MathsCBSE Class 12 · Board exam 2024, set 65/5/1 · 5 marks · Long answer

CBSE Class 12 Mathematics 2024: question 35

Set in the CBSE Class 12 board exam of 2024, paper 65/5/1, section D, question 35, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 5 marks · Long answerLinear Programming

Solve the following L.P.P. graphically :
Maximise $Z = 60 x + 40 y$
Subject to $x + 2 y \leq 12$ , $2 x + y \leq 12$ , $4 x + 5 y \geq 20$ , $x, y \geq 0$

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 65/5/1.

Hints

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

Hint 1

Plot each constraint line using its intercepts, then decide which side of each line is feasible.

Hint 2

The constraint $4 x + 5 y \geq 20$ cuts off the region near the origin; find where $x + 2 y = 12$ and $2 x + y = 12$ intersect.

Hint 3

Evaluate Z at every corner point of the bounded region and compare the values.

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. Draw the boundary lines:
- $x + 2 y = 12$ through $(12, 0)$ and $(0, 6)$
- $2 x + y = 12$ through $(6, 0)$ and $(0, 12)$
- $4 x + 5 y = 20$ through $(5, 0)$ and $(0, 4)$
Shade the region that satisfies all the constraints:
- below $x + 2 y = 12$ and below $2 x + y = 12$
- above $4 x + 5 y = 20$
- in the first quadrant
The two lines $x + 2 y = 12$ and $2 x + y = 12$ meet at $C (4, 4)$ . The feasible region is bounded, with corner points $A (0, 4)$ , $B (0, 6)$ , $C (4, 4)$ , $D (6, 0)$ and $E (5, 0)$ .
3½ marks
Evaluate $Z = 60 x + 40 y$ at each corner point:
Corner point $Z = 60 x + 40 y$
$A (0, 4)$ 160
$B (0, 6)$ 240
$C (4, 4)$ 400
$D (6, 0)$ 360
$E (5, 0)$ 300
1 mark
The largest value in the table is at $C$ . Hence the maximum value of $Z$ is $400$ , at $x = 4, y = 4$ .
½ mark

Corner point	$Z = 60 x + 40 y$
$A (0, 4)$	160
$B (0, 6)$	240
$C (4, 4)$	400
$D (6, 0)$	360
$E (5, 0)$	300

Answer: Max $Z = 400$ at $x = 4, y = 4$

Where the marks are

Correct graph with the feasible region shaded and corner points $(0, 4), (0, 6), (4, 4), (6, 0), (5, 0)$ (3½ marks)
Table of Z values at the corner points (1 mark)
Conclusion: max Z = 400 at (4,4) (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 paper2024 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 2024: question 35

Hints

Step-by-step solution

Revise the chapter

More questions from Linear Programming