CBSE Maths › Class 12 PYQs › Linear Programming

Linear Programming: 3 marks Questions (CBSE Class 12)

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

Q303 marksShort AnswerLinear Programming CBSE 2026 · 65/1/1

Solve the following linear programming problem graphically :
Minimize $Z = 13 x - 15 y$
Subject to constraints
$x + y \leq 7,$
$2 x - 3 y + 6 \geq 0,$
$x \geq 0, y \geq 0$

Show answer & solution

Answer: Minimum Z =

- 30

at (0, 2)

Lines: $x + y = 7$ and $2 x - 3 y + 6 = 0$ (through (0, 2) and (-3, 0)).
They meet where $2 (7 - y) - 3 y + 6 = 0 \Rightarrow y = 4, x = 3$ .
The feasible region is bounded with corner points O(0, 0), A(7, 0), B(3, 4), C(0, 2).
Z at these: 0, 91, 39 - 60 = -21, -30.
Minimum Z = $- 30$ at (0, 2).

Also asked in: 2026 65/1/2, 2026 65/1/3

Q303 marksShort AnswerLinear Programming CBSE 2026 · 65/2/1

Solve the following linear programming problem graphically :
Maximize $Z = 10500 x + 9000 y$
Subject to constraints
$x + y \leq 50$
$2 x + y \leq 80$
$x, y \geq 0$

Show answer & solution

Answer: Maximum

Z = 495000

x = 30

y = 20

Lines: $x + y = 50$ through $(50, 0)$ , $(0, 50)$ ; $2 x + y = 80$ through $(40, 0)$ , $(0, 80)$ ; they meet at $(30, 20)$ .
The feasible region (origin side of both lines, first quadrant) has corners $O (0, 0)$ , $A (40, 0)$ , $B (30, 20)$ , $C (0, 50)$ .
$Z (O) = 0$ , $Z (A) = 420000$ , $Z (B) = 315000 + 180000 = 495000$ , $Z (C) = 450000$ .
Maximum $Z = 495000$ at $(30, 20)$ .

Q283 marksShort AnswerLinear Programming CBSE 2026 · 65/2/2

Solve the following linear programming problem graphically :
Maximize $Z = 8000 x + 12000 y$
Subject to constraints
$3 x + 4 y \leq 60$
$x + 3 y \leq 30$
$x \geq 0$ , $y \geq 0$

Show answer & solution

Answer: Maximum

Z = 168000

x = 12

y = 6

Lines: $3 x + 4 y = 60$ through $(20, 0)$ , $(0, 15)$ ; $x + 3 y = 30$ through $(30, 0)$ , $(0, 10)$ ; they meet at $(12, 6)$ .
Feasible region corners: $O (0, 0)$ , $A (20, 0)$ , $B (12, 6)$ , $C (0, 10)$ .
$Z (O) = 0$ , $Z (A) = 160000$ , $Z (B) = 96000 + 72000 = 168000$ , $Z (C) = 120000$ .
Maximum $Z = 168000$ at $(12, 6)$ .

Q313 marksShort AnswerLinear Programming CBSE 2026 · 65/2/3

Solve the following linear programming problem graphically :
Maximize $Z = 4500 x + 5000 y$
Subject to constraints
$x + y \leq 250$
$25 x + 40 y \leq 7000$
$x \geq 0$ , $y \geq 0$

Show answer & solution

Answer: Maximum

Z = 1150000

x = 200

y = 50

Lines: $x + y = 250$ through $(250, 0)$ , $(0, 250)$ ; $25 x + 40 y = 7000$ through $(280, 0)$ , $(0, 175)$ ; they meet at $(200, 50)$ .
Feasible region corners: $O (0, 0)$ , $A (250, 0)$ , $B (200, 50)$ , $C (0, 175)$ .
$Z (O) = 0$ , $Z (A) = 1125000$ , $Z (B) = 900000 + 250000 = 1150000$ , $Z (C) = 875000$ .
Maximum $Z = 1150000$ at $(200, 50)$ .

Q293 marksShort AnswerLinear Programming CBSE 2026 · 65/4/1

Solve the following Linear Programming Problem graphically :
Maximise $Z = 200 x + 120 y$
subject to the constraints
$x + y \leq 300$
$3 x + y \leq 600$
$x - y \geq - 100$
$x, y \geq 0$

Show answer & solution

Answer: Maximum Z = 48000 at x = 150, y = 150

Draw $x + y = 300$ , $3 x + y = 600$ and $x - y = - 100$ in the first quadrant; the feasible region is the bounded region on the origin side of the first two lines and below $y = x + 100$ .
Corner points: O(0, 0), (200, 0), (150, 150) [from $x + y = 300$ , $3 x + y = 600$ ], (100, 200) [from $x + y = 300$ , $y = x + 100$ ], (0, 100).
Z at these points: 0, 40000, 48000, 44000, 12000.
Maximum $Z = 48000$ at $(150, 150)$ .

Q293 marksShort AnswerLinear Programming CBSE 2026 · 65/4/2

Solve the following Linear Programming Problem graphically :
Minimize $Z = 20 x + 10 y$
subject to
$x + 2 y \leq 40$
$3 x + y \geq 30$
$4 x + 3 y \geq 60$
$x, y \geq 0$

Show answer & solution

Answer: Minimum Z = 240 at x = 6, y = 12

Draw $x + 2 y = 40$ , $3 x + y = 30$ and $4 x + 3 y = 60$ ; the feasible region lies below the first line and above the other two, in the first quadrant (bounded).
Corner points: (15, 0), (40, 0), (4, 18) [from $x + 2 y = 40$ , $3 x + y = 30$ ], (6, 12) [from $3 x + y = 30$ , $4 x + 3 y = 60$ ].
Z at these points: 300, 800, 260, 240.
Minimum $Z = 240$ at $(6, 12)$ .

Q293 marksShort AnswerLinear Programming CBSE 2026 · 65/4/3

Solve the following Linear Programming Problem graphically :
Maximize $Z = 20 x + 10 y$
subject to constraints
$x + 2 y \leq 28$
$3 x + y \leq 24$
$x \geq 2$
$x, y \geq 0$

Show answer & solution

Answer: Maximum Z = 200 at x = 4, y = 12

Draw $x + 2 y = 28$ , $3 x + y = 24$ and $x = 2$ ; the feasible region is the bounded region right of $x = 2$ , above the x-axis and below both lines.
Corner points: (2, 0), (8, 0), (4, 12) [from $x + 2 y = 28$ , $3 x + y = 24$ ], (2, 13).
Z at these points: 40, 160, 200, 170.
Maximum $Z = 200$ at $(4, 12)$ .

Q293 marksShort AnswerLinear Programming CBSE 2026 · 65/5/1

Solve the following Linear Programming Problem graphically :
Maximize $Z = \frac{2 x}{5} + \frac{3 y}{10}$
subject to constraints
$2 x + y \leq 1000$
$x + y \leq 800$
$x, y \geq 0$ .

Show answer & solution

Answer: Maximum Z = 260 at x = 200, y = 600.

Draw $2 x + y = 1000$ through (500, 0), (0, 1000) and $x + y = 800$ through (800, 0), (0, 800); they meet at (200, 600).
Feasible region (towards the origin, in the first quadrant) has corners (0, 0), (500, 0), (200, 600), (0, 800).
Z values: 0, 200, 260, 240.
Maximum Z = 260 at (200, 600).

Also asked in: 2026 65/5/2, 2026 65/5/3

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