CBSE Maths › Class 12 PYQs › Linear Programming

Linear Programming: 5 marks Questions (CBSE Class 12)

3 different 5 marks questions on Linear Programming from CBSE Class 12 Maths board exams 2026, newest first.

Q335 marksLong AnswerLinear Programming CBSE 2026 · 65/3/1

Solve the following Linear Programming Problem graphically :
Maximise $Z = 600 x + 400 y$
subject to the constraints
$x + 2 y \leq 12$
$4 x + 5 y \geq 20$
$2 x + y \leq 12$
$x, y \geq 0$

Show answer & solution

Answer: Maximum

Z = 4000

x = 4

y = 4

Draw $x + 2 y = 12$ (through (12, 0), (0, 6)), $4 x + 5 y = 20$ (through (5, 0), (0, 4)) and $2 x + y = 12$ (through (6, 0), (0, 12)).
The feasible region lies in the first quadrant, above $4 x + 5 y = 20$ and below the other two lines; it is bounded.
$x + 2 y = 12$ and $2 x + y = 12$ meet at (4, 4).
Corner points: (5, 0), (6, 0), (4, 4), (0, 6), (0, 4).
Z: (5, 0) → 3000; (6, 0) → 3600; (4, 4) → 4000; (0, 6) → 2400; (0, 4) → 1600.
Maximum Z = 4000 at (4, 4).

Q355 marksLong AnswerLinear Programming CBSE 2026 · 65/3/2

Solve the following Linear Programming Problem graphically :
Maximise $Z = 12 x + 18 y$
subject to the constraints
$x + y \leq 1200$
$x - 2 y \geq 0$
$x + 3 y \geq 600$
$x \geq 0, y \geq 0$

Show answer & solution

Answer: Maximum

Z = 16800

x = 800

y = 400

Draw $x + y = 1200$ (through (1200, 0), (0, 1200)), $x = 2 y$ (through (0, 0), (800, 400)) and $x + 3 y = 600$ (through (600, 0), (0, 200)).
The feasible region lies on or below $x + y = 1200$ , on or below the line $x = 2 y$ (towards the x-axis), on or above $x + 3 y = 600$ , with $y \geq 0$ ; it is bounded.
Corner points: $x + 3 y = 600$ and $x = 2 y$ meet at (240, 120); $x + y = 1200$ and $x = 2 y$ meet at (800, 400); also (600, 0) and (1200, 0).
Z: (240, 120) → 5040; (600, 0) → 7200; (1200, 0) → 14400; (800, 400) → 16800.
Maximum Z = 16800 at (800, 400).

Q345 marksLong AnswerLinear Programming CBSE 2026 · 65/3/3

Solve the following Linear Programming Problem graphically :
Maximize $Z = 8 x + 8 y$
subject to the constraints
$x - 20 \leq 0$
$2 x + 3 y \leq 120$
$2 x + y \leq 60$
$x \geq 0, y \geq 0$

Show answer & solution

Answer: Maximum

Z = 360

x = 15

y = 30

Draw $x = 20$ , $2 x + 3 y = 120$ (through (60, 0), (0, 40)) and $2 x + y = 60$ (through (30, 0), (0, 60)).
The feasible region is the bounded region in the first quadrant to the left of $x = 20$ and below both lines.
$2 x + y = 60$ meets $x = 20$ at (20, 20) and meets $2 x + 3 y = 120$ at (15, 30).
Corner points: (0, 0), (20, 0), (20, 20), (15, 30), (0, 40).
Z: (0, 0) → 0; (20, 0) → 160; (20, 20) → 320; (15, 30) → 360; (0, 40) → 320.
Maximum Z = 360 at (15, 30).

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