CBSE Board Examination 2025 · Section E · 4 marks

Q36: Case study: a solar panel with a partition and 300 m of boundary material

Board 2025Ch 6 · Application of Derivatives

The question

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)

(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)

Exam answer

(i) The boundary has two sides of length $x$ and two of length $y$ ⁠, and the partition (parallel to the sides of length $y$ ⁠) has length $y$ ⁠.

Why?

Draw the rectangle with the partition across it. The partition is parallel to the sides of length $y$ ⁠, so it is also $y$ long. The material used is

x + x + y + y + y = 2x + 3y

(The marking scheme also accepts some other readings of this wording.)

$2x + 3y = 300$

(ii) From (i), $y = \frac{300 - 2x}{3}$ ⁠.

Why?

The area of a rectangle is length times breadth, $xy$ ⁠. To make it a function of $x$ alone, solve (i) for $y$ ⁠: $3y = 300 - 2x$ ⁠.

\begin{aligned} 2x + 3y &= 300 \\ 3y &= 300 - 2x \\ y &= \frac{300 - 2x}{3} \\[4pt] A = xy &= x\cdot\frac{300 - 2x}{3} \\ &= \frac{x}{3}(300 - 2x) \end{aligned}

For a real panel both sides must be positive, so $0 < x < 150$ ⁠.

$A(x) = xy = \frac{x}{3}(300 - 2x)$

(iii) (a)

\begin{aligned} A(x) &= \frac13(300x - 2x^2) \\ A'(x) &= \frac13(300 - 4x) \end{aligned}

Why?

Expand first, then differentiate term by term: $\frac{d}{dx}(300x) = 300$ and $\frac{d}{dx}(2x^2) = 4x$ ⁠. The constant $\frac13$ stays in front.

\begin{aligned} \frac{x}{3}(300 - 2x) &= \frac13\big(x \cdot 300 - x \cdot 2x\big) \\ &= \frac13(300x - 2x^2) \end{aligned}

A'(x) = 0 \Rightarrow x = 75

So the critical point is $x = 75$ ⁠.

Why?

A critical point is where $A'(x) = 0$ ⁠: $300 - 4x = 0$ gives $x = \frac{300}{4} = 75$ ⁠.

\begin{aligned} \frac13(300 - 4x) &= 0 \\ 300 - 4x &= 0 \quad (\times\, 3) \\ 300 &= 4x \\ x &= \frac{300}{4} = 75 \end{aligned}

A''(x) = -\frac43 < 0

So $A(x)$ is maximum at $x = 75$ (then $y = 50$ ⁠).

Why?

Second derivative test: if $A'(c) = 0$ and $A''(c) < 0$ ⁠, then $A$ has a maximum at $c$ ⁠. Here $A''(x) = \frac13(-4)$ for every $x$ ⁠, so it is negative at $x = 75$ ⁠.

\begin{aligned} A''(x) &= \frac{d}{dx}\left[\frac13(300 - 4x)\right] \\ &= \frac13(0 - 4) = -\frac43 \end{aligned}

$y = \frac{300 - 150}{3} = 50$ ⁠.

\begin{aligned} y &= \frac{300 - 2(75)}{3} \\ &= \frac{300 - 150}{3} = \frac{150}{3} = 50 \end{aligned}

A(75) = \frac{75}{3}(300 - 150) = 3750

Why?

$\frac{75}{3} = 25$ and $2\times75 = 150$ ⁠, so $A = 25\times150$ ⁠. The area is in square metres.

\begin{aligned} A(75) &= \frac{75}{3}\big(300 - 2(75)\big) \\ &= 25(300 - 150) \\ &= 25 \times 150 = 3750 \end{aligned}

Check: $xy = 75 \times 50 = 3750$ too.

$\therefore$ The maximum area is 3750 m².

(iii) (b)

A'(x) = \frac13(300 - 4x)

A'(x) = 0 \Rightarrow x = 75

Why?

The first derivative test also starts from the critical point: $A'(x) = 0$ at $x = 75$ ⁠, as in (iii) (a).

For $x < 75$ ⁠, $A'(x) > 0$ ⁠, and for $x > 75$ ⁠, $A'(x) < 0$ ⁠. So $A'(x)$ changes sign from positive to negative as $x$ passes through 75, and $x = 75$ is the point of maximum.

Why?

$300 - 4x > 0$ exactly when $x < 75$ ⁠. For example, $A'(74) = \frac43 > 0$ and $A'(76) = -\frac43 < 0$ ⁠.

\begin{aligned} A'(74) &= \frac13\big(300 - 4(74)\big) \\ &= \frac13(300 - 296) = \frac43 \\[4pt] A'(76) &= \frac13\big(300 - 4(76)\big) \\ &= \frac13(300 - 304) = -\frac43 \end{aligned}

First derivative test: if $A'$ changes from positive to negative as $x$ increases through $c$ ⁠, the area rises and then falls, so $A$ has a maximum at $c$ ⁠.

A(75) = \frac{75}{3}(300 - 150) = 3750

Why?

The same working as in (iii) (a):

\begin{aligned} A(75) &= \frac{75}{3}\big(300 - 2(75)\big) \\ &= 25(300 - 150) \\ &= 25 \times 150 = 3750 \end{aligned}

$\therefore$ The maximum area is 3750 m².

Where this comes from

Chapter 6, Application of Derivatives: Case study: maximum area with a partition.

Question type: Word problems

How to spot it: click to see

Type 10. Word problems: the biggest box, the least cost

Spot it: a real object and the words "maximum", "minimum", "largest" or "least".

Here: 300 m of material gives $2x + 3y = 300$ ⁠, so the area in $x$ alone is $A(x) = \frac{x}{3}(300 - 2x)$ ⁠. Its only critical point is $x = 75$ ⁠: (iii)(a) shows a maximum with $A'' = -\frac43$ ⁠, (iii)(b) with $A'$ going from $+$ to $-$ ⁠.

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