MathsCBSE Class 12 · Board exam 2023, set 65/5/1 · 5 marks · Long answer

CBSE Class 12 Mathematics 2023: question 34

Set in the CBSE Class 12 board exam of 2023, paper 65/5/1, section D, question 34, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34 · 5 marks · Long answerLinear Programming

Solve the following Linear Programming Problem graphically :

Maximize : $P = 70 x + 40 y$

subject to : $3 x + 2 y \leq 9$ , $3 x + y \leq 9$ , $x \geq 0, y \geq 0$

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 65/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Find the intercepts of each line $3 x + 2 y = 9$ and $3 x + y = 9$ on both axes, then decide which side of each line to shade using the origin.

Hint 2

Check where the two lines intersect. One constraint may lie entirely inside the other in the first quadrant.

Hint 3

List the corner points of the shaded region and substitute each into $P = 70 x + 40 y$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Graph of the constraints. Draw each constraint line and shade the side that satisfies the inequality.
- Line $3 x + 2 y = 9$ passes through $(3, 0)$ and $(0, 4.5)$ . The point $(0, 0)$ gives $0 \leq 9$ , so we shade the origin side.
- Line $3 x + y = 9$ passes through $(3, 0)$ and $(0, 9)$ . Again we shade the origin side.
- $x \geq 0, y \geq 0$ restricts the region to the first quadrant.
Both lines meet the $x$ -axis at the same point $C (3, 0)$ : subtracting the equations gives $y = 0$ , so $x = 3$ .
For $y \geq 0$ we have $3 x + y \leq 3 x + 2 y \leq 9$ . So the second constraint is automatically satisfied.
The feasible region is therefore the bounded triangle $O B C$ with vertices $O (0, 0)$ , $B (0, 4.5)$ and $C (3, 0)$ . It is shaded in the first quadrant below the line $3 x + 2 y = 9$ .
3 marks
Corner point method. Evaluate $P = 70 x + 40 y$ at each vertex of the feasible region.
Corner point $P = 70 x + 40 y$
$O (0, 0)$ $0$
$B (0, 4.5)$ $40 \times 4.5 = 180$
$C (3, 0)$ $70 \times 3 = 210$ (maximum)
1½ marks
The feasible region is bounded, so the maximum occurs at a corner point.
Hence the maximum value of $P$ is $210$ , attained at $x = 3, y = 0$ .
½ mark

Corner point	$P = 70 x + 40 y$
$O (0, 0)$	$0$
$B (0, 4.5)$	$40 \times 4.5 = 180$
$C (3, 0)$	$70 \times 3 = 210$ (maximum)

Answer: Maximum P = 210 at x = 3, y = 0

Where the marks are

Correct graph of both lines with the feasible triangle O(0,0), B(0,4.5), C(3,0) correctly shaded (3 marks)
Table of corner points with values of P: 0, 180, 210 (1½ marks)
States maximum P = 210 at (3, 0) (½ mark)

Revise the chapter

Chapter: Linear Programming (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Linear Programming, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 65/5/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Mathematics 2023: question 34

Hints

Step-by-step solution

Revise the chapter

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