Linear Programming

1

MCQ · Numerical

A dealer can invest at most ₹50,000. A table costs ₹2500 and a chair costs ₹500. If he buys x tables and y chairs, which of the following correctly represents the investment constraint?

A

2500x + 500y ≥ 50000

B

5x + y ≤ 100

C

x + y ≤ 50

D

500x + 2500y ≤ 50000

✔ Solution

Investment = 2500x + 500y ≤ 50000. Dividing throughout by 500 gives 5x + y ≤ 100.

Answer: B

2

MCQ · Numerical

For the objective function Z = 250x + 75y, evaluate Z at the corner point B(10, 50) of the feasible region. What is the value of Z?

A

₹4500

B

₹5000

C

₹6250

D

₹7000

✔ Solution

Z = 250(10) + 75(50) = 2500 + 3750 = 6250.

Answer: C — ₹6250

3

MCQ · Numerical

Maximise Z = 4x + y subject to x + y ≤ 50, 3x + y ≤ 90, x, y ≥ 0. What is the maximum value of Z?

A

50

B

110

C

120

D

130

✔ Solution

Corner points: O(0,0)→Z=0; A(30,0)→Z=120; B(20,30)→Z=110; C(0,50)→Z=50.

Maximum value of Z = 120 at point A(30, 0).

Answer: C — 120

4

MCQ · Numerical

Minimise Z = 200x + 500y subject to x + 2y ≥ 10, 3x + 4y ≤ 24, x, y ≥ 0. The minimum value of Z is:

A

2500

B

2300

C

3000

D

1800

✔ Solution

Corner points: A(0,5)→Z=2500; B(4,3)→Z=200(4)+500(3)=800+1500=2300; C(0,6)→Z=3000.

Minimum value = 2300 at B(4, 3).

Answer: B — 2300

5

MCQ · Concept

The optimal value of the objective function in a Linear Programming Problem is always attained at:

A

Any interior point of feasible region

B

A corner point (vertex) of the feasible region

C

The centre of the feasible region

D

The midpoint of any boundary line

✔ Solution

By Theorem 1 (Corner Point Theorem): When Z has an optimal value, it must occur at a corner point (vertex) of the feasible region.

Answer: B

6

MCQ · Numerical

For Z = 3x + 9y subject to x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x, y ≥ 0: the minimum value of Z is:

A

90

B

60

C

180

D

30

✔ Solution

Corner points: A(0,10)→90; B(5,5)→60; C(15,15)→180; D(0,20)→180.

Minimum value = 60 at B(5, 5).

Answer: B — 60

7

MCQ · Numerical

For Z = 3x + 9y (same problem as Q6), the maximum value of Z is:

A

90

B

120

C

180

D

200

✔ Solution

Maximum occurs at both C(15,15) and D(0,20), each giving Z = 180. This is a case of multiple optimal solutions.

Answer: C — 180

8

MCQ · Numerical

Maximise Z = 3x + 4y subject to x + y ≤ 4, x ≥ 0, y ≥ 0. The maximum value of Z is:

A

12

B

16

C

14

D

10

✔ Solution

Corner points: O(0,0)→0; A(4,0)→12; B(0,4)→16.

Maximum Z = 3(0)+4(4) = 16 at (0, 4).

Answer: B — 16

9

MCQ · Numerical

Minimise Z = –3x + 4y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x, y ≥ 0. The minimum value of Z is:

A

–12

B

0

C

4

D

8

✔ Solution

Corner points: O(0,0)→0; A(4,0)→–12; B(2,3)→–6+12=6; C(0,4)→16.

Minimum value = –12 at A(4, 0).

Answer: A — –12

10

MCQ · Numerical

Maximise Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0. The maximum value of Z is:

A

15

B

18

C

20

D

12

✔ Solution

Intersection of 3x+y=15 and x+2y=10: solving → x=4, y=3. Corner points: O(0,0)→0; A(5,0)→15; B(4,3)→12+6=18; C(0,5)→10.

Maximum Z = 18 at B(4, 3).

Answer: B — 18

11

MCQ · Numerical

A dealer buys only chairs (no tables) with ₹50,000. Each chair costs ₹500 but storage allows only 60 pieces. His maximum profit is (profit per chair = ₹75):

A

₹7500

B

₹5000

C

₹4500

D

₹6000

✔ Solution

He can buy 50000÷500 = 100 chairs but storage allows only 60. So profit = 60 × 75 = ₹4500.

Answer: C — ₹4500

12

MCQ · Numerical

For the LPP: Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x, y ≥ 0. The maximum value of Z is approximately:

A

9

B

235/19

C

15

D

10

✔ Solution

Solving 3x+5y=15 and 5x+2y=10: x=20/19, y=45/19. Z = 5(20/19)+3(45/19) = 100/19+135/19 = 235/19 ≈ 12.37.

Corner O(0,0)→0; A(2,0)→10; B(20/19,45/19)→235/19; C(0,3)→9.

Answer: B — 235/19

13

MCQ · Concept

If two corner points of a bounded feasible region yield the same maximum value of the objective function, then:

A

No maximum exists

B

Every point on the line segment joining them gives the same maximum value

C

Only the midpoint gives the maximum value

D

The problem has no solution

✔ Solution

This is the case of Multiple Optimal Solutions. Every point on the line segment joining the two corner points also gives the same maximum value of Z.

Answer: B

14

MCQ · Numerical

Minimise Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0. At which point does the minimum occur?

A

(0, 3)

B

(3, 0)

C

(6, 0)

D

Minimum occurs along the entire line segment from (0,3) to (6,0)

✔ Solution

At (0,3): Z=0+6=6. At (6,0): Z=6+0=6. Both yield Z=6, and since the objective line has the same slope as the constraint boundary x+2y=6, minimum occurs at every point of that segment.

Answer: D

15

MCQ · Numerical

A dealer buys 10 tables and 50 chairs. Profit per table = ₹250, profit per chair = ₹75. What is his total profit?

A

₹5000

B

₹4500

C

₹6250

D

₹7250

✔ Solution

Total profit = 250×10 + 75×50 = 2500 + 3750 = ₹6250.

Answer: C — ₹6250

16

One-Word Ans. · Numerical

If a dealer buys only tables (no chairs) with ₹50,000 at ₹2500 each, how many tables can he buy?

✔ Solution

Number of tables = 50000 ÷ 2500 = 20.

Answer: 20

17

One-Word Ans. · Numerical

For Z = 250x + 75y, evaluate Z at corner point O(0, 0).

✔ Solution

Z = 250(0) + 75(0) = 0.

Answer: 0

18

One-Word Ans. · Numerical

For Z = 4x + y subject to given constraints, the maximum occurs at corner point A. What is the x-coordinate of A?

✔ Solution

From Example 1 in NCERT: Maximum Z = 120 at A(30, 0). The x-coordinate of A is 30.

Answer: 30

19

One-Word Ans. · Numerical

How many corner points does the feasible region OABC have (as formed in the furniture dealer problem)?

✔ Solution

The feasible region OABC has vertices O, A, B, and C — a total of 4 corner points.

Answer: 4

20

One-Word Ans. · Numerical

For Z = 200x + 500y, compute Z at corner point B(4, 3). What is its value?

✔ Solution

Z = 200(4) + 500(3) = 800 + 1500 = 2300.

Answer: 2300

21

One-Word Ans. · Numerical

The storage constraint in the furniture dealer problem restricts the total number of items to at most how many pieces?

✔ Solution

The storage constraint is x + y ≤ 60.

Answer: 60

22

One-Word Ans. · Numerical

Maximise Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0. What is the y-coordinate of the optimal corner point?

✔ Solution

Optimal point is B(4, 3) where Z = 18 (maximum). The y-coordinate is 3.

Answer: 3

23

One-Word Ans. · Numerical

A dealer's profit per table is ₹250. If he buys 20 tables only, what is his total profit (in ₹)?

✔ Solution

Profit = 250 × 20 = ₹5000.

Answer: 5000

24

One-Word Ans. · Numerical

For Z = 3x + 9y subject to given constraints, what is the minimum value of Z at corner point B(5, 5)?

✔ Solution

Z = 3(5) + 9(5) = 15 + 45 = 60. This is the minimum value.

Answer: 60

25

One-Word Ans. · Numerical

In the furniture dealer LPP, the optimal strategy is to buy 10 tables and 50 chairs. What is the maximum profit (in ₹) the dealer earns?

✔ Solution

Z = 250(10) + 75(50) = 2500 + 3750 = ₹6250. This is the maximum profit.

Answer: 6250

Multiple Choice Questions

One-Word / One-Value Answer (Numerical)

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