CBSE Maths › Class 12 PYQs › Linear Programming

Linear Programming: 2 marks Questions (CBSE Class 12)

3 different 2 marks questions on Linear Programming from CBSE Class 12 Maths board exams 2024–2026, newest first.

1 mark (25)2 marks (3)3 marks (29)4 marks (1)5 marks (6)

Q242 marksVery Short AnswerLinear Programming CBSE 2025 · 65/7/1

In a Linear Programming Problem, the objective function $Z = 5 x + 4 y$ needs to be maximised under constraints $3 x + y \leq 6$ , $x \leq 1$ , $x, y \geq 0$ . Express the LPP on the graph and shade the feasible region and mark the corner points.

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Answer: Feasible region is the quadrilateral with corner points (0, 0), (1, 0), (1, 3) and (0, 6); (maximum Z = 24 at (0, 6)).

Draw the lines $3 x + y = 6$ (through (2, 0) and (0, 6)) and $x = 1$ .
The feasible region lies on the origin side of both lines in the first quadrant.
$x = 1$ meets $3 x + y = 6$ at (1, 3).
Corner points: O(0, 0), A(1, 0), B(1, 3), C(0, 6); shade OABC.
Z at these points: 0, 5, 17, 24, so the maximum is 24 at (0, 6).

Q242 marksVery Short AnswerLinear Programming CBSE 2025 · 65/7/2

In a Linear Programming Program (LPP) for objective function $Z = 14 x - 10 y$
subject to constraints
$x + y \leq 8$
$3 x - 2 y \geq - 6$
$x, y \geq 0$
shade the feasible region and mark the corner points in a neatly drawn graph.

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Answer: Feasible region is the quadrilateral with corner points (0, 0), (8, 0), (2, 6) and (0, 3).

Draw $x + y = 8$ through (8, 0) and (0, 8); draw $3 x - 2 y = - 6$ through (−2, 0) and (0, 3).
(0, 0) satisfies both inequalities, so the feasible region is on the origin side of both lines, in the first quadrant.
The lines meet where $3 x - 2 (8 - x) = - 6$ , i.e. $x = 2$ , $y = 6$ .
Corner points: (0, 0), (8, 0), (2, 6), (0, 3); shade the quadrilateral formed by them.

Q242 marksVery Short AnswerLinear Programming CBSE 2025 · 65/7/3

For a Linear Programming Problem, find min $Z = 5 x + 3 y$ (where Z is the objective function) for the feasible region shaded in the given figure.
(Note : The figure is not to scale)

Diagram for CBSE 2025 Class 12 Maths question 24

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Answer: Minimum Z = 9 at C(0, 3)

B is where $x + y = 5$ meets the y-axis: B(0, 5); C is where $x + 3 y = 9$ meets the y-axis: C(0, 3).
A: solving $x + y = 5$ and $x + 3 y = 9$ gives $2 y = 4$ , so A(3, 2).
Z at B = 15, at C = 9, at A = 15 + 6 = 21.
The region is bounded, so min Z = 9 at C(0, 3).

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