CBSE Class 12 · Chapter 12 · Linear Programming · 2026-27

Linear Programming Class 12: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceCorner points and optimal values

A bounded feasible region has corner points $(0, 0)$ , $(7, 0)$ , $(5, 4)$ and $(0, 6)$ . The maximum of $Z = 3 x + 4 y$ over the region is

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Answer: (a)

31

Why: Evaluate Z at each corner: 0, 21, 31, 24.

$Z (0, 0) = 0$ , $Z (7, 0) = 21$ , $Z (5, 4) = 15 + 16 = 31$ , $Z (0, 6) = 24$ . Maximum $= 31$ at $(5, 4)$ .

Q2

·1 mark·Multiple choiceCorner points and optimal values

Over the same region (corners $(0, 0)$ , $(7, 0)$ , $(5, 4)$ , $(0, 6)$ ), the least value of $Z = x - 2 y$ is

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Answer: (b)

- 12

Why: Z values: 0, 7, −3, −12.

$Z (0, 0) = 0$ , $Z (7, 0) = 7$ , $Z (5, 4) = 5 - 8 = - 3$ , $Z (0, 6) = - 12$ . Minimum $= - 12$ at $(0, 6)$ .

Q3

·1 mark·Multiple choiceFeasible regions

Which point lies in the feasible region of $2 x + y \leq 10$ , $x + 3 y \leq 15$ , $x \geq 0$ , $y \geq 0$ ?

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Answer: (c)

(3, 3)

Why: (3, 3): 9 ≤ 10 and 12 ≤ 15, with x, y ≥ 0.

$(3, 3)$ : $6 + 3 = 9 \leq 10$ , $3 + 9 = 12 \leq 15$ ✓. $(4, 3)$ : $11 > 10$ ; $(1, 5)$ : $16 > 15$ ; $(- 1, 2)$ : $x < 0$ .

Q4

·1 mark·Multiple choiceTerminology and formulation

The feasible region of an LPP is bounded, and $Z = a x + b y$ takes its maximum value at two adjacent corner points $P$ and $Q$ . Then $Z$ is maximum

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Answer: (a) at every point of the segment

P Q

Why: Z is linear, so along the edge PQ it takes every value between Z(P) and Z(Q), which are equal.

On the segment $P Q$ , a point is $(1 - t) P + t Q$ with $0 \leq t \leq 1$ , and since $Z$ is linear, $Z = (1 - t) Z (P) + t Z (Q) = Z (P)$ . So every point of $P Q$ gives the same maximum; there are infinitely many optimal solutions.

Q5

·1 mark·Multiple choiceFeasible regions

The corner points of the feasible region given by $x + y \leq 4$ , $x \geq 0$ , $y \geq 0$ are

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Answer: (a)

(0, 0), (4, 0), (0, 4)

Why: The region is the triangle cut off by x + y = 4 and the axes.

The boundary lines $x + y = 4$ , $x = 0$ , $y = 0$ meet in pairs at $(0, 0)$ , $(4, 0)$ , $(0, 4)$ , all feasible: a triangle.

Q6

·1 mark·Multiple choiceCorner points and optimal values

The maximum value of $Z = 5 x + 3 y$ subject to $x + y \leq 4$ , $x \geq 0$ , $y \geq 0$ is

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Answer: (b)

20

Why: Corners (0, 0), (4, 0), (0, 4) give 0, 20, 12.

$Z (0, 0) = 0$ , $Z (4, 0) = 20$ , $Z (0, 4) = 12$ . Maximum $= 20$ at $(4, 0)$ .

Q7

·1 mark·Multiple choiceTerminology and formulation

If the feasible region of an LPP is bounded, the maximum and minimum of the objective function $Z = a x + b y$ occur

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Answer: (c) at corner points of the region

Why: Corner point theorem.

For a bounded feasible region, $Z$ has both a maximum and a minimum, and each occurs at a corner (vertex) of the region.

Q8

·1 mark·Multiple choiceCorner points and optimal values

The maximum value of $Z = x + y$ subject to $x + 3 y \leq 12$ , $3 x + y \leq 12$ , $x \geq 0$ , $y \geq 0$ is

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Answer: (d)

6

Why: Corners (0, 0), (4, 0), (3, 3), (0, 4): Z = 0, 4, 6, 4.

$x + 3 y = 12$ and $3 x + y = 12$ meet at $(3, 3)$ . Corners $(0, 0), (4, 0), (3, 3), (0, 4)$ give $Z = 0, 4, 6, 4$ . Maximum $= 6$ .

Q9

·1 mark·Multiple choiceCorner points and optimal values

The minimum value of $Z = 3 x + 2 y$ subject to $x + y \geq 6$ , $x \leq 4$ , $y \leq 5$ , $x \geq 0$ , $y \geq 0$ is

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Answer: (a)

13

Why: Corners (1, 5), (4, 2), (4, 5): Z = 13, 16, 22.

The feasible region is the triangle with corners $(1, 5)$ , $(4, 2)$ , $(4, 5)$ . $Z = 13, 16, 22$ ; minimum $= 13$ at $(1, 5)$ .

Q10

·1 mark·Multiple choiceCorner points and optimal values

The maximum value of $Z = 2 x + 5 y$ subject to $0 \leq x \leq 3$ , $0 \leq y \leq 2$ is

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Answer: (c)

16

Why: The region is a rectangle; Z is largest at (3, 2).

Corners $(0, 0), (3, 0), (3, 2), (0, 2)$ : $Z = 0, 6, 16, 10$ . Maximum $= 16$ at $(3, 2)$ .

Q11

·1 mark·Multiple choiceTerminology and formulation

A point $(x, y)$ that satisfies all the constraints of an LPP (including $x, y \geq 0$ ) is called

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Answer: (b) a feasible solution

Why: Feasible = satisfies every constraint; optimal = feasible and best.

Every point of the feasible region is a feasible solution. An optimal solution is a feasible solution that gives the best value of $Z$ ; a corner point is a vertex of the region.

Q12

·1 mark·Multiple choiceTerminology and formulation

A tailor has $30$ m of cloth. A shirt needs $2$ m and a pair of trousers needs $3$ m. If he makes $x$ shirts and $y$ pairs of trousers, the cloth constraint is

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Answer: (a)

2 x + 3 y \leq 30

Why: Cloth used = 2x + 3y and cannot exceed 30.

Cloth used $= 2 x + 3 y$ metres, which cannot exceed the $30$ m available: $2 x + 3 y \leq 30$ .

Q13

·1 mark·Multiple choiceCorner points and optimal values

The corner points of the feasible region of an LPP are $(0, 2)$ , $(3, 0)$ , $(6, 0)$ , $(6, 8)$ and $(0, 5)$ . The minimum value of $Z = 4 x + 6 y$ is

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Answer: (b)

12

Why: Z values: 12, 12, 24, 72, 30; the minimum 12 occurs at two corners.

$Z (0, 2) = 12$ , $Z (3, 0) = 12$ , $Z (6, 0) = 24$ , $Z (6, 8) = 72$ , $Z (0, 5) = 30$ . Minimum $= 12$ , attained at both $(0, 2)$ and $(3, 0)$ , hence at every point of the segment joining them. ( $(0, 0)$ is not a corner of this region.)

Q14

·1 mark·Multiple choiceCorner points and optimal values

The corner points of a feasible region are $(0, 0)$ , $(4, 0)$ , $(3, 3)$ and $(0, 5)$ . For which objective function is the maximum attained at $(3, 3)$ ?

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Answer: (d)

Z = x + y

Why: x + y gives 0, 4, 6, 5 — largest at (3, 3).

$Z = x + y$ : $0, 4, 6, 5$ , maximum at $(3, 3)$ . $Z = x$ is largest at $(4, 0)$ ; $Z = y$ at $(0, 5)$ ; $Z = x - y$ at $(4, 0)$ .

Q15

·1 mark·Multiple choiceFeasible regions

For the constraints $x + y \geq 2$ , $x \geq 0$ , $y \geq 0$ , the objective function $Z = x + y$

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Answer: (c) has no maximum value

Why: The region is unbounded upwards and Z grows without limit along it.

The feasible region is unbounded; e.g. $(n, 0)$ is feasible for every $n \geq 2$ with $Z = n$ . So $Z$ has no maximum (its minimum is $2$ , on the segment from $(2, 0)$ to $(0, 2)$ ).

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 12 case study questions by chapter.

Case study 1: Wooden toys and boxes (4 marks)

A craftsman makes wooden toys and jewellery boxes. A toy takes $2$ hours of carving and $1$ kg of wood; a box takes $1$ hour and $2$ kg. Each week he has at most $40$ hours and $50$ kg of wood. He earns ₹ $30$ profit on a toy and ₹ $20$ on a box. Let him make $x$ toys and $y$ boxes a week.

(i) Write the constraints (other than

x, y \geq 0

). [1 mark]

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Answer:

2 x + y \leq 40

x + 2 y \leq 50

Hours:

2 x + y \leq 40

; wood:

x + 2 y \leq 50

. A1

(ii) Find the corner point where the two constraint lines meet. [1 mark]

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Answer:

(10, 20)

Solving

2 x + y = 40

x + 2 y = 50

x = 10

y = 20

. A1

(iii) Find the maximum weekly profit and the numbers of toys and boxes that give it. [2 marks]

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Answer: ₹

700

from

10

toys and

20

boxes

Corners

(0, 0), (20, 0), (10, 20), (0, 25)

give

Z = 30 x + 20 y = 0, 600, 700, 500

M1. Maximum ₹

700

(10, 20)

. A1

OR If the profit on a box rises to ₹

40

, find the new maximum profit. [2 marks]

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Answer: ₹

1100

Z = 30 x + 40 y

at the corners:

0, 600, 1100, 1000

M1; maximum ₹

1100

(10, 20)

. A1

Case study 2: Weekly study plan (4 marks)

Ananya plans $x$ hours of mathematics and $y$ hours of physics revision a week. She can study at most $12$ hours in all, wants at most $8$ hours of mathematics and at least $2$ hours of physics. Her teacher estimates the gain in test score as $Z = 5 x + 4 y$ .

(i) Is the plan

x = 6

y = 5

feasible? Justify. [1 mark]

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Answer: Yes

6 + 5 = 11 \leq 12

6 \leq 8

5 \geq 2

(and both are non-negative): feasible. A1

(ii) Write the corner points of the feasible region. [1 mark]

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Answer:

(0, 2), (8, 2), (8, 4), (0, 12)

(0, 2)

(8, 2)

(8, 4)

(0, 12)

. A1

(iii) Find the plan that maximises

Z

and the maximum value. [2 marks]

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Answer:

x = 8

y = 4

;

Z = 56

Z = 8, 48, 56, 48

at the four corners M1; maximum

56

(8, 4)

. A1

OR Find the minimum value of

Z

over the feasible region and where it occurs. [2 marks]

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Answer:

8

(0, 2)

Z = 8, 48, 56, 48

at the corners M1; minimum

8

(0, 2)

. A1

Next steps for Linear Programming

This free set is separate from the chapter's question bank. On the Linear Programming chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 49 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Linear Programming: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.