Questions and Answers (Board Exam)

Board Exam Preparation: Linear Programming

This section contains 30 board-style questions on Linear Programming. For board exams, step marks are very important. Always:

1. Define the decision variables clearly.
2. Write the objective function separately.
3. Write every constraint carefully from the wording.
4. Mention the non-negativity restrictions explicitly.
5. While solving graphically, draw neat boundary lines, shade the feasible region, list the corner points, evaluate the objective function systematically, and write the final conclusion in words.

Question 1 [CBSE 2026]

Solve the following LPP graphically: Maximize $Z = 3x + 2y$ subject to $x + 2y \le 10$, $3x + y \le 15$, $x, y \ge 0$.

Solution: Step 1: Draw the line $x + 2y = 10$. Its intercepts are $(10,0)$ and $(0,5)$. Since the constraint is $x + 2y \le 10$, the feasible side is the region below this line.

Step 2: Draw the line $3x + y = 15$. Its intercepts are $(5,0)$ and $(0,15)$. Since the constraint is $3x + y \le 15$, the feasible side is again the region below this line.

Step 3: Because $x \ge 0$ and $y \ge 0$, the feasible region lies in the first quadrant.

Step 4: Find the point of intersection of the two lines: Multiply $x + 2y = 10$ by 3:

$$3x + 6y = 30$$

Subtract $3x + y = 15$:

$$5y = 15 \Rightarrow y = 3$$

Substitute into $x + 2y = 10$:

$$x + 6 = 10 \Rightarrow x = 4$$

So the intersection point is $B(4,3)$.

Step 5: The corner points of the feasible region are:

$$O(0,0),\ A(5,0),\ B(4,3),\ C(0,5)$$

Step 6: Evaluate $Z = 3x + 2y$ at each corner point:

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

$$Z(5,0) = 15$$

$$Z(4,3) = 12 + 6 = 18$$

$$Z(0,5) = 10$$

Step 7: The maximum value is 18.

Answer: Maximum $Z = 18$ at $x = 4, y = 3$.

Question 2 [CBSE 2025]

A diet is to contain at least 80 units of Vitamin A and 100 units of minerals. Two foods $F_1$ and $F_2$ are available. $F_1$ costs Rs 4 per unit and $F_2$ costs Rs 6 per unit. One unit of $F_1$ contains 3 units of Vitamin A and 4 units of minerals. One unit of $F_2$ contains 6 units of Vitamin A and 3 units of minerals. Formulate this as a linear programming problem to minimize the cost.

Solution: Step 1: Let $x$ be the number of units of food $F_1$ and $y$ be the number of units of food $F_2$.

Step 2: Since the aim is to minimize total cost, the objective function is:

$$Z = 4x + 6y$$

which is to be minimized.

Step 3: Vitamin A requirement: Food $F_1$ contributes 3 units per unit and food $F_2$ contributes 6 units per unit. Hence total Vitamin A is $3x + 6y$. Since at least 80 units are required,

$$3x + 6y \ge 80$$

Step 4: Mineral requirement: Food $F_1$ contributes 4 units per unit and food $F_2$ contributes 3 units per unit. Hence total minerals are $4x + 3y$. Since at least 100 units are required,

$$4x + 3y \ge 100$$

Step 5: Non-negativity restrictions:

$$x \ge 0, \quad y \ge 0$$

Answer: Minimize $Z = 4x + 6y$ subject to $3x + 6y \ge 80$, $4x + 3y \ge 100$, $x, y \ge 0$.

Question 3 [CBSE 2024]

Solve graphically: Minimize $Z = x + 2y$ subject to $2x + y \ge 3$, $x + 2y \ge 6$, $x, y \ge 0$.

Solution: Step 1: Draw the line $2x + y = 3$. Its intercepts are $(1.5,0)$ and $(0,3)$. Since the inequality is $2x + y \ge 3$, the feasible side is the region above this line.

Step 2: Draw the line $x + 2y = 6$. Its intercepts are $(6,0)$ and $(0,3)$. Since the inequality is $x + 2y \ge 6$, the feasible side is again the region above this line.

Step 3: Find the point of intersection of the two lines:

$$2x + y = 3$$

$$x + 2y = 6$$

From the first equation,

$$y = 3 - 2x$$

Substitute into the second:

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

$$x + 6 - 4x = 6$$

$$-3x = 0 \Rightarrow x = 0$$

Then,

$$y = 3$$

So the lines intersect at $(0,3)$.

Step 4: The feasible region lies in the first quadrant and is unbounded. The boundary corner points are $(6,0)$ and $(0,3)$.

Step 5: Evaluate the objective function:

$$Z(6,0) = 6$$

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

Step 6: Since the same minimum value is obtained at two adjacent corner points, every point on the line segment joining $(6,0)$ and $(0,3)$ gives the same value, provided it is feasible.

Step 7: In fact, the whole segment joining $(6,0)$ and $(0,3)$ lies on the line $x + 2y = 6$, and every feasible point satisfies $x + 2y \ge 6$. Therefore the minimum possible value of $Z = x + 2y$ is exactly 6.

Answer: Minimum $Z = 6$ at every point on the line segment joining $(6,0)$ and $(0,3)$.

Question 4 [CBSE 2026]

A cooperative society of farmers has 50 hectares of land to grow two crops X and Y. The profit from crops X and Y per hectare are Rs 10,500 and Rs 9,000 respectively. To control weeds, a herbicide must be used at rates of 20 litres and 10 litres per hectare for X and Y. No more than 800 litres of herbicide should be used. Formulate to maximize profit.

Solution: Step 1: Let $x$ be the number of hectares used for crop X and $y$ be the number of hectares used for crop Y.

Step 2: Profit from crop X is Rs 10,500 per hectare and from crop Y is Rs 9,000 per hectare. Therefore the objective function is:

$$Z = 10500x + 9000y$$

which is to be maximized.

Step 3: Land is limited to 50 hectares, so:

$$x + y \le 50$$

Step 4: Herbicide usage is 20 litres per hectare for X and 10 litres per hectare for Y, with a maximum of 800 litres available. Hence:

$$20x + 10y \le 800$$

Dividing by 10:

$$2x + y \le 80$$

Step 5: Non-negativity restrictions:

$$x \ge 0, \quad y \ge 0$$

Answer: Maximize $Z = 10500x + 9000y$ subject to $x + y \le 50$, $2x + y \le 80$, $x, y \ge 0$.

Question 5 [CBSE 2025]

Solve graphically: Maximize $Z = -x + 2y$, subject to $x \ge 3$, $x + y \ge 5$, $x + 2y \ge 6$, $y \ge 0$.

Solution: Step 1: Rewrite the constraints in convenient form:

$$x \ge 3$$

$$y \ge 5 - x$$

$$y \ge \frac{6 - x}{2}$$

$$y \ge 0$$

So the feasible region lies to the right of the line $x = 3$ and above all the other boundary lines.

Step 2: Find the important intersection points:

• Intersection of $x = 3$ and $x + y = 5$ gives $(3,2)$.
• Intersection of $x + y = 5$ and $x + 2y = 6$:

$$(x + 2y) - (x + y) = 6 - 5 \Rightarrow y = 1$$

Then,

$$x = 4$$

So the point is $(4,1)$.

• Intersection of $x + 2y = 6$ with $y = 0$ gives $(6,0)$.

Step 3: These points form the lower boundary corner points of the feasible region:

$$(3,2),\ (4,1),\ (6,0)$$

The region is unbounded upward and to the right.

Step 4: Evaluate $Z = -x + 2y$ at these corner points:

$$Z(3,2) = -3 + 4 = 1$$

$$Z(4,1) = -4 + 2 = -2$$

$$Z(6,0) = -6$$

Step 5: The largest value among these corner points is 1 at $(3,2)$. But since the region is unbounded, we must check whether larger values are possible.

Step 6: Take a feasible point such as $(3,3)$. It satisfies:

$$x \ge 3,$$

$$x+y = 6 \ge 5,$$

$$x+2y = 9 \ge 6,$$

$$y \ge 0.$$

Then

$$Z(3,3) = -3 + 6 = 3 > 1$$

So values larger than 1 are possible.

Step 7: In fact, if we keep $x=3$ fixed and increase $y$, then

$$Z = -3 + 2y$$

can be made arbitrarily large. Hence no maximum exists.

Answer: No maximum value exists.

Question 6 [CBSE 2023]

A company manufactures two types of sweaters: type A and type B. It costs Rs 360 to make a type A sweater and Rs 120 to make a type B sweater. The company can make at most 300 sweaters and spend at most Rs 72,000 a day. The number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100. The company makes a profit of Rs 200 on each type A sweater and Rs 120 on each type B sweater. Formulate this problem as an LPP.

Solution: Step 1: Let $x$ be the number of type A sweaters and $y$ be the number of type B sweaters.

Step 2: Profit function:

$$Z = 200x + 120y$$

which is to be maximized.

Step 3: Total number of sweaters cannot exceed 300:

$$x + y \le 300$$

Step 4: Total production cost cannot exceed Rs 72,000:

$$360x + 120y \le 72000$$

Divide by 120:

$$3x + y \le 600$$

Step 5: Type B sweaters cannot exceed type A sweaters by more than 100:

$$y - x \le 100$$

Step 6: Non-negativity restrictions:

$$x \ge 0, \quad y \ge 0$$

Answer: Maximize $Z = 200x + 120y$ subject to $x+y \le 300$, $3x+y \le 600$, $y-x \le 100$, $x,y \ge 0$.

Question 7 [CBSE 2026]

Solve the LPP graphically: Maximize $Z = 4x + y$ subject to $x + y \le 50$, $3x + y \le 90$, $x, y \ge 0$.

Solution: Step 1: Draw the lines $x + y = 50$ and $3x + y = 90$. Since both constraints are of type $\le$, the feasible region lies below both lines in the first quadrant.

Step 2: Find the point of intersection: Subtract $x + y = 50$ from $3x + y = 90$:

$$2x = 40 \Rightarrow x = 20$$

Then

$$y = 30$$

So the intersection point is $(20,30)$.

Step 3: The corner points of the feasible region are:

$$O(0,0),\ A(30,0),\ B(20,30),\ C(0,50)$$

Step 4: Evaluate the objective function:

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

$$Z(30,0) = 4(30) = 120$$

$$Z(20,30) = 80 + 30 = 110$$

$$Z(0,50) = 50$$

Step 5: The largest value is 120.

Answer: Maximum $Z = 120$ at $(30,0)$.

Question 8 [CBSE 2024]

A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs 17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day to maximize his profit, if he operates his machines for at the most 12 hours a day? Formulate the LPP.

Solution: Step 1: Let $x$ be the number of packages of nuts and $y$ be the number of packages of bolts.

Step 2: Profit function:

$$Z = 17.5x + 7y$$

which is to be maximized.

Step 3: Machine A is available for at most 12 hours. Nuts require 1 hour and bolts require 3 hours, so:

$$x + 3y \le 12$$

Step 4: Machine B is available for at most 12 hours. Nuts require 3 hours and bolts require 1 hour, so:

$$3x + y \le 12$$

Step 5: Non-negativity restrictions:

$$x \ge 0, \quad y \ge 0$$

Answer: Maximize $Z = 17.5x + 7y$ subject to $x + 3y \le 12$, $3x + y \le 12$, $x, y \ge 0$.

Question 9 [CBSE 2025]

Solve graphically: Minimize $Z = 3x + 2y$ subject to $x + y \ge 8$, $3x + 5y \le 15$, $x, y \ge 0$.

Solution: Step 1: The line $x + y = 8$ cuts the axes at $(8,0)$ and $(0,8)$. Since the inequality is $x + y \ge 8$, the feasible side is above this line.

Step 2: The line $3x + 5y = 15$ cuts the axes at $(5,0)$ and $(0,3)$. Since the inequality is $3x + 5y \le 15$, the feasible side is below this line.

Step 3: In the first quadrant, the whole region satisfying $3x + 5y \le 15$ lies much closer to the origin than the region satisfying $x + y \ge 8$.

Step 4: Therefore the two half-planes have no common point in the first quadrant. So the feasible region is empty.

Answer: No feasible solution exists.

Question 10 [CBSE 2022]

Find the maximum and minimum values of $Z = 5x + 2y$ subject to $x - 2y \le 2$, $3x + 2y \le 12$, $-3x + 2y \le 3$, $x \ge 0, y \ge 0$.

Solution: Step 1: Rewrite the constraints in boundary-line form:

$$x - 2y \le 2$$

$$3x + 2y \le 12$$

$$-3x + 2y \le 3$$

with $x, y \ge 0$. The feasible region is bounded in the first quadrant.

Step 2: Find important intersection points:

• Intersection of $x - 2y = 2$ and $3x + 2y = 12$: Add the equations:

$$4x = 14 \Rightarrow x = 3.5$$

Then

$$3.5 - 2y = 2 \Rightarrow 2y = 1.5 \Rightarrow y = 0.75$$

So point $A(3.5,0.75)$.

• Intersection of $3x + 2y = 12$ and $-3x + 2y = 3$: Add the equations:

$$4y = 15 \Rightarrow y = 3.75$$

Then

$$3x + 7.5 = 12 \Rightarrow 3x = 4.5 \Rightarrow x = 1.5$$

So point $B(1.5,3.75)$.

• On the x-axis ($y=0$), the condition $x - 2y \le 2$ gives $x \le 2$, so $(2,0)$ is a corner point.
• On the y-axis ($x=0$), the condition $-3x + 2y \le 3$ gives $y \le 1.5$, so $(0,1.5)$ is a corner point.
• The origin $(0,0)$ is also feasible.

Step 3: Thus the corner points are:

$$(0,0),\ (2,0),\ (3.5,0.75),\ (1.5,3.75),\ (0,1.5)$$

Step 4: Evaluate $Z = 5x + 2y$ at each point:

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

$$Z(2,0) = 10$$

$$Z(3.5,0.75) = 17.5 + 1.5 = 19$$

$$Z(1.5,3.75) = 7.5 + 7.5 = 15$$

$$Z(0,1.5) = 3$$

Step 5: Therefore:

$$\text{Maximum value} = 19$$

$$\text{Minimum value} = 0$$

Answer: Maximum $Z = 19$ at $(3.5,0.75)$ and minimum $Z = 0$ at $(0,0)$.

Question 11 [CBSE 2026]

A firm manufactures two types of products, A and B, and sells them at a profit of Rs 2 on type A and Rs 3 on type B. Each product is processed on two machines $M_1$ and $M_2$. Type A requires 1 minute of processing time on $M_1$ and 2 minutes on $M_2$. Type B requires 1 minute on $M_1$ and 1 minute on $M_2$. The machine $M_1$ is available for not more than 6 hours 40 minutes, while machine $M_2$ is available for 10 hours during any working day. Formulate the LPP to maximize profit.

Solution: Step 1: Let $x$ be the number of units of product A and $y$ be the number of units of product B.

Step 2: Profit function:

$$Z = 2x + 3y$$

which is to be maximized.

Step 3: Convert working times into minutes:

$$6\text{ hours }40\text{ minutes} = 6\times 60 + 40 = 400\text{ minutes}$$

$$10\text{ hours} = 600\text{ minutes}$$

Step 4: Machine $M_1$ time: Type A takes 1 minute and type B takes 1 minute. So,

$$x + y \le 400$$

Step 5: Machine $M_2$ time: Type A takes 2 minutes and type B takes 1 minute. So,

$$2x + y \le 600$$

Step 6: Non-negativity restrictions:

$$x \ge 0, \quad y \ge 0$$

Answer: Maximize $Z = 2x + 3y$ subject to $x + y \le 400$, $2x + y \le 600$, $x, y \ge 0$.

Question 12 [CBSE 2024]

Solve the LPP graphically: Maximize $Z = 3x + 2y$ subject to $x + 2y \le 10$, $3x + y \le 15$, $x, y \ge 0$.

Solution: Step 1: Draw the lines $x + 2y = 10$ and $3x + y = 15$. The feasible region lies below both lines in the first quadrant.

Step 2: Find their intersection: Multiply $x + 2y = 10$ by 3:

$$3x + 6y = 30$$

Subtract $3x + y = 15$:

$$5y = 15 \Rightarrow y = 3$$

Then

$$x = 4$$

So the intersection point is $(4,3)$.

Step 3: The corner points are:

$$(0,0),\ (5,0),\ (4,3),\ (0,5)$$

Step 4: Evaluate $Z = 3x + 2y$:

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

$$Z(5,0) = 15$$

$$Z(4,3) = 18$$

$$Z(0,5) = 10$$

Step 5: The maximum value is 18.

Answer: Maximum $Z = 18$ at $(4,3)$.

Question 13 [CBSE 2025]

A merchant plans to sell two types of personal computers: a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computer which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000. Formulate as LPP.

Solution: Step 1: Let $x$ be the number of desktop models and $y$ be the number of portable models.

Step 2: Profit function:

$$Z = 4500x + 5000y$$

which is to be maximized.

Step 3: Total monthly demand does not exceed 250 units:

$$x + y \le 250$$

Step 4: Total investment should not exceed Rs 70,00,000:

$$25000x + 40000y \le 7000000$$

Divide by 5000:

$$5x + 8y \le 1400$$

Step 5: Non-negativity restrictions:

$$x \ge 0, \quad y \ge 0$$

Answer: Maximize $Z = 4500x + 5000y$ subject to $x + y \le 250$, $5x + 8y \le 1400$, $x, y \ge 0$.

Question 14 [CBSE 2023]

Solve graphically: Minimize $Z = -3x + 4y$ subject to $x + 2y \le 8$, $3x + 2y \le 12$, $x \ge 0$, $y \ge 0$.

Solution: Step 1: Draw the lines $x + 2y = 8$ and $3x + 2y = 12$. Since both inequalities are of type $\le$, the feasible region lies below both lines in the first quadrant.

Step 2: Find the intersection point: Subtract the first equation from the second:

$$2x = 4 \Rightarrow x = 2$$

Then

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

So the point is $(2,3)$.

Step 3: The corner points are:

$$(0,0),\ (4,0),\ (2,3),\ (0,4)$$

Step 4: Evaluate $Z = -3x + 4y$:

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

$$Z(4,0) = -12$$

$$Z(2,3) = -6 + 12 = 6$$

$$Z(0,4) = 16$$

Step 5: The minimum value is $-12$.

Answer: Minimum $Z = -12$ at $(4,0)$.

Question 15 [CBSE 2026]

A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B, and 8 units of vitamin C. The vitamin contents of one kg food is given below: Food X contains 1, 2, 3 units of Vitamin A, B, C respectively. Food Y contains 2, 2, 1 units respectively. One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Formulate LPP to minimize the cost.

Solution: Step 1: Let $x$ kg of food X and $y$ kg of food Y be used.

Step 2: Since the aim is to minimize cost,

$$Z = 16x + 20y$$

Step 3: Vitamin A requirement:

$$x + 2y \ge 10$$

Step 4: Vitamin B requirement:

$$2x + 2y \ge 12$$

Simplify:

$$x + y \ge 6$$

Step 5: Vitamin C requirement:

$$3x + y \ge 8$$

Step 6: Non-negativity restrictions:

$$x \ge 0, \quad y \ge 0$$

Answer: Minimize $Z = 16x + 20y$ subject to $x + 2y \ge 10$, $x + y \ge 6$, $3x + y \ge 8$, $x, y \ge 0$.

Question 16 [CBSE 2024]

Solve graphically: Maximize $Z = 5x + 3y$ subject to $3x + 5y \le 15$, $5x + 2y \le 10$, $x \ge 0$, $y \ge 0$.

Solution: Step 1: Draw the lines $3x + 5y = 15$ and $5x + 2y = 10$. The feasible region lies below both lines in the first quadrant.

Step 2: Find the point of intersection: Multiply $3x + 5y = 15$ by 5:

$$15x + 25y = 75$$

Multiply $5x + 2y = 10$ by 3:

$$15x + 6y = 30$$

Subtract:

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

Substitute into $5x + 2y = 10$:

$$5x + \frac{90}{19} = 10 = \frac{190}{19}$$

$$5x = \frac{100}{19}$$

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

So intersection point is $\left(\frac{20}{19}, \frac{45}{19}\right)$.

Step 3: The other corner points are the intercepts nearest the origin:

$$(2,0) \text{ from } 5x + 2y = 10$$

$$(0,3) \text{ from } 3x + 5y = 15$$

Thus the corner points are $(0,0)$, $(2,0)$, $\left(\frac{20}{19},\frac{45}{19}\right)$, $(0,3)$.

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

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

$$Z(2,0) = 10$$

$$Z\left(\frac{20}{19},\frac{45}{19}\right) = \frac{100}{19} + \frac{135}{19} = \frac{235}{19}$$

$$Z(0,3) = 9$$

Step 5: The maximum value is $\frac{235}{19}$.

Answer: Maximum $Z = \frac{235}{19}$ at $\left(\frac{20}{19},\frac{45}{19}\right)$.

Question 17 [CBSE 2025]

A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman's time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftsman's time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman's time. What number of rackets and bats must be made if the profit on a racket and on a bat is Rs 20 and Rs 10 respectively? Formulate LPP.

Solution: Step 1: Let $x$ be the number of tennis rackets and $y$ be the number of cricket bats.

Step 2: Profit function:

$$Z = 20x + 10y$$

which is to be maximized.

Step 3: Machine time constraint:

$$1.5x + 3y \le 42$$

Multiply by 2:

$$3x + 6y \le 84$$

Divide by 3:

$$x + 2y \le 28$$

Step 4: Craftsman's time constraint:

$$3x + y \le 24$$

Step 5: Non-negativity restrictions:

$$x \ge 0, \quad y \ge 0$$

Answer: Maximize $Z = 20x + 10y$ subject to $x + 2y \le 28$, $3x + y \le 24$, $x, y \ge 0$.

Question 18 [CBSE 2022]

Solve graphically: Minimize $Z = 3x + 5y$ subject to $x + 3y \ge 3$, $x + y \ge 2$, $x, y \ge 0$.

Solution: Step 1: Draw the lines $x + 3y = 3$ and $x + y = 2$. Since both inequalities are of type $\ge$, the feasible region lies above both lines in the first quadrant, so it is unbounded.

Step 2: Find their intersection: Subtract the second equation from the first:

$$2y = 1 \Rightarrow y = 0.5$$

Then

$$x = 1.5$$

So the point is $(1.5,0.5)$.

Step 3: The other corner points on the boundary are $(3,0)$ and $(0,2)$.

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

$$Z(3,0) = 9$$

$$Z(1.5,0.5) = 4.5 + 2.5 = 7$$

$$Z(0,2) = 10$$

Step 5: The least value among corner points is 7. Because the region is unbounded, we verify that no feasible point can make $3x + 5y < 7$. The half-plane $3x + 5y < 7$ has no common point with the feasible region.

Answer: Minimum $Z = 7$ at $(1.5,0.5)$.

Question 19 [CBSE 2026]

A manufacturer produces two types of steel trunks. He has two machines, A and B. The first type of trunk requires 3 hours on machine A and 3 hours on machine B. The second type requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs 30 and Rs 25 per trunk on the first and second type respectively. Formulate LPP to maximize profit.

Solution: Step 1: Let $x$ be the number of trunks of type 1 and $y$ be the number of trunks of type 2.

Step 2: Profit function:

$$Z = 30x + 25y$$

which is to be maximized.

Step 3: Machine A can work for at most 18 hours:

$$3x + 3y \le 18$$

Divide by 3:

$$x + y \le 6$$

Step 4: Machine B can work for at most 15 hours:

$$3x + 2y \le 15$$

Step 5: Non-negativity restrictions:

$$x \ge 0, \quad y \ge 0$$

Answer: Maximize $Z = 30x + 25y$ subject to $x + y \le 6$, $3x + 2y \le 15$, $x, y \ge 0$.

Question 20 [CBSE 2024]

Solve the LPP graphically: Maximize $Z = 5x + 7y$ subject to $x + y \le 4$, $3x + 8y \le 24$, $10x + 7y \le 35$, $x, y \ge 0$.

Solution: Step 1: Draw the three boundary lines:

$$x + y = 4$$

$$3x + 8y = 24$$

$$10x + 7y = 35$$

The feasible region lies below all three lines in the first quadrant.

Step 2: Find the useful pairwise intersections.

• Intersection of $x + y = 4$ and $10x + 7y = 35$: Substitute $y = 4 - x$ into the second equation:

$$10x + 7(4 - x) = 35$$

$$10x + 28 - 7x = 35$$

$$3x = 7 \Rightarrow x = \frac{7}{3}$$

Then

$$y = 4 - \frac{7}{3} = \frac{5}{3}$$

So one corner point is $A\left(\frac{7}{3},\frac{5}{3}\right)$.

• Intersection of $x + y = 4$ and $3x + 8y = 24$: Substitute $y = 4 - x$:

$$3x + 8(4 - x) = 24$$

$$3x + 32 - 8x = 24$$

$$-5x = -8 \Rightarrow x = \frac{8}{5}$$

Then

$$y = 4 - \frac{8}{5} = \frac{12}{5}$$

So another corner point is $B\left(\frac{8}{5},\frac{12}{5}\right)$.

• Intersection of $3x + 8y = 24$ and $10x + 7y = 35$: Multiply the first equation by 10 and the second by 3:

$$30x + 80y = 240$$

$$30x + 21y = 105$$

Subtracting gives

$$59y = 135 \Rightarrow y = \frac{135}{59}$$

Then

$$x = \frac{112}{59}$$

But for this point,

$$x + y = \frac{112 + 135}{59} = \frac{247}{59} > 4$$

so it does not satisfy the constraint $x + y \le 4$. Hence this point is not feasible and is not a corner point of the feasible region.

Step 3: The axis intercept corner points are:

• On x-axis ($y=0$): $x + y \le 4$ gives $x \le 4$, $3x + 8y \le 24$ gives $x \le 8$, $10x + 7y \le 35$ gives $x \le 3.5$. So the feasible x-axis corner point is $(3.5,0)$.
• On y-axis ($x=0$): $x + y \le 4$ gives $y \le 4$, $3x + 8y \le 24$ gives $y \le 3$, $10x + 7y \le 35$ gives $y \le 5$. So the feasible y-axis corner point is $(0,3)$.
• The origin $(0,0)$ is also feasible.

Step 4: Therefore the corner points of the feasible region are:

$$(0,0),\ (3.5,0),\ \left(\frac{7}{3},\frac{5}{3}\right),\ \left(\frac{8}{5},\frac{12}{5}\right),\ (0,3)$$

Step 5: Evaluate $Z = 5x + 7y$:

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

$$Z(3.5,0) = 17.5$$

$$Z\left(\frac{7}{3},\frac{5}{3}\right) = \frac{35}{3} + \frac{35}{3} = \frac{70}{3} \approx 23.33$$

$$Z\left(\frac{8}{5},\frac{12}{5}\right) = 8 + \frac{84}{5} = \frac{124}{5} = 24.8$$

$$Z(0,3) = 21$$

Step 6: The maximum value is $\frac{124}{5}$.

Answer: Maximum $Z = \frac{124}{5}$ at $\left(\frac{8}{5},\frac{12}{5}\right)$.

Question 21 [CBSE 2025]

A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is at most 24. It takes 1 hour to make a ring and 30 minutes to make a chain. The maximum number of hours available per day is 16. If the profit on a ring is Rs 300 and on a chain is Rs 190, find the number of rings and chains that should be manufactured to maximize profit. Formulate LPP.

Solution: Step 1: Let $x$ be the number of gold rings and $y$ be the number of gold chains.

Step 2: Profit function:

$$Z = 300x + 190y$$

which is to be maximized.

Step 3: Total number of items cannot exceed 24:

$$x + y \le 24$$

Step 4: Time available per day is 16 hours. One ring takes 1 hour and one chain takes 0.5 hour, so:

$$x + 0.5y \le 16$$

Multiply by 2:

$$2x + y \le 32$$

Step 5: Non-negativity restrictions:

$$x \ge 0, \quad y \ge 0$$

Answer: Maximize $Z = 300x + 190y$ subject to $x + y \le 24$, $2x + y \le 32$, $x, y \ge 0$.

Question 22 [CBSE 2023]

Solve graphically: Minimize $Z = 200x + 500y$ subject to $x + 2y \ge 10$, $3x + 4y \le 24$, $x, y \ge 0$.

Solution: Step 1: Draw the lines $x + 2y = 10$ and $3x + 4y = 24$. The feasible region lies above the first line and below the second line in the first quadrant.

Step 2: Find the intersection point: Multiply $x + 2y = 10$ by 2:

$$2x + 4y = 20$$

Subtract from $3x + 4y = 24$:

$$x = 4$$

Then

$$4 + 2y = 10 \Rightarrow y = 3$$

So the point is $(4,3)$.

Step 3: On the y-axis, the two lines give $(0,5)$ and $(0,6)$, and both satisfy the inequalities on the boundary. Hence the corner points are $(0,5)$, $(0,6)$, $(4,3)$.

Step 4: Evaluate $Z = 200x + 500y$:

$$Z(0,5) = 2500$$

$$Z(0,6) = 3000$$

$$Z(4,3) = 800 + 1500 = 2300$$

Step 5: The minimum value is 2300.

Answer: Minimum $Z = 2300$ at $(4,3)$.

Question 23 [CBSE 2026]

A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3g of silver and 1g of gold while that of type B requires 1g of silver and 2g of gold. The company can use at most 9g of silver and 8g of gold. If each unit of type A brings a profit of Rs 40 and that of type B Rs 50, formulate LPP to maximize profit.

Solution: Step 1: Let $x$ be the number of units of A and $y$ be the number of units of B.

Step 2: Profit function:

$$Z = 40x + 50y$$

which is to be maximized.

Step 3: Silver restriction: Type A requires 3g silver and type B requires 1g silver. Total silver available is 9g. Hence,

$$3x + y \le 9$$

Step 4: Gold restriction: Type A requires 1g gold and type B requires 2g gold. Total gold available is 8g. Hence,

$$x + 2y \le 8$$

Step 5: Non-negativity restrictions:

$$x \ge 0, \quad y \ge 0$$

Answer: Maximize $Z = 40x + 50y$ subject to $3x + y \le 9$, $x + 2y \le 8$, $x, y \ge 0$.

Question 24 [CBSE 2022]

Determine graphically the minimum value of the objective function $Z = -50x + 20y$ subject to the constraints $2x - y \ge -5$, $3x + y \ge 3$, $2x - 3y \le 12$, $x, y \ge 0$.

Solution: Step 1: Rewrite the inequalities in a clearer form:

$$2x - y \ge -5 \Rightarrow y \le 2x + 5$$

$$3x + y \ge 3 \Rightarrow y \ge 3 - 3x$$

$$2x - 3y \le 12 \Rightarrow y \ge \frac{2x - 12}{3}$$

with $x \ge 0, y \ge 0$.

Step 2: So the feasible region is the set of points in the first quadrant lying below $y = 2x + 5$ and above both $y = 3 - 3x$ and $y = \frac{2x - 12}{3}$. This region is unbounded to the right.

Step 3: Identify the visible corner points in the first quadrant:

• On the y-axis, $3x + y = 3$ gives $(0,3)$.
• On the y-axis, $y = 2x + 5$ gives $(0,5)$.
• On the x-axis, $3x + y = 3$ gives $(1,0)$.
• On the x-axis, $2x - 3y = 12$ gives $(6,0)$. So relevant boundary corner points are $(0,3)$, $(0,5)$, $(1,0)$, $(6,0)$.

Step 4: Evaluate the objective function at these points:

$$Z(0,3) = 60$$

$$Z(0,5) = 100$$

$$Z(1,0) = -50$$

$$Z(6,0) = -300$$

Step 5: The least value among these corner points is $-300$ at $(6,0)$.

Step 6: However, since the feasible region is unbounded, we must check whether even smaller values are possible. Take a feasible point on the line $y = \frac{2x - 12}{3}$, for example $x = 12$. Then

$$y = \frac{24 - 12}{3} = 4$$

So $(12,4)$ is feasible. Now,

$$Z(12,4) = -50(12) + 20(4) = -600 + 80 = -520 < -300$$

Hence values smaller than $-300$ occur in the feasible region.

Step 7: In fact, moving further to the right along the feasible region makes $Z$ decrease without bound.

Answer: No minimum value exists.

Question 25 [CBSE 2024]

A dealer wishes to purchase a number of fans and sewing machines. He has only Rs 5,760 to invest and has space for at most 20 items. A fan costs him Rs 360 and a sewing machine Rs 240. His expectation is that he can sell a fan at a profit of Rs 22 and a sewing machine at a profit of Rs 18. Assuming that he can sell all the items that he can buy, how should he invest his money to maximize profit? Formulate as LPP.

Solution: Step 1: Let $x$ be the number of fans and $y$ be the number of sewing machines.

Step 2: Profit function:

$$Z = 22x + 18y$$

which is to be maximized.

Step 3: Space restriction: total number of items cannot exceed 20:

$$x + y \le 20$$

Step 4: Investment restriction:

$$360x + 240y \le 5760$$

Divide by 120:

$$3x + 2y \le 48$$

Step 5: Non-negativity restrictions:

$$x \ge 0, \quad y \ge 0$$

Answer: Maximize $Z = 22x + 18y$ subject to $x + y \le 20$, $3x + 2y \le 48$, $x, y \ge 0$.

Question 26 [CBSE 2025]

Solve graphically: Maximize $Z = x + 2y$ subject to $x - y \le 10$, $x + y \le 20$, $x, y \ge 0$.

Solution: Step 1: Draw the lines $x - y = 10$ and $x + y = 20$ in the first quadrant. The inequalities are:

$$x - y \le 10 \Rightarrow y \ge x - 10$$

$$x + y \le 20$$

with $x, y \ge 0$.

Step 2: Find the intersection of the two lines:

$$x - y = 10$$

$$x + y = 20$$

Add them:

$$2x = 30 \Rightarrow x = 15$$

Then

$$y = 5$$

So the intersection point is $(15,5)$.

Step 3: The corner points of the feasible region are:

$$(0,0),\ (10,0),\ (15,5),\ (0,20)$$

Step 4: Evaluate $Z = x + 2y$:

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

$$Z(10,0) = 10$$

$$Z(15,5) = 15 + 10 = 25$$

$$Z(0,20) = 40$$

Step 5: The maximum value is 40.

Answer: Maximum $Z = 40$ at $(0,20)$.

Question 27 [CBSE 2023]

A retired person wants to invest an amount of up to Rs 50,000. His broker recommends investing in two types of bonds A and B yielding 10% and 9% return respectively on the invested amount. He decides to invest at least Rs 20,000 in bond A and at least Rs 10,000 in bond B. He also wants to invest at least as much in bond A as in bond B. Solve this LPP graphically to maximize his returns.

Solution: Step 1: Let $x$ be the amount invested in bond A and $y$ be the amount invested in bond B.

Step 2: Return function:

$$Z = 0.10x + 0.09y$$

which is to be maximized.

Step 3: Constraints: Total investment at most Rs 50,000:

$$x + y \le 50000$$

At least Rs 20,000 in bond A:

$$x \ge 20000$$

At least Rs 10,000 in bond B:

$$y \ge 10000$$

Bond A investment at least as much as bond B investment:

$$x \ge y$$

Step 4: Corner points of the feasible region:

• $(20000,10000)$
• $(40000,10000)$ from $y = 10000$ and $x + y = 50000$
• $(25000,25000)$ from $x = y$ and $x + y = 50000$
• $(20000,20000)$ from $x = 20000$ and $x = y$

Step 5: Evaluate $Z$:

$$Z(20000,10000) = 2000 + 900 = 2900$$

$$Z(40000,10000) = 4000 + 900 = 4900$$

$$Z(25000,25000) = 2500 + 2250 = 4750$$

$$Z(20000,20000) = 2000 + 1800 = 3800$$

Step 6: Maximum return is 4900.

Answer: Maximum return is Rs 4900 when Rs 40,000 is invested in bond A and Rs 10,000 in bond B.