A technical company is designing a rectangular solar panel installation on a roof using 300 metres of boundary material. The design includes a partition running parallel to one of the sides dividing the area (roof) into two sections. Let the length of the side perpendicular to the partition be x metres and with parallel to the partition be y metres. Based on this information, answer the following questions : (i) Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of x and y. (ii) Write the area of the solar panel as a function of x. (iii)(a) Find the critical points of the area function. Use second derivative test to determine critical points at the maximum area. Also, find the maximum area. (iii)(b) Using first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition.

(i) 2x+3y=300; (ii) A=x/3(300-2x); (iii) critical point x=75, maximum area 3750 m²

MathsCBSE Class 12 · Board exam 2025, set 65/1/1 · 4 marks · Case-based

CBSE Class 12 Mathematics 2025: question 36

Set in the CBSE Class 12 board exam of 2025, paper 65/1/1, section E, question 36, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 4 marks · Case-basedCompetency-basedApplication of Derivatives

A technical company is designing a rectangular solar panel installation on a roof using 300 metres of boundary material. The design includes a partition running parallel to one of the sides dividing the area (roof) into two sections.
Let the length of the side perpendicular to the partition be $x$ metres and with parallel to the partition be y metres.

Based on this information, answer the following questions :

(i)
Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of $x$ and y.
1 mark
(ii)
Write the area of the solar panel as a function of $x$ .
1 mark
(iii)(a)
Find the critical points of the area function. Use second derivative test to determine critical points at the maximum area. Also, find the maximum area.
2 marks
OR
(iii)(b)
Using first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition.
2 marks

Figure in the paper: Photograph of two rectangular solar panels installed side by side on a roof. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 65/1/1.

Hints

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

Hint 1

Count how many lengths of x and how many of y are used, remembering that the partition is parallel to the y-sides.

Hint 2

Use the boundary equation to write y in terms of x, then form A = xy.

Hint 3

Differentiate A, set it to zero, and then check the second derivative (or the sign change of A′).

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.

(i) The two sides perpendicular to the partition use $2 x$ m of material. The two sides parallel to the partition, plus the partition itself, use $3 y$ m. Hence
$2 x + 3 y = 300$
1 mark
(ii) From (i), $y = \frac{300 - 2 x}{3}$ . So
$A (x) = x y = \frac{x}{3} (300 - 2 x) = \frac{1}{3} (300 x - 2 x^{2})$
1 mark
(iii)(a) $\frac{d A}{d x} = \frac{1}{3} (300 - 4 x)$
½ mark
(iii)(a) For critical points, $\frac{d A}{d x} = 0 \Rightarrow x = 75$ .
½ mark
(iii)(a) $\frac{d ^{2} A}{d x ^{2}} = - \frac{4}{3} < 0$ , so $A$ is maximum at $x = 75$ .
½ mark
(iii)(a) Maximum area $= \frac{75}{3} (300 - 150) = 25 \times 150 = 3750 m^{2}$ (with $y = 50$ m).
½ mark
(iii)(b) $A = \frac{1}{3} (300 x - 2 x^{2})$ , so $\frac{d A}{d x} = \frac{1}{3} (300 - 4 x)$ . Setting $\frac{d A}{d x} = 0$ gives the critical point $x = 75$ .
1 mark
(iii)(b) For $x < 75$ , $\frac{d A}{d x} > 0$ , and for $x > 75$ , $\frac{d A}{d x} < 0$ . Since $\frac{d A}{d x}$ changes sign from positive to negative at $x = 75$ , the area is maximum there [first derivative test].
½ mark
(iii)(b) Maximum area $= \frac{75}{3} (300 - 150) = 3750 m^{2}$ .
½ mark

Answer: (i) $2 x + 3 y = 300$ ; (ii) $A = \frac{x}{3} (300 - 2 x)$ ; (iii) critical point $x = 75$ , maximum area $3750 m^{2}$

Where the marks are

(i) Writes $2 x + 3 y = 300$ (1 mark)
(ii) Writes $A = \frac{x}{3} (300 - 2 x)$ (1 mark)
(iii)(a) Finds $A^{'} (x)$ (½ mark), $x = 75$ (½ mark), $A^{''} < 0$ (½ mark), maximum area 3750 m² (½ mark)
(iii)(b) Finds $A^{'} (x)$ and $x = 75$ (½ + ½ mark), uses the sign change of $A^{'}$ (½ mark), maximum area 3750 m² (½ mark)

Revise the chapter

Chapter: Application of Derivatives (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Application of Derivatives, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 65/1/1: every question, by section.

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CBSE Class 12 Mathematics 2025: question 36

Hints

Step-by-step solution

Revise the chapter

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