four lines on the coordinate plane and shade the region that satisfies all constraints — it will form a rectangle with the above corner points.

Given Inequality:

• $x + 2y \leq 6$
• $x \geq 0$
• $y \geq 0$

Step 1: Convert the inequality into equation

Consider the equation $x + 2y = 6$ to draw the boundary line.

Find two points on this line by substituting suitable values:

• Let $x = 0 \Rightarrow 2y = 6 \Rightarrow y = 3 \Rightarrow$ Point $(0, 3)$
• Let $y = 0 \Rightarrow x = 6 \Rightarrow$ Point $(6, 0)$

Step 2: Draw the line

Plot the line through points $(0, 3)$ and $(6, 0)$. Use a solid line as the inequality includes equality (≤).

Step 3: Identify the feasible region

Since the inequality is $x + 2y \leq 6$, shade the region below or on the line $x + 2y = 6$ and also within the first quadrant (i.e., where $x \geq 0$ and $y \geq 0$).

Corner Points of the Feasible Region:

• $(0, 0)$
• $(0, 3)$
• $(6, 0)$

This forms a triangular region in the first quadrant bounded by the axes and the line $x + 2y = 6$.

Definition:

In the context of a Linear Programming Problem (LPP), the feasible region is the set of all possible solutions that satisfy the given system of constraints.

Bounded Feasible Region:

A feasible region is said to be bounded if it is enclosed within a finite area. This means the feasible region lies within a closed polygon on the graph and all the points (solutions) are restricted within certain limits.

Example: Constraints like $x \ge 0$, $y \ge 0$, $x \le 4$, $y \le 3$ form a bounded region in the first quadrant.

Unbounded Feasible Region:

A feasible region is unbounded if it is not enclosed and extends infinitely in at least one direction. In such a case, there is no limit to the values of the variables in that direction.

Example: Constraints like $x \ge 0$, $y \ge 0$, $x + y \ge 5$ form an unbounded region extending infinitely outward from the line.

Key Difference:

• In a bounded region, the maximum and minimum values of the objective function always exist.
• In an unbounded region, the minimum value may exist, but the maximum value may not exist (or vice versa), depending on the direction of optimization.

Definition:

An optimization problem in the context of a Linear Programming Problem (LPP) refers to the task of maximizing or minimizing a linear objective function, subject to a set of linear inequalities or equations known as constraints.

Objective Function:

This is a linear function of the form $Z = ax + by$ (or similar), where the goal is to find the values of variables $x$ and $y$ that either:

• Maximize the value of $Z$ (e.g., profit, output), or
• Minimize the value of $Z$ (e.g., cost, distance)

Constraints:

These are the restrictions or conditions given in the form of linear inequalities, such as:

• $x + y \leq 5$
• $x \geq 0$, $y \geq 0$

The solution must lie within the feasible region defined by these constraints.

Conclusion:

Thus, an optimization problem in LPP aims to find the best possible outcome (maximum or minimum value of the objective function) within a feasible region that satisfies all the constraints.

Definition of Feasible Region:

A feasible region is the set of all points that satisfy all the constraints of a Linear Programming Problem (LPP). It represents all the possible solutions that are allowed by the given inequalities or equations.

Definition of Infeasible Region:

An infeasible region refers to a situation where there is no common solution that satisfies all the given constraints of the LPP simultaneously. In this case, the feasible region does not exist.

Key Differences:

Example:

• Constraints like $x + y \geq 5$ and $x + y \leq 3$ cannot be satisfied simultaneously.
• Hence, these form an infeasible system, and no feasible region is formed.

Meaning of Empty Feasible Region:

If the feasible region of a Linear Programming Problem (LPP) is empty, it means that there is no point that satisfies all the constraints of the problem simultaneously.

Implication:

This implies that the LPP has no solution. Such a problem is referred to as an infeasible problem.

Reason:

The constraints of the LPP are inconsistent or contradictory, so they do not intersect at any common region on the graph.

Example:

• Consider the constraints:
• $x + y \geq 6$
• $x + y \leq 4$
• No value of $x$ and $y$ can satisfy both inequalities at the same time.

Conclusion:

Therefore, if the feasible region is empty, the LPP has no feasible solution, and optimization (maximization or minimization) cannot be performed.

Two common types of Linear Programming Problems (LPPs) based on their application areas are:

1. Diet Problem:

This type of problem involves determining the most economical combination of different foods that will satisfy all nutritional requirements such as calories, proteins, vitamins, etc.

Application: Used in planning low-cost, nutritious diets in health science and military planning.

2. Transportation Problem:

This type deals with finding the most efficient way to transport goods from several suppliers to several consumers in such a way that the total transportation cost is minimized.

Application: Widely used in logistics, supply chain management, and distribution networks.

Understanding Non-Negativity Constraint:

In Linear Programming, a non-negativity constraint ensures that the decision variables cannot take negative values, as negative production of items is not meaningful in real-life situations.

Let:

• $x$ = number of units of item A produced
• $y$ = number of units of item B produced

Non-Negativity Constraints:

Conclusion:

Hence, the non-negativity constraint ensures that the number of items produced is zero or more.

Understanding the Objective Function:

In a Linear Programming Problem (LPP), the objective function is the expression that needs to be either maximized or minimized based on the problem context.

Here, the objective function is:

Meaning of the Expression:

• $x$ and $y$ are decision variables (e.g., number of units of two products).
• $3x$ and $4y$ represent the contribution of each unit of $x$ and $y$ respectively to the total value $Z$.
• $Z$ represents the total value of the objective function — usually profit, revenue, or utility.

Goal:

The goal is to maximize the value of $Z$ by choosing suitable values of $x$ and $y$ that satisfy all constraints of the LPP. This means finding the combination of $x$ and $y$ within the feasible region that gives the highest possible value of $Z$.

Definition:

In a Linear Programming Problem (LPP), decision variables are the variables that represent quantities to be determined for optimizing the objective function.

Role of Decision Variables:

• They represent the unknowns in the problem, such as the number of units to produce, amounts to allocate, etc.
• The values assigned to these variables determine the value of the objective function (e.g., profit or cost).
• They are subject to constraints defined in the problem (e.g., resource limitations, non-negativity, etc.).
• The optimal solution to the LPP is the set of values of these decision variables that maximize or minimize the objective function while satisfying all the constraints.

Example:

In an LPP with the objective function $Z = 5x + 3y$, the variables $x$ and $y$ are the decision variables. Finding their optimal values is the goal of solving the LPP.

Step 1: Consider the corresponding equation:

This represents a straight line which we will use to define the boundary of the inequality.

Step 2: Find the intercepts of the line:

• When $x = 0$: $4y = 12 \Rightarrow y = 3$
• When $y = 0$: $3x = 12 \Rightarrow x = 4$

So, the line passes through points $(0, 3)$ and $(4, 0)$.

Step 3: Draw the line:

Plot the line joining points $(0, 3)$ and $(4, 0)$. This line divides the plane into two regions.

Step 4: Determine the region satisfying the inequality:

Choose a test point not on the line, like $(0, 0)$:

Substitute into the inequality: $3(0) + 4(0) = 0$

Since $0 \not\ge 12$, the point does not satisfy the inequality.

So, the region that satisfies the inequality lies above the line $3x + 4y = 12$.

Step 5: Final Feasible Region:

Shade the region that lies above the line and in the first quadrant (i.e., $x \geq 0$, $y \geq 0$).

Note: The boundary line $3x + 4y = 12$ is included in the solution set since the inequality is $\geq$.

Example of a Resource Constraint:

Suppose a factory can allocate a total of at most 100 hours of labor. Producing one unit of product $x$ takes 4 hours, and one unit of product $y$ takes 5 hours. If $x$ and $y$ represent the number of units produced, the constraint can be expressed as:

Explanation:

This inequality limits the total amount of labor used by both products $x$ and $y$ to be no more than 100 hours. This type of inequality is typical in Linear Programming Problems where available resources like time, money, or materials are limited.

Definition:

The graphical method for solving a Linear Programming Problem (LPP) involves representing all constraints and the objective function on a graph to find the optimal solution visually.

Steps Involved:

1. Graph the system of inequalities (constraints) on the coordinate plane.
2. Identify the feasible region, which is the common region satisfying all the constraints.
3. Plot the objective function, usually in the form $Z = ax + by$.
4. Find the coordinates of all corner points (vertices) of the feasible region.
5. Evaluate the objective function at each corner point to find the maximum or minimum value, depending on the problem.

Applicability:

The graphical method is applicable only when there are two variables ($x$ and $y$). If an LPP involves more than two variables, other methods like the Simplex Method are used instead.

Given Constraints:

• $x + y \leq 4$
• $x \geq 0$
• $y \geq 0$

To Find: The vertices (corner points) of the feasible region formed by the above inequalities.

Step-by-step:

Let us find the points of intersection of the boundary lines of the constraints:

• Line 1: $x + y = 4$ (boundary of inequality)
• Line 2: $x = 0$ (y-axis)
• Line 3: $y = 0$ (x-axis)

Now find the intersections:

• Intersection of $x + y = 4$ and $x = 0$: Put $x = 0$ in $x + y = 4$ ⇒ $y = 4$ ⇒ Point $(0, 4)$
• Intersection of $x + y = 4$ and $y = 0$: Put $y = 0$ ⇒ $x = 4$ ⇒ Point $(4, 0)$
• Intersection of $x = 0$ and $y = 0$: Origin ⇒ Point $(0, 0)$

Vertices of the Feasible Region:

• $(0, 0)$
• $(0, 4)$
• $(4, 0)$

Hence, the feasible region is a triangle bounded by the coordinate axes and the line $x + y = 4$.

No, it is not possible for a Linear Programming Problem (LPP) to have exactly two optimal feasible solutions.

Explanation:

If an LPP has more than one optimal solution, it must have infinitely many optimal solutions.

This happens when the objective function line is parallel to one of the edges of the feasible region and overlaps it, resulting in the objective function having the same value at all points along that edge (which includes more than two points).

So, if the optimum occurs at two distinct corner points, then all the points on the line segment joining these two points will also yield the same optimal value, making the set of optimal solutions infinite.

Conclusion:

Exactly two optimal solutions cannot occur in LPP. The number of optimal solutions is either:

• One (unique corner point)
• Infinitely many (all points on a line segment in the feasible region)

Example:

A factory produces two types of products: Product A and Product B. The profit per unit of Product A is $\textsf{₹}$30 and for Product B is $\textsf{₹}$40.

Constraints:

• Each unit of Product A requires 2 hours of machine time and 3 kg of raw material.
• Each unit of Product B requires 1 hour of machine time and 2 kg of raw material.
• There are a total of 100 machine hours and 120 kg of raw material available.

Objective: Maximize total profit

Let $x$ be the number of units of Product A, and $y$ be the number of units of Product B.

Then the Linear Programming formulation will be:

Maximize $Z = 30x + 40y$

Subject to:

• $2x + y \leq 100$            (Machine hours constraint)
• $3x + 2y \leq 120$            (Raw material constraint)
• $x \geq 0,\; y \geq 0$            (Non-negativity constraint)

Thus, this production problem can be formulated and solved using Linear Programming methods.

Constraints in a Linear Programming Problem (LPP) are usually represented as linear inequalities because:

1. Representation of Real-world Limits:

They model real-life limitations on resources such as time, labor, material, or budget. For example, if a factory has a maximum of 100 machine hours, this can be expressed as a linear inequality like:

2. Simplicity in Mathematical Modeling:

Linear inequalities allow for simple graphical and algebraic solutions, making it easier to identify the feasible region (a polygon or polyhedron formed by such constraints).

3. Definition of Feasible Region:

Linear inequalities divide the coordinate plane into half-planes and their intersection forms a feasible region where the solution must lie.

4. Optimization is Easier:

With linear constraints, optimization of a linear objective function becomes mathematically manageable using methods like graphical method (for two variables) or Simplex method.

Conclusion:

Linear inequalities are used in LPP to realistically and effectively model constraints while keeping the problem solvable using linear methods.

We are given the following constraints:

• $x + y \geq 6$
• $x \geq 0$
• $y \geq 0$

We need to check which of the points satisfy all three constraints.

1. Point (2, 2):

• $x + y = 2 + 2 = 4$ ⟶ Does not satisfy $x + y \geq 6$ ❌
• $x = 2 \geq 0$, $y = 2 \geq 0$ ⟶ Satisfies ✅

Hence, (2, 2) is not a feasible solution.

2. Point (3, 4):

• $x + y = 3 + 4 = 7$ ⟶ Satisfies $x + y \geq 6$ ✅
• $x = 3 \geq 0$, $y = 4 \geq 0$ ⟶ Satisfies ✅

Hence, (3, 4) is a feasible solution.

3. Point (5, 0):

• $x + y = 5 + 0 = 5$ ⟶ Does not satisfy $x + y \geq 6$ ❌
• $x = 5 \geq 0$, $y = 0 \geq 0$ ⟶ Satisfies ✅

Hence, (5, 0) is not a feasible solution.

Final Answer: Only point (3, 4) is a feasible solution.

The vertices (or corner points) of the feasible region in a Linear Programming Problem (LPP) play a crucial role in determining the optimal solution.

Key Significance:

1. Optimal Value Lies at a Vertex:

If there is an optimal (maximum or minimum) value of the objective function, it occurs at one of the vertices (or corner points) of the feasible region formed by the constraints.

2. Finite Number of Evaluations:

Instead of testing all the infinite points in the feasible region, we only need to evaluate the objective function at the finite number of corner points (vertices).

3. Multiple Optimal Solutions:

If the objective function is parallel to a constraint boundary segment, the optimum may occur at multiple vertices lying on that segment.

Conclusion:

The vertices of the feasible region are the only candidates where the maximum or minimum value of the objective function in an LPP can occur.

Yes, a Linear Programming Problem (LPP) can have an unbounded feasible region and still have a bounded optimal solution.

Explanation with Example:

Consider the following LPP:

Maximize $Z = 3x + 2y$

Subject to:

• $x \geq 0$
• $y \geq 0$
• $x + y \geq 5$
• $x \leq 6$

Observation:

• The feasible region is unbounded towards the top (large values of $y$).
• However, due to the constraint $x \leq 6$, $x$ is bounded.
• The objective function $Z = 3x + 2y$ will have a maximum value at one of the corner points of this feasible region, say at $(x, y) = (6, y_0)$ satisfying all constraints.

Conclusion:

Even if the feasible region is unbounded, the objective function can attain a bounded optimal value as long as it does not increase indefinitely in the unbounded direction.

Let:

$x$ = Number of pens of type A produced

$y$ = Number of pens of type B produced

Given:

• Time available on Machine 1 = 6 hours = 360 minutes
• Time available on Machine 2 = 5 hours = 300 minutes
• Type A requires:
  • 2 minutes on Machine 1
  • 3 minutes on Machine 2
• Type B requires:
  • 3 minutes on Machine 1
  • 2 minutes on Machine 2

Constraints:

Conclusion:

The constraints for the LPP are:

• $2x + 3y \leq 360$
• $3x + 2y \leq 300$
• $x \geq 0,\;\; y \geq 0$

A feasible region in a Linear Programming Problem (LPP) is said to be convex if for any two points within the region, the entire line segment joining them also lies entirely within the region.

Mathematically: If $P$ and $Q$ are any two points in the feasible region, then every point $R$ on the line segment joining $P$ and $Q$, given by:

also lies within the feasible region.

Conclusion: In linear programming, the feasible region formed by a system of linear inequalities is always convex. This property ensures that if an optimal solution exists, it lies at one of the corner (extreme) points of the feasible region.

Given Constraints:

• $x \geq 2$
• $y \geq 1$
• $x \geq 0$
• $y \geq 0$

Graphical Interpretation:

We are required to graph the inequalities in the coordinate plane and shade the region that satisfies all the given conditions.

• The line $x = 2$ is a vertical line. The region to the right of this line satisfies $x \geq 2$.
• The line $y = 1$ is a horizontal line. The region above this line satisfies $y \geq 1$.
• $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

Feasible Region:

The feasible region is the part of the first quadrant that lies to the right of the line $x = 2$ and above the line $y = 1$. This region is unbounded and lies in the intersection of all the above conditions.

Graphical Sketch:

A rough sketch would look like this:

• Draw the vertical line $x = 2$ and shade the region to the right.
• Draw the horizontal line $y = 1$ and shade the region above.
• The overlapping region in the first quadrant bounded below by $y = 1$ and to the left by $x = 2$ is the required feasible region.

Objective: The aim is to minimize the cost using a linear objective function within a set of constraints forming a feasible region.

Interpretation: In a Linear Programming Problem (LPP), if the goal is to minimize the cost function, then we must search for the minimum value of the objective function $Z$ over the entire feasible region.

The objective function is generally written as:

To minimize $Z$, we evaluate its value at all corner points of the feasible region and select the one with the lowest value.

Conclusion: Since the goal is to minimize cost, we look for the minimum value of the objective function within the feasible region.

Given Constraints:

• $x - y \leq 1$
• $x \geq 0$
• $y \geq 0$

Step 1: Draw the line $x - y = 1$

This is the boundary line for the inequality. To plot it, find two points:

• When $x = 0$, $0 - y = 1 \Rightarrow y = -1$
• When $y = 0$, $x - 0 = 1 \Rightarrow x = 1$

So the line passes through $(0, -1)$ and $(1, 0)$. But since we are concerned with the first quadrant ($x \geq 0, y \geq 0$), we will only consider the region starting from the point $(1, 0)$ and moving upwards and rightward where the inequality is satisfied.

Step 2: Determine the region for $x - y \leq 1$

We test a point, say $(0, 0)$:

Substitute into the inequality: $0 - 0 \leq 1 \Rightarrow 0 \leq 1$ → True

So the region below or on the line $x - y = 1$ (i.e., $x - y \leq 1$) is the valid region.

Step 3: Include non-negativity conditions

The feasible region also lies in the first quadrant due to $x \geq 0$ and $y \geq 0$. Hence, we only shade the part of the region that satisfies all three conditions:

• Lies below or on the line $x - y = 1$
• Lies to the right of the $y$-axis ($x \geq 0$)
• Lies above the $x$-axis ($y \geq 0$)

Conclusion: The required region is the portion of the first quadrant lying below the line $x - y = 1$ (including the line itself).

Fundamental Theorem of Linear Programming:

The fundamental theorem of Linear Programming states that:

If there exists an optimal solution to a linear programming problem (LPP), then at least one optimal solution occurs at a corner point (also called a vertex) of the feasible region.

Explanation:

The feasible region in an LPP is a convex polygon (or polyhedron in higher dimensions) formed by the intersection of linear inequalities. The objective function is a linear expression, typically written as:

The value of $Z$ changes linearly as we move within the feasible region. Because the function is linear, its maximum or minimum value will always occur at one of the vertices (corner points) of the feasible region, not in the interior.

Conclusion: To find the optimal solution (maximum or minimum value of $Z$), we only need to evaluate $Z$ at the corner points of the feasible region. This greatly simplifies the process of solving LPPs.

In a diet problem in Linear Programming:

Decision Variables:

The decision variables typically represent the quantity (in units such as kg, grams, or servings) of different foods to be included in the diet. For example, if there are two foods involved:

• Let $x$ = quantity of food X (e.g., in kg)
• Let $y$ = quantity of food Y (e.g., in kg)

Objective Function:

The goal is to minimize the total cost of the selected foods while meeting nutritional constraints. If the cost per unit of food X is $a$ and the cost per unit of food Y is $b$, then the objective function becomes:

Conclusion: In such problems:

• Decision variables represent the quantities of different food items.
• Objective function represents the total cost to be minimized, written as a linear combination of the decision variables.

To find the vertices (corner points) of a polygon-shaped feasible region in Linear Programming, follow these steps:

Step 1: Graph all the constraints

Plot each inequality as an equation on the coordinate plane (e.g., $x + y = 4$). Use solid lines for $\le$ or $\ge$ and dashed lines for $<$ or $>$ (if needed).

Step 2: Identify the feasible region

The feasible region is the common area that satisfies all the inequalities. It will be a convex polygon (if bounded).

Step 3: Find the points of intersection of the boundary lines

To determine the vertices, solve the equations of two boundary lines at a time. These intersection points form the vertices of the feasible region.

For example, solving:

Solving them simultaneously gives a vertex of the feasible region.

Step 4: Verify if each intersection point lies within the feasible region

Check that each point satisfies all original inequalities. Only those points are considered valid vertices.

Conclusion: The vertices of the feasible region are obtained by solving pairs of constraint equations and selecting only those points that lie in the common shaded region.

Definition:

An infeasible solution in a Linear Programming Problem (LPP) is a solution that does not satisfy one or more of the given constraints.

Given Constraints:

• $x + y \leq 5$
• $x \geq 0$
• $y \geq 0$

Example of an infeasible solution:

Consider the point $(4, 3)$.

Check the constraints:

• $x + y = 4 + 3 = 7$ ⟶ violates $x + y \leq 5$
• $x = 4 \geq 0$ ⟶ satisfies
• $y = 3 \geq 0$ ⟶ satisfies

Since $(4, 3)$ does not satisfy all constraints, it is an infeasible solution.

Conclusion: An infeasible solution lies outside the feasible region and does not fulfill all the constraints of the problem.

Step 1: Convert inequalities into equations to plot the boundary lines.

• $x + y = 3$
• $x + 2y = 6$
• $x = 0$ (y-axis)
• $y = 0$ (x-axis)

Step 2: Find the intercepts of each line to help draw them.

For $x + y = 3$:

• When $x = 0$, $y = 3$ ⟶ point (0, 3)
• When $y = 0$, $x = 3$ ⟶ point (3, 0)

For $x + 2y = 6$:

• When $x = 0$, $2y = 6 \Rightarrow y = 3$ ⟶ point (0, 3)
• When $y = 0$, $x = 6$ ⟶ point (6, 0)

Step 3: Draw the region satisfying all the constraints.

• Shade the region above or on the line $x + y = 3$.
• Shade the region below or on the line $x + 2y = 6$.
• Restrict the region to the first quadrant where $x \geq 0$ and $y \geq 0$.

Step 4: Identify the points of intersection (vertices).

• Intersection of $x + y = 3$ and $x + 2y = 6$:

Solving simultaneously:

Subtract (i) from (ii):

$(x + 2y) - (x + y) = 6 - 3 \Rightarrow y = 3$

Put $y = 3$ in (i): $x + 3 = 3 \Rightarrow x = 0$

So, point of intersection is $(0, 3)$

Other relevant points (from intercepts and axes):

• $(3, 0)$ ⟶ on $x + y = 3$
• $(6, 0)$ ⟶ on $x + 2y = 6$

Conclusion: The feasible region is bounded by the lines $x + y = 3$, $x + 2y = 6$, $x = 0$, and $y = 0$. It lies in the first quadrant and includes the points $(0, 3)$, $(3, 0)$, and $(6, 0)$ as boundary intersections.

Explanation:

If the objective function line is parallel to one of the boundary lines of the feasible region, then there are two possibilities for the optimal solution depending on whether the boundary line is part of the feasible region or not.

Case 1: The boundary line is a part of the feasible region

In this case, the objective function will attain the same optimal value at every point along the line segment of intersection between the feasible region and the boundary line.

Hence, there are infinitely many optimal solutions.

Case 2: The boundary line is not part of the feasible region

Then the objective function will touch the feasible region at only one corner point (vertex), and hence the optimal solution will be unique.

Conclusion: If the objective function is parallel to a boundary line of the feasible region and that boundary is part of the feasible region, then the LPP has infinitely many optimal solutions along that boundary segment.

Let:

• $x_1$ = number of units of product P1 produced
• $x_2$ = number of units of product P2 produced

Labour Constraint:

Each unit of P1 requires 4 hours and each unit of P2 requires 3 hours of labour. The total available labour hours are 120.

Raw Material Constraint:

Each unit of P1 requires 2 kg and each unit of P2 requires 3 kg of raw material. The total available raw material is 150 kg.

Also, the number of units cannot be negative:

In the context of a Linear Programming Problem (LPP):

1. Unique Optimal Solution:

An LPP has a unique optimal solution when the objective function reaches its maximum or minimum value at exactly one point in the feasible region, typically at a corner (vertex) of the feasible region.

This occurs when the objective function line intersects the feasible region at a single vertex only.

Graphically: The objective function line is not parallel to any boundary of the feasible region and touches only one corner point.

2. Multiple Optimal Solutions:

An LPP has multiple optimal solutions when the objective function has the same optimal value at more than one point within the feasible region.

This occurs when the objective function line is parallel to a boundary line (edge) of the feasible region and lies along it for some segment.

Hence, every point on that segment provides the same optimal value.

Summary Table:

Given Inequality: $y \geq x - 1$

Also given: $x \geq 0$, $y \geq 0$

Step 1: Draw the line $y = x - 1$

To plot this line, we find two points:

• When $x = 1$, $y = 0$
• When $x = 2$, $y = 1$

Plot these points and draw a straight line through them.

This line passes through the point $(1, 0)$ and $(2, 1)$.

Step 2: Use solid line

Since the inequality includes equality (≥), draw the line as a solid line to show points on the line are included in the solution set.

Step 3: Shade the solution region

The inequality is $y \geq x - 1$, so we shade the region above the line $y = x - 1$.

Also, restrict the shaded region to the first quadrant only since $x \geq 0$ and $y \geq 0$.

Final Region:

The required solution region lies in the first quadrant and includes all points on or above the line $y = x - 1$.

Corner Point Method: The corner point method is a graphical technique used to find the optimal solution (maximum or minimum value) of a Linear Programming Problem (LPP).

Role and Steps:

1. Plot the Constraints: Graph the system of linear inequalities (constraints) on the Cartesian plane to form a feasible region.

2. Identify the Feasible Region: The feasible region is the common region that satisfies all the given constraints.

3. Locate the Corner Points: Determine the coordinates of all the vertices (corner points) of the feasible region.

4. Evaluate the Objective Function: Substitute the coordinates of each corner point into the objective function (e.g., $Z = ax + by$) to calculate its value.

5. Select the Optimum Value: Choose the point that gives the maximum or minimum value of the objective function, depending on whether the goal is to maximize or minimize $Z$.

Conclusion: The corner point method works because the optimal value of a linear objective function in a convex feasible region always occurs at a vertex (corner point) of the region.

Given Constraints:

• $x \leq 5$
• $y \geq 2$
• $x \geq 0$
• $y \geq 0$

Step 1: Interpret each constraint graphically

• $x \leq 5$ is a vertical line passing through $x = 5$ and includes all points to the left of it.
• $y \geq 2$ is a horizontal line passing through $y = 2$ and includes all points above it.
• $x \geq 0$ and $y \geq 0$ imply that the solution lies in the first quadrant.

Step 2: Determine the Feasible Region

The feasible region is the overlapping area that satisfies all the given constraints. This is the region:

• to the left of the line $x = 5$
• above the line $y = 2$
• in the first quadrant ($x \geq 0, y \geq 0$)

Step 3: Find the Vertices (Corner Points)

The region is bounded by the lines $x = 0$, $x = 5$, $y = 2$, and the $x$-axis and $y$-axis.

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

• Point A: $(0, 2)$ → intersection of $x = 0$ and $y = 2$
• Point B: $(5, 2)$ → intersection of $x = 5$ and $y = 2$
• Point C: $(5, \infty)$ → extends infinitely upward
• Point D: $(0, \infty)$ → extends infinitely upward

Hence, the region is unbounded above in the vertical direction but has vertices at: $(0, 2)$ and $(5, 2)$.

Yes, an LPP (Linear Programming Problem) with an unbounded feasible region can still have an optimal solution for a maximization problem, but not always.

Explanation:

An unbounded feasible region means that the region defined by the constraints extends infinitely in at least one direction.

However, even if the feasible region is unbounded, the objective function (e.g., $Z = ax + by$) may still attain a maximum value at a specific corner point within that region.

Two cases arise:

• Case 1: If the objective function increases without bound in the direction of the unbounded region, then there is no maximum value.
• Case 2: If the objective function does not increase in the direction of the unbounded region, the maximum occurs at a corner point, and the LPP has an optimal solution.

Conclusion: An unbounded feasible region may have an optimal solution for a maximization problem, but it is not guaranteed. One must analyze the behavior of the objective function to confirm whether a maximum exists.

Example:

A company has three factories located in different cities that manufacture goods, and it needs to transport these goods to four warehouses in other locations. Each factory has a certain supply capacity, and each warehouse has a specific demand requirement.

The goal is to determine the quantity of goods to be transported from each factory to each warehouse so that the total transportation cost is minimized, while satisfying the supply and demand constraints.

This is a classic example of a transportation problem in linear programming.

Step 1: Understand the inequalities

• $x + y \geq 2$  →  Region on or above the line $x + y = 2$
• $x + y \leq 5$  →  Region on or below the line $x + y = 5$
• $x \geq 0$  →  Region to the right of $y$-axis
• $y \geq 0$  →  Region above the $x$-axis

Step 2: Sketching the region

The region bounded by the lines $x + y = 2$ and $x + y = 5$ in the first quadrant (where both $x \geq 0$ and $y \geq 0$) forms a trapezium-like shape. The region lies between the two lines and above the $x$-axis and to the right of the $y$-axis.

Step 3: Is this a feasible region of a typical LPP?

Yes, this is a feasible region of a typical Linear Programming Problem (LPP) because:

• It is a closed and bounded region (defined by inequalities including $\geq$ and $\leq$).
• It lies in the first quadrant, which is typical for LPPs as variables often represent non-negative quantities.
• The feasible region contains corner points (vertices) where the optimal value of the objective function (if provided) can occur.

Conclusion: The shaded region represents a valid feasible region for an LPP as it satisfies all necessary conditions: bounded, closed, and located in the first quadrant.

Purpose of Shading the Region in Graphical Method of LPP:

In the graphical method of solving Linear Programming Problems (LPPs), shading the region helps to visually identify the feasible region, which represents all possible solutions that satisfy all the given constraints simultaneously.

Specifically, shading helps to:

• Locate the feasible region: The common overlapped shaded region from all inequalities represents feasible solutions.
• Identify boundary lines: The lines where constraints change from inequality to equality (e.g., $x + y = 5$ from $x + y \leq 5$) help in locating vertices.
• Determine corner points: These points (vertices of the shaded region) are essential for evaluating the objective function.
• Visualize unbounded or bounded regions: Shading helps see whether the feasible region is bounded (closed) or unbounded (open), which affects the existence of an optimal solution.

Conclusion: Shading is crucial to graphically represent and analyze the feasible solution space in order to find the optimal solution of an LPP.

No, the optimal solution of a Linear Programming Problem (LPP) cannot lie on the boundary without being at a vertex, unless there are multiple optimal solutions.

Explanation:

According to the fundamental theorem of Linear Programming, if an optimal solution exists for a linear programming problem, then it occurs at a vertex (corner point) of the feasible region.

• If there is a unique optimal solution, it will be found at a single vertex of the feasible region.
• If there are multiple optimal solutions, the objective function will have the same optimal value at two or more vertices. In such a case, every point on the line segment joining those vertices (which lies on the boundary) will also be an optimal solution.

Conclusion: An optimal solution may lie on the boundary but only if it is at a vertex or lies on the line segment between two optimal vertices. In such cases, there are infinitely many optimal solutions, all of which lie on that boundary segment.

Using Iso-profit/Iso-cost Lines to Find the Optimal Solution:

To solve an LPP graphically using the objective function $Z = 2x - 3y$, we use the concept of iso-profit lines (for maximization) or iso-cost lines (for minimization).

Steps Involved:

• Step 1: Plot all constraints on the graph and identify the feasible region (shaded area that satisfies all constraints).
• Step 2: Choose an arbitrary value of $Z$ (say $Z = 0$ or any constant) and write the objective function as an equation:

  $Z = 2x - 3y$

  (Objective Function)

  For example, set $Z = 0$, then:

  $2x - 3y = 0$

  (Line for initial iso-profit)

  Plot this line on the graph.
• Step 3: Draw several iso-profit lines (by assigning different values of $Z$ such as $Z = 3, 6, 9, \dots$), which are parallel lines of the objective function.
• Step 4: Move the iso-profit line parallelly towards the direction of increasing (or decreasing) $Z$ value depending on whether the goal is to maximize or minimize $Z$.
• Step 5: The last point of contact of the iso-profit line with the feasible region (before leaving it) gives the optimal solution.

Conclusion: The iso-profit/iso-cost method helps in graphically determining the maximum or minimum value of the objective function by sliding a line parallel to itself until it touches the optimal point in the feasible region.

Let:

$x$ = number of units of Toy A produced per week

$y$ = number of units of Toy B produced per week

Objective:

To maximize total profit $Z$.

Constraints:

• Assembly time constraint: Each unit of A takes 3 hours and each unit of B takes 2 hours. Total assembly time ≤ 120 hours:

  $3x + 2y \leq 120$

  (Assembly time)

• Finishing time constraint: Each unit of both A and B takes 1 hour. Total finishing time ≤ 50 hours:

  $x + y \leq 50$

  (Finishing time)

• Non-negativity constraints:

  $x \geq 0, \; y \geq 0$

  (Cannot produce negative units)

Linear Programming Formulation:

Maximize $Z = 50x + 40y$

Subject to:

$3x + 2y \leq 120$

$x + y \leq 50$

$x \geq 0, \; y \geq 0$

Step 1: Plot the inequalities on a graph.

Convert inequalities to equations for plotting boundary lines:

Step 2: Identify the feasible region.

The feasible region lies in the first quadrant and is bounded by the lines defined in equations (i), (ii), (iii), and (iv). Shade the region satisfying all inequalities.

Step 3: Find the corner points of the feasible region.

We find points of intersection of the lines:

• Intersection of (i) and (ii):

From (ii), $y = 15 - x$

Substitute in (i): $2x + (15 - x) = 24 \Rightarrow x = 9$, $y = 6$ ⇒ Point $(9, 6)$

• Intersection of (ii) and (iii):

Put $x = 2$ in (ii): $2 + y = 15 \Rightarrow y = 13$ ⇒ Point $(2, 13)$

• Intersection of (i) and (iii):

Put $x = 2$ in (i): $2(2) + y = 24 \Rightarrow y = 20$ ⇒ Point $(2, 20)$

• Intersection of (ii) and $y = 0$:

Put $y = 0$ in (ii): $x = 15$ ⇒ Point $(15, 0)$

• Intersection of (i) and $y = 0$:

Put $y = 0$ in (i): $2x = 24 \Rightarrow x = 12$ ⇒ Point $(12, 0)$

• Intersection of (iii) and (iv):

$x = 2$, $y = 0$ ⇒ Point $(2, 0)$

Feasible corner points: $(2, 0)$, $(2, 13)$, $(9, 6)$, $(12, 0)$

Step 4: Evaluate objective function $Z = 6x + 4y$ at each corner point:

Step 5: Conclusion

The maximum value of $Z = 78$ occurs at point $(9, 6)$.

Optimal solution: $x = 9,\; y = 6,\; Z = 78$

Let:

$x$ = number of sponge cakes made

$y$ = number of chocolate cakes made

Objective:

To maximize the total profit $Z$

Constraints:

• Flour constraint: 1 sponge cake uses 200g, and 1 chocolate cake uses 100g. Total flour available is 5kg = 5000g

  $200x + 100y \leq 5000$

  (Flour constraint)

  Simplifying by dividing by 100:

  $2x + y \leq 50$

  …(i)

• Fat constraint: 1 sponge cake uses 25g, and 1 chocolate cake uses 50g. Total fat available is 1kg = 1000g

  $25x + 50y \leq 1000$

  (Fat constraint)

  Simplifying by dividing by 25:

  $x + 2y \leq 40$

  …(ii)

• Non-negativity constraints:

  $x \geq 0, \; y \geq 0$

  (Cannot make negative cakes)

Linear Programming Formulation:

Maximize $Z = 10x + 15y$

Subject to:

$2x + y \leq 50$

$x + 2y \leq 40$

$x \geq 0$, $y \geq 0$

Graphical Representation:

To graph the feasible region, draw the lines for the constraints:

• $2x + y = 50$
• $x + 2y = 40$
• $x = 0$ (y-axis)
• $y = 0$ (x-axis)

The feasible region is the area bounded by these lines in the first quadrant, representing all possible combinations of $x$ and $y$ that satisfy the constraints.

Note: To solve graphically, plot the lines, identify corner points of the feasible region, and evaluate $Z = 10x + 15y$ at those points to determine the maximum profit.

Step 1: Convert the inequalities to equations to plot the boundary lines.

Also include: $x \geq 0$ and $y \geq 0$ (first quadrant)

Step 2: Plot the lines on the graph.

• For $3x + y = 9$:

  When $x = 0$, $y = 9$

  When $y = 0$, $x = 3$

• For $x + y = 5$:

  When $x = 0$, $y = 5$

  When $y = 0$, $x = 5$

Shade the region satisfying the inequalities $3x + y \geq 9$ and $x + y \geq 5$ in the first quadrant.

Step 3: Identify the feasible region.

The feasible region lies above both lines and in the first quadrant. This region is unbounded in the upward direction.

Step 4: Find the corner points of the feasible region.

• Intersection of lines (i) and (ii):

From (ii), $y = 5 - x$

Substitute into (i): $3x + (5 - x) = 9 \Rightarrow 2x = 4 \Rightarrow x = 2$, $y = 3$

So, intersection point is $(2, 3)$

• Other vertices on axes do not lie in the feasible region due to the $\geq$ constraints.

Only corner point in feasible region: $(2, 3)$

Step 5: Evaluate $Z = 10x + 7y$ at the corner point.

At $(2, 3)$, $Z = 10(2) + 7(3) = 20 + 21 = 41$

Step 6: Conclusion

The feasible region is unbounded and lies above the lines. In an unbounded region, a minimum may or may not exist.

However, since the objective function is linear and increases as we move away from the origin, and the feasible region extends infinitely in that direction, no smaller value of $Z$ exists within the region than at $(2, 3)$.

Thus, the minimum value of $Z = 41$ exists at point $(2, 3)$.

Let:

$x$ = land in square meters allocated for potatoes

$y$ = land in square meters allocated for onions

Objective:

To maximize total profit $Z$

Constraints:

• Land constraint:

Potatoes require 4 m² per unit, so land used for potatoes = $4x$

Onions require 3 m² per unit, so land used for onions = $3y$

Total land available = 100 m²

$4x + 3y \leq 100$

…(i)

• Fertilizer constraint:

Potatoes require 2 kg per unit, so fertilizer used for potatoes = $2x$

Onions require 3 kg per unit, so fertilizer used for onions = $3y$

Total fertilizer available = 90 kg

$2x + 3y \leq 90$

…(ii)

• Non-negativity constraints:

$x \geq 0, \; y \geq 0$

(Cannot allocate negative land)

Linear Programming Problem (LPP):

Maximize $Z = 50x + 60y$

Subject to:

$4x + 3y \leq 100$

$2x + 3y \leq 90$

$x \geq 0$, $y \geq 0$

Graphical Solution:

Convert constraints into equations to draw the boundary lines:

• For (i): $4x + 3y = 100$

When $x = 0 \Rightarrow y = \frac{100}{3} \approx 33.33$

When $y = 0 \Rightarrow x = 25$

• For (ii): $2x + 3y = 90$

When $x = 0 \Rightarrow y = 30$

When $y = 0 \Rightarrow x = 45$

Plot the lines and shade the feasible region bounded by these constraints in the first quadrant.

Corner Points of Feasible Region:

• Intersection of $4x + 3y = 100$ and $2x + 3y = 90$:

  Subtracting the second from the first:

  $(4x + 3y) - (2x + 3y) = 100 - 90 \Rightarrow 2x = 10 \Rightarrow x = 5$

  Substitute into (ii): $2(5) + 3y = 90 \Rightarrow 10 + 3y = 90 \Rightarrow y = 26.67$

  Point: $(5, 26.67)$

• Intersection of $4x + 3y = 100$ and $y = 0$:

$4x = 100 \Rightarrow x = 25$ → Point: $(25, 0)$

• Intersection of $2x + 3y = 90$ and $y = 0$:

$2x = 90 \Rightarrow x = 45$ → Point: $(45, 0)$

• Intersection of both lines with $x = 0$:

From (i): $y = 33.33$, from (ii): $y = 30$

Take the smaller value: $(0, 30)$

Evaluate $Z = 50x + 60y$ at corner points:

• At $(0, 30)$: $Z = 50(0) + 60(30) = 1800$
• At $(5, 26.67)$: $Z = 50(5) + 60(26.67) = 250 + 1600.2 = 1850.2$
• At $(25, 0)$: $Z = 50(25) + 60(0) = 1250$
• $(45, 0)$ is not feasible for both constraints

Conclusion:

The maximum profit of $\textsf{₹}$1850.20 occurs when the farmer allocates 5 m² for potatoes and approximately 26.67 m² for onions.

Step 1: Plot the boundary lines by converting inequalities to equations.

Also include: $x \geq 0$ and $y \geq 0$ (first quadrant)

Step 2: Find intersection points (vertices) of the feasible region.

• Intersection of (i) and (ii):

From (ii): $x = 3 - y$

Substitute into (i): $(3 - y) + 2y = 10 \Rightarrow 3 + y = 10 \Rightarrow y = 7$, then $x = 3 - 7 = -4$

Point $(-4, 7)$ is not feasible (since $x < 0$)

• Use graphical method instead:

Find where the lines intersect with axes:

• For $x + 2y = 10$:

  When $x = 0$, $2y = 10 \Rightarrow y = 5$

  When $y = 0$, $x = 10$

  So, points: $(0, 5)$ and $(10, 0)$

• For $x + y = 3$:

  When $x = 0$, $y = 3$

  When $y = 0$, $x = 3$

  So, points: $(0, 3)$ and $(3, 0)$

• Find point of intersection of both lines:

Equating (i) and (ii):

From (ii): $y = 3 - x$

Substitute into (i): $x + 2(3 - x) = 10 \Rightarrow x + 6 - 2x = 10 \Rightarrow -x = 4 \Rightarrow x = -4$

Again, $x = -4$ is not feasible.

So, we identify vertices of feasible region graphically by plotting and shading the region that satisfies all constraints.

Step 3: Feasible Region (bounded and convex)

From the graph, the feasible region is bounded by the following intersection points:

• $(0, 3)$ → intersection of $x + y = 3$ and $y$-axis
• $(0, 5)$ → intersection of $x + 2y = 10$ and $y$-axis
• $(2, 4)$ → point of intersection of both lines (found below)
• $(3, 0)$ → intersection of $x + y = 3$ and $x$-axis
• $(10, 0)$ → intersection of $x + 2y = 10$ and $x$-axis

Now find correct feasible points by solving the equations properly:

Solving $x + 2y = 10$ and $x + y = 3$ again:

From $x + y = 3 \Rightarrow y = 3 - x$

Substitute in $x + 2(3 - x) = 10 \Rightarrow x + 6 - 2x = 10 \Rightarrow -x = 4 \Rightarrow x = -4$

Still infeasible, hence vertices of the feasible region are:

• $(0, 3)$
• $(0, 5)$
• $(2, 4)$ — point of intersection of $x + 2y = 10$ and $x = 2$ (from substitution)
• $(3, 0)$

Step 4: Evaluate $Z = 3x + 5y$ at each vertex

Step 5: Conclusion

Maximum value of $Z$ = 26 at point $(2, 4)$

Minimum value of $Z$ = 9 at point $(3, 0)$

Let:

$x$ = number of units of product A

$y$ = number of units of product B

Objective:

To maximize the total profit $Z$

Constraints:

• Labour constraint: Each A needs 1 hour, B needs 3 hours. Max 100 hours.

$x + 3y \leq 100$

…(i)

• Raw material constraint: A needs 2 units, B needs 1 unit. Max 50 units.

$2x + y \leq 50$

…(ii)

• Market demand constraint: At least 10 units of product A

$x \geq 10$

…(iii)

• Non-negativity constraints:

$x \geq 0, \; y \geq 0$

(Feasible in first quadrant)

Linear Programming Problem (LPP):

Maximize $Z = 150x + 200y$

Subject to:

$x + 3y \leq 100$

$2x + y \leq 50$

$x \geq 10$, $x \geq 0$, $y \geq 0$

Graphical Solution:

• Plot $x + 3y = 100$:

When $x = 0$, $y = \frac{100}{3} \approx 33.33$

When $y = 0$, $x = 100$

Points: $(0, 33.33)$ and $(100, 0)$

• Plot $2x + y = 50$:

When $x = 0$, $y = 50$

When $y = 0$, $x = 25$

Points: $(0, 50)$ and $(25, 0)$

• Plot $x = 10$: Vertical line passing through $x = 10$

Shade the region satisfying all inequalities and lying in the first quadrant.

Find corner points of feasible region:

• Point A: Intersection of $x = 10$ and $x + 3y = 100$

$10 + 3y = 100 \Rightarrow y = 30$

→ Point A = $(10, 30)$

• Point B: Intersection of $x = 10$ and $2x + y = 50$

$2(10) + y = 50 \Rightarrow y = 30$

→ Point B = $(10, 30)$

• Point C: Intersection of $x + 3y = 100$ and $2x + y = 50$

From $x + 3y = 100$ → $x = 100 - 3y$

Substitute into $2x + y = 50$:

$2(100 - 3y) + y = 50 \Rightarrow 200 - 6y + y = 50 \Rightarrow -5y = -150 \Rightarrow y = 30$

Then $x = 100 - 3(30) = 10$

→ Point C = $(10, 30)$ (same point again)

• Point D: Intersection of $2x + y = 50$ and $y = 0$

$2x = 50 \Rightarrow x = 25$

→ Point D = $(25, 0)$

• Point E: Intersection of $x + 3y = 100$ and $y = 0$

$x = 100$

→ Point E = $(100, 0)$

• However, we need $x \geq 10$, so ignore any point where $x < 10$.

Feasible vertices: $(10, 30)$ and $(25, 0)$

Evaluate $Z = 150x + 200y$ at each vertex:

Conclusion:

The maximum profit of $\textsf{₹}$7500 is obtained when the factory produces 10 units of product A and 30 units of product B.

Let:

$x$ = kilograms of Ingredient I

$y$ = kilograms of Ingredient II

Objective: To minimize the cost of the mixture

Constraints:

• Vitamin X constraint: Each kg of I provides 3 units, and II provides 2 units. Minimum required = 30

$3x + 2y \geq 30$

…(i)

• Vitamin Y constraint: I provides 2 units/kg and II provides 3 units/kg. Minimum required = 25

$2x + 3y \geq 25$

…(ii)

• Non-negativity constraints: $x \geq 0$, $y \geq 0$

Linear Programming Problem (LPP):

Minimize $Z = 5x + 7y$

Subject to:

$3x + 2y \geq 30$

$2x + 3y \geq 25$

$x \geq 0$, $y \geq 0$

Graphical Solution:

• Plot $3x + 2y = 30$:

When $x = 0$, $2y = 30 \Rightarrow y = 15$

When $y = 0$, $x = 10$

Points: $(0, 15)$ and $(10, 0)$

• Plot $2x + 3y = 25$:

When $x = 0$, $3y = 25 \Rightarrow y = \frac{25}{3} \approx 8.33$

When $y = 0$, $x = 12.5$

Points: $(0, 8.33)$ and $(12.5, 0)$

The feasible region lies in the first quadrant and above both lines (as inequalities are $\geq$).

Find point of intersection of the two lines:

Solve:
Equation (i): $3x + 2y = 30$
Equation (ii): $2x + 3y = 25$

Multiply (i) by 3 and (ii) by 2 to eliminate $y$:

Subtract (iv) from (iii):

$(9x + 6y) - (4x + 6y) = 90 - 50 \Rightarrow 5x = 40 \Rightarrow x = 8$

Substitute into (i): $3(8) + 2y = 30 \Rightarrow 24 + 2y = 30 \Rightarrow y = 3$

Point of intersection: $(8, 3)$

Evaluate $Z = 5x + 7y$ at corner points of feasible region:

• $(8, 3)$: $Z = 5(8) + 7(3) = 40 + 21 = 61$
• $(0, 15)$: $Z = 5(0) + 7(15) = 0 + 105 = 105$
• $(12.5, 0)$: $Z = 5(12.5) + 7(0) = 62.5$

Conclusion:

The minimum cost is $\textsf{₹}$61 when 8 kg of Ingredient I and 3 kg of Ingredient II are used.

Given: A system of inequalities that defines a feasible region

Objective:

Maximize $Z = 5x + 7y$ over the feasible region

Step 1: Graph the inequalities to identify the feasible region:

• $x + y \leq 6$ → Line passes through $(0,6)$ and $(6,0)$
• $2x + y \leq 8$ → Line passes through $(0,8)$ and $(4,0)$
• $x \leq 3$ → Vertical line $x = 3$
• $x \geq 0$, $y \geq 0$ → First quadrant

The feasible region is bounded and lies in the first quadrant, limited by the above lines.

Step 2: Find vertices of the feasible region

• Intersection of $x + y = 6$ and $2x + y = 8$:

Subtract equations:

$2x + y - (x + y) = 8 - 6 \Rightarrow x = 2$

Substitute in $x + y = 6 \Rightarrow 2 + y = 6 \Rightarrow y = 4$

Vertex A: $(2, 4)$

• Intersection of $x + y = 6$ and $x = 3$:

$3 + y = 6 \Rightarrow y = 3$

Vertex B: $(3, 3)$

• Intersection of $2x + y = 8$ and $x = 3$:

$2(3) + y = 8 \Rightarrow y = 2$

Vertex C: $(3, 2)$

• Intersection of $2x + y = 8$ and $x$-axis ($y = 0$):

$2x = 8 \Rightarrow x = 4$

Point $(4, 0)$ is outside the region defined by $x \leq 3$ → Not feasible

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

$x = 6$ → Not feasible since $x \leq 3$

• Intersection of $x = 0$ and $2x + y = 8$:

$y = 8$ → Not feasible since $x + y = 6$ would be violated

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

$y = 6$

Vertex D: $(0, 6)$

• Intersection of $x = 0$ and $y = 0$:

Vertex E: $(0, 0)$

Feasible vertices: $(0, 0)$, $(0, 6)$, $(2, 4)$, $(3, 3)$, $(3, 2)$

Step 3: Evaluate objective function $Z = 5x + 7y$ at each vertex

Conclusion:

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

Let:

$x$ = number of hectares allocated for crop A

$y$ = number of hectares allocated for crop B

Objective: To maximize total profit

Constraints:

• Land constraint: Total land available is 50 hectares

$x + y \leq 50$

…(i)

• Herbicide constraint: A uses 20 L/ha and B uses 10 L/ha; total available = 800 L

$20x + 10y \leq 800$

…(ii)

• Cost constraint: A costs $\textsf{₹}$2,000/ha and B costs $\textsf{₹}$1,000/ha; budget = $\textsf{₹}$4,00,000

$2000x + 1000y \leq 400000$

…(iii)

• Non-negativity constraints: $x \geq 0$, $y \geq 0$

Linear Programming Problem (LPP):

Maximize $Z = 10000x + 9000y$

Subject to:

$x + y \leq 50$

$20x + 10y \leq 800$

$2000x + 1000y \leq 400000$

$x \geq 0$, $y \geq 0$

Step 1: Simplify constraints (ii) and (iii):

Since $2x + y \leq 80$ is stricter than $2x + y \leq 400$, we only need to consider $2x + y \leq 80$.

Step 2: Plot the constraints:

• $x + y = 50$ → Intercepts: $(0, 50)$ and $(50, 0)$
• $2x + y = 80$ → Intercepts: $(0, 80)$ and $(40, 0)$

The feasible region is bounded and lies below both lines and in the first quadrant.

Step 3: Find corner points of feasible region:

• Intersection of $x + y = 50$ and $2x + y = 80$:

Subtracting equations: $(2x + y) - (x + y) = 80 - 50 \Rightarrow x = 30$

Substitute $x = 30$ in $x + y = 50 \Rightarrow y = 20$

Point A: $(30, 20)$

• Other points from intercepts:
• Point B: $(0, 50)$
• Point C: $(40, 0)$
• Point D: $(0, 80)$ is not valid due to land constraint $x + y \leq 50$

Feasible region vertices: $(0, 50)$, $(30, 20)$, $(40, 0)$

Step 4: Evaluate objective function at each vertex:

Conclusion:

The maximum profit is $\textsf{₹}$4,80,000 when 30 hectares are used for crop A and 20 hectares for crop B.

Let:

$x$ = number of Gel pens produced

$y$ = number of Roller pens produced

Objective:

Maximize profit $Z = 5x + 7y$

Constraints:

Machine P is available for 20 hours = 20 × 60 = 1200 minutes

Machine Q is available for 10 hours = 10 × 60 = 600 minutes

• Machine P constraint: $4x + 3y \leq 1200$
• Machine Q constraint: $2x + 3y \leq 600$
• Non-negativity constraints: $x \geq 0$, $y \geq 0$

LPP Formulation:

Maximize $Z = 5x + 7y$

Subject to:

$4x + 3y \leq 1200$

$2x + 3y \leq 600$

$x \geq 0$, $y \geq 0$

Step 1: Graph the constraints

• $4x + 3y = 1200$ → Intercepts: $(0, 400)$ and $(300, 0)$
• $2x + 3y = 600$ → Intercepts: $(0, 200)$ and $(300, 0)$

The feasible region lies in the first quadrant and below both lines.

Step 2: Find points of intersection

Intersection of $4x + 3y = 1200$ and $2x + 3y = 600$

Subtracting the equations:

So, $x = 300$

Substitute in $2x + 3y = 600$:

$2(300) + 3y = 600 \Rightarrow 600 + 3y = 600 \Rightarrow y = 0$

Intersection point: $(300, 0)$

Other points:

• $(0, 400)$ from $4x + 3y = 1200$
• $(0, 200)$ from $2x + 3y = 600$

Feasible region vertices: $(0, 200)$, $(0, 400)$ (not feasible), $(300, 0)$

But the common feasible region lies between lines and their intersection.

Find point of intersection of the two constraints:

Already found: $(300, 0)$

Let’s find point of intersection of $4x + 3y = 1200$ and $2x + 3y = 600$ → Already done

Also find point of intersection with axes:

• From $2x + 3y = 600$, if $x = 0$, then $y = 200$
• From $4x + 3y = 1200$, if $x = 0$, then $y = 400$ (but this lies outside the $2x + 3y \leq 600$ constraint)

So valid vertices are: $(0, 200)$, $(150, 100)$, $(300, 0)$

Step 3: Evaluate $Z = 5x + 7y$ at each vertex

Conclusion:

The maximum profit of $\textsf{₹}$1500 occurs at $(300, 0)$, i.e., by producing 300 Gel pens and 0 Roller pens.

Given:

Objective function: Minimize $Z = 4x + 5y$

Subject to the constraints:

• $2x + y \geq 7$
• $2x + 3y \geq 15$
• $x \geq 0$
• $y \geq 0$

Step 1: Convert inequalities to equations to find boundary lines:

• $2x + y = 7$ → Intercepts: $(0,7)$ and $(3.5, 0)$
• $2x + 3y = 15$ → Intercepts: $(0,5)$ and $(7.5, 0)$

The feasible region will lie in the first quadrant and above both lines.

Step 2: Find point of intersection of $2x + y = 7$ and $2x + 3y = 15$

From $2x + y = 7$, we get:

Substitute (i) into $2x + 3y = 15$:

$2x + 3(7 - 2x) = 15$

$2x + 21 - 6x = 15 \Rightarrow -4x = -6 \Rightarrow x = 1.5$

Substitute $x = 1.5$ into (i): $y = 7 - 2(1.5) = 4$

Intersection point: $(1.5, 4)$

Step 3: Identify corner points (vertices) of feasible region

• Point A: Intersection of $2x + y = 7$ with $x$-axis: $(3.5, 0)$
• Point B: Intersection point: $(1.5, 4)$
• Point C: Intersection of $2x + 3y = 15$ with $y$-axis: $(0, 5)$

Step 4: Evaluate objective function $Z = 4x + 5y$ at each vertex

Conclusion:

The minimum value of $Z = 14$ occurs at point $(3.5, 0)$

Let:

$x$ = number of tables purchased

$y$ = number of chairs purchased

Objective:

Maximize profit $Z = 250x + 75y$

Constraints:

• Investment constraint: $2500x + 500y \leq 50000$
• Storage constraint: $x + y \leq 35$
• Non-negativity constraints: $x \geq 0$, $y \geq 0$

Step 1: Simplify the investment constraint

Divide entire inequality $2500x + 500y \leq 50000$ by 500:

Also, from the storage constraint:

Step 2: Find the corner points of the feasible region

• From (i): $5x + y = 100$ → Intercepts: $(0,100)$ and $(20, 0)$
• From (ii): $x + y = 35$ → Intercepts: $(0,35)$ and $(35, 0)$

Find the point of intersection of the two lines:

Subtract (ii) from (i):

$(5x + y) - (x + y) = 100 - 35 \Rightarrow 4x = 65 \Rightarrow x = 16.25$

Substitute into (ii): $16.25 + y = 35 \Rightarrow y = 18.75$

Intersection point: $(16.25, 18.75)$

Step 3: Evaluate objective function $Z = 250x + 75y$ at all corner points

Conclusion:

Maximum profit of $\textsf{₹}$5468.75 is obtained when dealer purchases 16.25 tables and 18.75 chairs.

Note: Since number of items cannot be fractional in practice, the dealer must choose nearest valid integer values respecting all constraints.

Objective:

Maximize $Z = 3x + 2y$

Subject to the constraints:

• $x + 2y \leq 8$
• $x + y \leq 5$
• $x - y \leq 2$
• $x \geq 0$
• $y \geq 0$

Step 1: Convert inequalities to equations for plotting

• $x + 2y = 8$ → Intercepts: $(0,4)$ and $(8,0)$
• $x + y = 5$ → Intercepts: $(0,5)$ and $(5,0)$
• $x - y = 2$ → Intercepts: $(2,0)$ and $(0,-2)$ (only consider $y \geq 0$)

Step 2: Find the corner points of feasible region

Intersection of $x + 2y = 8$ and $x + y = 5$:

Substitute into $x + 2y = 8$:

$x + 2(5 - x) = 8 \Rightarrow x + 10 - 2x = 8 \Rightarrow -x = -2 \Rightarrow x = 2$

$y = 5 - 2 = 3$

Point A: $(2, 3)$

Intersection of $x + y = 5$ and $x - y = 2$:

Add both: $x + y + x - y = 5 + 2 \Rightarrow 2x = 7 \Rightarrow x = 3.5$

Substitute into $x + y = 5$: $3.5 + y = 5 \Rightarrow y = 1.5$

Point B: $(3.5, 1.5)$

Intersection of $x + 2y = 8$ and $x - y = 2$:

From $x - y = 2 \Rightarrow x = y + 2$

Substitute into $x + 2y = 8$: $(y + 2) + 2y = 8 \Rightarrow 3y = 6 \Rightarrow y = 2$

$x = 4$

Point C: $(4, 2)$

Intersection of $x + y = 5$ and $x$-axis ($y = 0$):

$x + 0 = 5 \Rightarrow x = 5$ → Point D: $(5, 0)$

Intersection of $x - y = 2$ and $x$-axis ($y = 0$):

$x - 0 = 2 \Rightarrow x = 2$ → Point E: $(2, 0)$

Intersection of $x + 2y = 8$ and $y$-axis ($x = 0$):

$0 + 2y = 8 \Rightarrow y = 4$ → Point F: $(0, 4)$

Intersection of $x + y = 5$ and $y$-axis ($x = 0$):

$0 + y = 5 \Rightarrow y = 5$ → Point G: $(0, 5)$

Step 3: Determine feasible corner points from graph:

• $(0, 0)$
• $(2, 0)$
• $(3.5, 1.5)$
• $(2, 3)$

Step 4: Evaluate $Z = 3x + 2y$ at each corner point

Conclusion:

The maximum value of $Z = 13.5$ occurs at point $(3.5, 1.5)$

Let:

$x$ = number of units of product P1 produced

$y$ = number of units of product P2 produced

Objective:

Maximize profit:

Subject to constraints:

• Cutting time constraint: $2x + y \leq 100$
• Assembly time constraint: $x + 2y \leq 140$
• Finishing time constraint: $x + y \leq 70$
• Non-negativity constraints: $x \geq 0,\; y \geq 0$

Challenge in solving graphically:

To solve a Linear Programming Problem graphically, we must plot the feasible region in a two-dimensional space. In this case, since we only have two variables ($x$ and $y$), the graphical method is applicable. However, the challenge arises in:

• Precisely identifying the intersection points of the three constraint lines, especially when they do not intersect at integer values.
• Accurately calculating the objective function value $Z$ at those intersection points to determine the maximum.
• Graphing multiple inequalities simultaneously and determining the feasible region can become complex when boundaries are close to each other or nearly overlapping.

Note: Although graphical method is feasible for two-variable LPPs, precision is often better achieved through algebraic or computational methods in such multi-constraint cases.

Objective:

Subject to the constraints:

• $x + y \geq 4$
• $x + y \leq 2$
• $x \geq 0$
• $y \geq 0$

Step 1: Analyze the constraints geometrically

Consider the lines:

• $x + y = 4$ → Intercepts: $(0,4)$ and $(4,0)$
• $x + y = 2$ → Intercepts: $(0,2)$ and $(2,0)$

The inequality $x + y \geq 4$ represents the region *above* the line $x + y = 4$.

The inequality $x + y \leq 2$ represents the region *below* the line $x + y = 2$.

Also, the feasible values of $x$ and $y$ must lie in the first quadrant due to $x \geq 0$ and $y \geq 0$.

Step 2: Determine the feasible region

The constraint $x + y \geq 4$ defines a region that lies entirely above the line $x + y = 4$.

The constraint $x + y \leq 2$ defines a region that lies entirely below the line $x + y = 2$.

Hence, the two inequalities are mutually exclusive — no point can simultaneously satisfy both $x + y \geq 4$ and $x + y \leq 2$.

Conclusion:

There is no feasible region that satisfies all the given constraints at the same time. Since the feasible region is empty, there exists:

• No corner points
• No bounded or unbounded region
• No optimal solution to the LPP

Therefore, the problem has no solution due to contradictory constraints.

Let:

$x$ = quantity (in kg) of Food X used

$y$ = quantity (in kg) of Food Y used

Objective:

Minimize the total cost:

Subject to the constraints:

• Vitamin A requirement: $2x + 4y \geq 8$
• Vitamin B requirement: $3x + 2y \geq 10$
• Non-negativity constraints: $x \geq 0,\; y \geq 0$

Step 1: Convert inequalities to equations for graphing

• $2x + 4y = 8$ → Divide by 2: $x + 2y = 4$ → Intercepts: $(0, 2)$ and $(4, 0)$
• $3x + 2y = 10$ → Intercepts: $(0, 5)$ and $(\frac{10}{3}, 0)$

Shade the region which satisfies both:

• $2x + 4y \geq 8$ → Region above or on the line $x + 2y = 4$
• $3x + 2y \geq 10$ → Region above or on the line $3x + 2y = 10$

Also consider $x \geq 0$ and $y \geq 0$ to restrict to the first quadrant.

Step 2: Find corner points of the feasible region

Solving equations:

From: $x + 2y = 4$ and $3x + 2y = 10$

Substitute $x = 3$ into $x + 2y = 4$:

$3 + 2y = 4 \Rightarrow 2y = 1 \Rightarrow y = \frac{1}{2}$

Intersection point: $(3, \frac{1}{2})$

Other vertices (from intersection with axes):

• From $x + 2y = 4$ ⇒ $(0, 2)$ and $(4, 0)$
• From $3x + 2y = 10$ ⇒ $(0, 5)$ and $(\frac{10}{3}, 0)$

Feasible region lies in the first quadrant and above both lines.

Corner points to test: Intersection points inside the feasible region: $(3, \frac{1}{2})$, $(0, 5)$, $(\frac{10}{3}, 0)$

Step 3: Evaluate $Z = 10x + 15y$ at corner points

• At $(3, \frac{1}{2})$: $Z = 10(3) + 15(\frac{1}{2}) = 30 + 7.5 = 37.5$
• At $(0, 5)$: $Z = 10(0) + 15(5) = 75$
• At $(\frac{10}{3}, 0)$: $Z = 10(\frac{10}{3}) + 15(0) = \frac{100}{3} \approx 33.33$

Conclusion:

Minimum cost is approximately $\textsf{₹}$33.33 at the point $\left(\frac{10}{3}, 0\right)$.

Objective:

Subject to constraints:

• $x + 2y \leq 12$
• $3x + y \geq 9$
• $x - y \geq -2$
• $x \geq 0$
• $y \geq 0$

Step 1: Convert inequalities to equations to find boundary lines

• $x + 2y = 12$ → Intercepts: $(0,6)$ and $(12,0)$
• $3x + y = 9$ → Intercepts: $(0,9)$ and $(3,0)$
• $x - y = -2$ → Rewritten as $x = y - 2$ or $y = x + 2$ → Intercepts: $(0,2)$ and $(2,4)$

Graph these lines and shade the region satisfying:

• Below or on $x + 2y = 12$
• Above or on $3x + y = 9$
• Above or on $x - y = -2$ (i.e., region to the right of the line)
• In the first quadrant ($x \geq 0,\; y \geq 0$)

Step 2: Find vertices of the feasible region (intersection points)

• Intersection of $x + 2y = 12$ and $3x + y = 9$:

From $x + 2y = 12$ → $x = 12 - 2y$

Substitute into $3x + y = 9$:

$3(12 - 2y) + y = 9 \Rightarrow 36 - 6y + y = 9 \Rightarrow -5y = -27 \Rightarrow y = \frac{27}{5},\; x = 12 - 2 \cdot \frac{27}{5} = \frac{6}{5}$

Point A: $\left(\frac{6}{5}, \frac{27}{5}\right)$

• Intersection of $x + 2y = 12$ and $x - y = -2$:

From $x - y = -2$ → $x = y - 2$

Substitute into $x + 2y = 12$:

$(y - 2) + 2y = 12 \Rightarrow 3y = 14 \Rightarrow y = \frac{14}{3},\; x = \frac{14}{3} - 2 = \frac{8}{3}$

Point B: $\left(\frac{8}{3}, \frac{14}{3}\right)$

• Intersection of $3x + y = 9$ and $x - y = -2$:

From $x - y = -2$ → $x = y - 2$

Substitute into $3x + y = 9$:

$3(y - 2) + y = 9 \Rightarrow 3y - 6 + y = 9 \Rightarrow 4y = 15 \Rightarrow y = \frac{15}{4},\; x = \frac{15}{4} - 2 = \frac{7}{4}$

Point C: $\left(\frac{7}{4}, \frac{15}{4}\right)$

Step 3: Evaluate $Z = 4x + 6y$ at each vertex

• At A: $Z = 4 \cdot \frac{6}{5} + 6 \cdot \frac{27}{5} = \frac{24}{5} + \frac{162}{5} = \frac{186}{5} = 37.2$
• At B: $Z = 4 \cdot \frac{8}{3} + 6 \cdot \frac{14}{3} = \frac{32}{3} + \frac{84}{3} = \frac{116}{3} \approx 38.67$
• At C: $Z = 4 \cdot \frac{7}{4} + 6 \cdot \frac{15}{4} = 7 + \frac{90}{4} = 7 + 22.5 = 29.5$

Conclusion:

Maximum value of $Z$ is approximately 38.67 at point $\left(\frac{8}{3}, \frac{14}{3}\right)$.

Minimum value of $Z$ is 29.5 at point $\left(\frac{7}{4}, \frac{15}{4}\right)$.

Let:

$x$ = number of fans to be purchased

$y$ = number of sewing machines to be purchased

Objective:

Maximize the total profit:

Subject to the constraints:

• Investment constraint: $360x + 240y \leq 5760$
• Space constraint: $x + y \leq 20$
• Non-negativity constraints: $x \geq 0$, $y \geq 0$

Simplifying the investment constraint:

Divide entire inequality by 120:

Step 1: Plot the constraints

• From (ii): $3x + 2y = 48$ → Intercepts: $(0,24)$ and $(16,0)$
• From $x + y = 20$ → Intercepts: $(0,20)$ and $(20,0)$

Graph the two lines and shade the region satisfying both inequalities, and bounded in the first quadrant ($x \geq 0,\; y \geq 0$).

Step 2: Find the corner points of feasible region

• Intersection of $3x + 2y = 48$ and $x + y = 20$

From $x + y = 20$ ⇒ $y = 20 - x$

Substitute in $3x + 2y = 48$:

$3x + 2(20 - x) = 48 \Rightarrow 3x + 40 - 2x = 48 \Rightarrow x = 8,\; y = 12$

Intersection point: $(8, 12)$

Other points from axis intercepts:

• $(0, 20)$ from $x + y = 20$
• $(0, 24)$ from $3x + 2y = 48$
• $(16, 0)$ from $3x + 2y = 48$
• $(20, 0)$ from $x + y = 20$

Feasible region bounded by the lines: check common region for:

• $(0, 20)$
• $(8, 12)$
• $(16, 0)$

Step 3: Evaluate $Z = 22x + 18y$ at each vertex

• At $(0, 20)$: $Z = 22(0) + 18(20) = 360$
• At $(8, 12)$: $Z = 22(8) + 18(12) = 176 + 216 = 392$
• At $(16, 0)$: $Z = 22(16) + 18(0) = 352$

Conclusion:

Maximum profit is $\textsf{₹}$392 when the dealer buys 8 fans and 12 sewing machines.

Let:

$x$ = units of one variable

$y$ = units of another variable

Objective:

Subject to the constraints:

• $x + 2y \leq 40$
• $3x + y \geq 30$
• $4x + 3y \geq 60$
• $x \geq 0$, $y \geq 0$

Step 1: Graph the constraints

• Line 1: $x + 2y = 40$ → Intercepts: $(0, 20)$ and $(40, 0)$
• Line 2: $3x + y = 30$ → Intercepts: $(0, 30)$ and $(10, 0)$
• Line 3: $4x + 3y = 60$ → Intercepts: $(0, 20)$ and $(15, 0)$

Shade the region that satisfies:

• Below or on $x + 2y = 40$
• Above or on $3x + y = 30$
• Above or on $4x + 3y = 60$
• In the first quadrant: $x \geq 0$, $y \geq 0$

Step 2: Find corner points (vertices) of feasible region

• Intersection of $x + 2y = 40$ and $3x + y = 30$:

From $3x + y = 30$ → $y = 30 - 3x$

Substitute in $x + 2y = 40$:

$x + 2(30 - 3x) = 40 \Rightarrow x + 60 - 6x = 40 \Rightarrow -5x = -20 \Rightarrow x = 4,\; y = 30 - 12 = 18$

Point A: $(4, 18)$

• Intersection of $x + 2y = 40$ and $4x + 3y = 60$:

From $x + 2y = 40$ → $x = 40 - 2y$

Substitute in $4x + 3y = 60$:

$4(40 - 2y) + 3y = 60 \Rightarrow 160 - 8y + 3y = 60 \Rightarrow -5y = -100 \Rightarrow y = 20,\; x = 0$

Point B: $(0, 20)$

• Intersection of $3x + y = 30$ and $4x + 3y = 60$:

From $3x + y = 30$ → $y = 30 - 3x$

Substitute in $4x + 3y = 60$:

$4x + 3(30 - 3x) = 60 \Rightarrow 4x + 90 - 9x = 60 \Rightarrow -5x = -30 \Rightarrow x = 6,\; y = 30 - 18 = 12$

Point C: $(6, 12)$

Step 3: Evaluate $Z = 20x + 10y$ at each vertex

• At A $(4, 18)$: $Z = 20(4) + 10(18) = 80 + 180 = 260$
• At B $(0, 20)$: $Z = 20(0) + 10(20) = 0 + 200 = 200$
• At C $(6, 12)$: $Z = 20(6) + 10(12) = 120 + 120 = 240$

Conclusion:

Minimum value of $Z$ is 200 at point $(0, 20)$.

Let:

$x$ = number of units of item X to be produced

$y$ = number of units of item Y to be produced

Objective:

Maximize profit:

Subject to the constraints:

• Grinding time: $10x + 2y \leq 200$
• Polishing time: $2x + 4y \leq 100$
• Market condition for X: $x \geq 10$
• Market condition for Y: $y \leq 40$
• Non-negativity: $x \geq 0$, $y \geq 0$

Step 1: Simplify and graph the constraints

• $10x + 2y \leq 200$ → divide by 2: $5x + y \leq 100$
• $2x + 4y \leq 100$ → divide by 2: $x + 2y \leq 50$
• Other constraints: $x \geq 10$, $y \leq 40$, $x \geq 0$, $y \geq 0$

Line 1: $5x + y = 100$ → Intercepts: $(0, 100)$ and $(20, 0)$

Line 2: $x + 2y = 50$ → Intercepts: $(0, 25)$ and $(50, 0)$

Line 3: $x = 10$

Line 4: $y = 40$

Step 2: Find corner points of the feasible region

• Intersection of $5x + y = 100$ and $x + 2y = 50$:

From $x + 2y = 50$ → $x = 50 - 2y$

Substitute into $5x + y = 100$:

$5(50 - 2y) + y = 100 \Rightarrow 250 - 10y + y = 100 \Rightarrow -9y = -150 \Rightarrow y = 16.67,\; x = 50 - 33.33 = 16.67$

Point A: $(16.67, 16.67)$

• Intersection of $5x + y = 100$ and $y = 40$:

$5x + 40 = 100 \Rightarrow x = 12$

Point B: $(12, 40)$

• Intersection of $x = 10$ and $x + 2y = 50$:

$10 + 2y = 50 \Rightarrow y = 20$

Point C: $(10, 20)$

• Intersection of $x = 10$ and $y = 40$:

Point D: $(10, 40)$

Step 3: Evaluate objective function $Z = 10x + 15y$ at each point

• At A $(16.67, 16.67)$: $Z = 10(16.67) + 15(16.67) = 166.7 + 250.05 = 416.75$
• At B $(12, 40)$: $Z = 120 + 600 = 720$
• At C $(10, 20)$: $Z = 100 + 300 = 400$
• At D $(10, 40)$: $Z = 100 + 600 = 700$

Conclusion:

Maximum profit of $\textsf{₹}$720 is achieved when the firm produces 12 units of item X and 40 units of item Y.

Let:

$x$ = units of a product

$y$ = units of another product

Objective Function:

Subject to the constraints:

• $x + y \leq 7$
• $x + y \geq 3$
• $2x + y \leq 10$
• $x \geq 0$, $y \geq 0$

Step 1: Plot the lines

• Line 1: $x + y = 7$ → Intercepts: $(0,7)$ and $(7,0)$
• Line 2: $x + y = 3$ → Intercepts: $(0,3)$ and $(3,0)$
• Line 3: $2x + y = 10$ → Intercepts: $(0,10)$ and $(5,0)$

Shade the region:

• Below or on $x + y = 7$
• Above or on $x + y = 3$
• Below or on $2x + y = 10$
• In the first quadrant: $x \geq 0$, $y \geq 0$

Step 2: Find the corner points of the feasible region

• Intersection of $x + y = 3$ and $2x + y = 10$:

From $x + y = 3$ → $y = 3 - x$

Substitute into $2x + y = 10$:

$2x + (3 - x) = 10 \Rightarrow x + 3 = 10 \Rightarrow x = 7,\; y = 3 - 7 = -4$

Discard (not feasible since $y < 0$)

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

$x = 0 \Rightarrow y = 3$

Point A: $(0, 3)$

• Intersection of $x + y = 7$ and $x = 0$:

$x = 0 \Rightarrow y = 7$

Point B: $(0, 7)$

• Intersection of $x + y = 7$ and $2x + y = 10$:

From $x + y = 7$ → $y = 7 - x$

Substitute into $2x + y = 10$:

$2x + (7 - x) = 10 \Rightarrow x + 7 = 10 \Rightarrow x = 3,\; y = 4$

Point C: $(3, 4)$

• Intersection of $x + y = 3$ and $2x + y = 10$ → invalid, already checked
• Intersection of $x + y = 3$ and $y = 0$:

$y = 0 \Rightarrow x = 3$

Point D: $(3, 0)$

• Intersection of $2x + y = 10$ and $y = 0$:

$y = 0 \Rightarrow 2x = 10 \Rightarrow x = 5$

Point E: $(5, 0)$

Feasible vertices: A $(0,3)$, C $(3,4)$, B $(0,7)$, D $(3,0)$, E $(5,0)$

Step 3: Evaluate $Z = 5x + 2y$ at each vertex

• At A $(0, 3)$: $Z = 0 + 6 = 6$
• At B $(0, 7)$: $Z = 0 + 14 = 14$
• At C $(3, 4)$: $Z = 15 + 8 = 23$
• At D $(3, 0)$: $Z = 15 + 0 = 15$
• At E $(5, 0)$: $Z = 25 + 0 = 25$

Conclusion:

The maximum value of $Z = 25$ occurs at point $(5, 0)$.

Let:

$x$ = number of units of Product A to be produced

$y$ = number of units of Product B to be produced

Objective Function:

Maximize profit:

Subject to the constraints:

• Raw material: $3x + 4y \leq 180$
• Labour: $2x + y \leq 90$
• Minimum production of B: $y \geq 10$
• Non-negativity: $x \geq 0,\; y \geq 0$

Step 1: Find the intercepts of each constraint line

• Line 1: $3x + 4y = 180$ → intercepts: $(0, 45)$ and $(60, 0)$
• Line 2: $2x + y = 90$ → intercepts: $(0, 90)$ and $(45, 0)$
• Line 3: $y = 10$ → a horizontal line passing through $y = 10$

Step 2: Find points of intersection

• Intersection of $3x + 4y = 180$ and $2x + y = 90$

From $2x + y = 90$ → $y = 90 - 2x$

Substitute into $3x + 4y = 180$:

$3x + 4(90 - 2x) = 180 \Rightarrow 3x + 360 - 8x = 180$

$-5x = -180 \Rightarrow x = 36,\; y = 90 - 72 = 18$

Point A: $(36, 18)$

• Intersection of $3x + 4y = 180$ and $y = 10$

$3x + 4(10) = 180 \Rightarrow 3x = 140 \Rightarrow x = \frac{140}{3} \approx 46.67$

Point B: $(46.67, 10)$

• Intersection of $2x + y = 90$ and $y = 10$

$2x + 10 = 90 \Rightarrow x = 40$

Point C: $(40, 10)$

Step 3: Evaluate $Z = 300x + 200y$ at each point

• At A $(36, 18)$: $Z = 300(36) + 200(18) = 10800 + 3600 = \textsf{₹}14400$
• At B $(46.67, 10)$: $Z = 300(46.67) + 200(10) \approx 14000 + 2000 = \textsf{₹}16000$
• At C $(40, 10)$: $Z = 300(40) + 200(10) = 12000 + 2000 = \textsf{₹}14000$

Conclusion:

Maximum profit of approximately $\textsf{₹}16000$ occurs at $(x, y) = \left(\frac{140}{3},\ 10\right)$ or approximately $(46.67,\ 10)$.

Hence, the manufacturer should produce approximately 47 units of A and 10 units of B.

Objective Function:

Subject to constraints:

• $x + 2y \geq 8$
• $3x + y \geq 9$
• $x \geq 0$
• $y \geq 0$

Step 1: Plot the boundary lines

• Line 1: $x + 2y = 8$ → intercepts: $(0, 4)$ and $(8, 0)$
• Line 2: $3x + y = 9$ → intercepts: $(0, 9)$ and $(3, 0)$

Since the inequalities are $\geq$, the feasible region lies above both the lines and in the first quadrant.

Step 2: Find the point of intersection

To find the intersection of $x + 2y = 8$ and $3x + y = 9$:

From $3x + y = 9$ → $y = 9 - 3x$

Substitute into $x + 2y = 8$:

$x + 2(9 - 3x) = 8 \Rightarrow x + 18 - 6x = 8$

$-5x = -10 \Rightarrow x = 2$

$y = 9 - 3(2) = 3$

Intersection Point: $(2, 3)$

Step 3: Analyze feasible region

The region is bounded by the lines $x + 2y = 8$ and $3x + y = 9$ from below and is open in the direction away from origin. As $x$ and $y$ increase, all constraints continue to be satisfied.

Hence, the feasible region is unbounded.

Step 4: Evaluate $Z = 3x + 2y$ at corner points

• At $(0, 9)$: $Z = 3(0) + 2(9) = 18$
• At $(2, 3)$: $Z = 3(2) + 2(3) = 6 + 6 = 12$
• At $(8, 0)$: $Z = 3(8) + 2(0) = 24$

Step 5: Check for minimum value existence

The region is unbounded and extends infinitely, but since $Z = 3x + 2y$ increases as $x$ and $y$ increase, and the direction of optimization is minimization, we must ensure that the minimum value doesn't occur at infinity.

We observe that the lowest value of $Z$ within the feasible region is at point $(2, 3)$ which is $Z = 12$, and as we move further away, $Z$ increases. Therefore,

The minimum value exists and is $Z = 12$ at point $(2, 3)$.

Let:

$x$ = number of units of Product P

$y$ = number of units of Product Q

Objective Function:

Maximize profit:

Subject to the constraints:

• Cutting time: $x + 2y \leq 80$
• Assembly time: $2x + y \leq 100$
• Product Q restriction: $y \leq 30$
• Non-negativity: $x \geq 0,\; y \geq 0$

Step 1: Plot boundary lines and intercepts

• Constraint 1: $x + 2y = 80$ → intercepts: $(0, 40)$ and $(80, 0)$
• Constraint 2: $2x + y = 100$ → intercepts: $(0, 100)$ and $(50, 0)$
• Constraint 3: $y = 30$ → a horizontal line through $y = 30$

The feasible region lies in the first quadrant and is bounded by the intersection of these inequalities.

Step 2: Find vertices of the feasible region

• Intersection of $x + 2y = 80$ and $y = 30$:

$x + 2(30) = 80 \Rightarrow x = 20$ → Point A: $(20, 30)$

• Intersection of $2x + y = 100$ and $y = 30$:

$2x + 30 = 100 \Rightarrow x = 35$ → Point B: $(35, 30)$

• Intersection of $x + 2y = 80$ and $2x + y = 100$:

Multiply $x + 2y = 80$ by 2 → $2x + 4y = 160$

Subtract: $(2x + 4y) - (2x + y) = 160 - 100$ → $3y = 60 \Rightarrow y = 20$

$x = 80 - 2(20) = 40$ → Point C: $(40, 20)$

• Intersection of $2x + y = 100$ with $x$-axis: $y = 0 \Rightarrow x = 50$ → Point D: $(50, 0)$
• Intersection of $x + 2y = 80$ with $x$-axis: $y = 0 \Rightarrow x = 80$ → Point E: $(80, 0)$

Feasible corner points to evaluate: $(20, 30),\ (35, 30),\ (40, 20)$

Step 3: Evaluate $Z = 15x + 20y$ at each point

• At $(20, 30)$: $Z = 15(20) + 20(30) = 300 + 600 = \textsf{₹}900$
• At $(35, 30)$: $Z = 15(35) + 20(30) = 525 + 600 = \textsf{₹}1125$
• At $(40, 20)$: $Z = 15(40) + 20(20) = 600 + 400 = \textsf{₹}1000$

Conclusion:

Maximum profit of $\textsf{₹}1125$ occurs at $(x, y) = (35, 30)$.

Hence, the factory should produce 35 units of P and 30 units of Q to maximize profit.

Objective Function:

Subject to the constraints:

• $x + y \leq 9$
• $2x + y \leq 12$
• $x \leq 4$
• $x \geq 0$
• $y \geq 0$

Step 1: Plot boundary lines and find intercepts

• Line 1: $x + y = 9$ → intercepts: $(0, 9)$ and $(9, 0)$
• Line 2: $2x + y = 12$ → intercepts: $(0, 12)$ and $(6, 0)$
• Line 3: $x = 4$ → vertical line passing through $x = 4$

The feasible region lies in the first quadrant, bounded by the lines mentioned above.

Step 2: Find vertices of the feasible region

• Intersection of $x + y = 9$ and $x = 4$: $4 + y = 9 \Rightarrow y = 5$ → Point A: $(4, 5)$
• Intersection of $2x + y = 12$ and $x = 4$: $2(4) + y = 12 \Rightarrow y = 4$ → Point B: $(4, 4)$
• Intersection of $x + y = 9$ and $2x + y = 12$:

Subtracting: $(2x + y) - (x + y) = 12 - 9$ → $x = 3$

$y = 9 - x = 6$ → Point C: $(3, 6)$

• Intersection of $2x + y = 12$ and $y$-axis: $x = 0 \Rightarrow y = 12$ (not in feasible region)
• Intersection of $x + y = 9$ and $x$-axis: $y = 0 \Rightarrow x = 9$ (but $x \leq 4$ so not feasible)
• Check Point: Origin $(0, 0)$ (satisfies all constraints)

Feasible region vertices: $(0, 0),\ (3, 6),\ (4, 5),\ (4, 4)$

Step 3: Evaluate $Z = 6x + 5y$ at the vertices

• At $(0, 0)$: $Z = 6(0) + 5(0) = 0$
• At $(3, 6)$: $Z = 6(3) + 5(6) = 18 + 30 = 48$
• At $(4, 5)$: $Z = 6(4) + 5(5) = 24 + 25 = 49$
• At $(4, 4)$: $Z = 6(4) + 5(4) = 24 + 20 = 44$

Conclusion:

Maximum value of $Z$ is $49$ at point $(4, 5)$.

Hence, the optimal solution is $x = 4,\; y = 5$ with maximum profit $Z = 49$.

Let:

$x$ = number of units of solar panel A

$y$ = number of units of solar panel B

Objective Function:

Maximize total profit:

Subject to the constraints:

• Machine time: $2x + 3y \leq 150$
• Labour hours: $3x + 2y \leq 120$
• Production minimum: $x + y \geq 20$
• Non-negativity: $x \geq 0,\; y \geq 0$

Step 1: Convert inequalities into equations for graphing:

• $2x + 3y = 150$ → intercepts: $(0, 50)$ and $(75, 0)$
• $3x + 2y = 120$ → intercepts: $(0, 60)$ and $(40, 0)$
• $x + y = 20$ → intercepts: $(0, 20)$ and $(20, 0)$

Step 2: Find points of intersection

• Intersection of $2x + 3y = 150$ and $3x + 2y = 120$:

Multiply first equation by 3: $6x + 9y = 450$

Multiply second equation by 2: $6x + 4y = 240$

Subtract: $5y = 210 \Rightarrow y = 42$

Substitute in $2x + 3y = 150$:

$2x + 3(42) = 150 \Rightarrow 2x = 24 \Rightarrow x = 12$ → Point A: $(12, 42)$

• Intersection of $2x + 3y = 150$ and $x + y = 20$:

From $x + y = 20 \Rightarrow y = 20 - x$

Substitute in $2x + 3y = 150$:

$2x + 3(20 - x) = 150 \Rightarrow 2x + 60 - 3x = 150$

$-x = 90 \Rightarrow x = -90$ → not feasible

• Intersection of $3x + 2y = 120$ and $x + y = 20$:

From $x + y = 20 \Rightarrow y = 20 - x$

Substitute in $3x + 2y = 120$:

$3x + 2(20 - x) = 120 \Rightarrow 3x + 40 - 2x = 120$

$x = 80,\; y = -60$ → not feasible

Now, test feasible region bounded by:

Vertices: $(0, 40),\ (30, 0),\ (0, 20),\ (20, 0),\ (12, 42)$

From constraints, only points satisfying all constraints and $x + y \geq 20$ will be considered.

Step 3: Evaluate $Z = 2000x + 2500y$ at feasible points

• At $(0, 40)$: $Z = 0 + 2500(40) = \textsf{₹}100000$
• At $(30, 0)$: $Z = 2000(30) + 0 = \textsf{₹}60000$
• At $(20, 0)$: $Z = 2000(20) = \textsf{₹}40000$
• At $(0, 20)$: $Z = 2500(20) = \textsf{₹}50000$
• At $(12, 42)$: $Z = 2000(12) + 2500(42) = 24000 + 105000 = \textsf{₹}129000$

Conclusion:

The maximum profit is $\textsf{₹}129000$ at $(x, y) = (12, 42)$.

Hence, the company should produce 12 units of A and 42 units of B to maximize profit.