CBSE 2025 · Region 1 · Set 3 · Q38 · 4 marks
A technical company is designing a rectangular solar panel installation on a roof using $\displaystyle 300$ 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 $\displaystyle x$ 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 $\displaystyle x$ and $\displaystyle y$.
A technical company is designing a rectangular solar panel installation on a roof using $\displaystyle 300$ 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 $\displaystyle x$ 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 $\displaystyle x$ and $\displaystyle y$.
Marking-scheme solution
(i)
$\displaystyle 2 x+3 y=300$
(ii)
$\displaystyle A=x y=\frac{x}{3}(300-2 x)$
$\displaystyle (a) A=\frac{x}{3}(300-2 x)=\frac{1}{3}\left(300 x-2 x^{2}\right)$
$\displaystyle \Rightarrow \frac{d A}{d x}=\frac{1}{3}(300-4 x)$
For critical points, put $\displaystyle \frac{d \boldsymbol{A}}{d \boldsymbol{x}}=\mathbf{0} \Rightarrow x=75$
Also, $\displaystyle \frac{d^{2} A}{d x^{2}}=-\frac{4}{3}<0$. So, $\displaystyle A$ is maximum at $\displaystyle x=75$
Also, maximum area is $\displaystyle A=\frac{75}{3}(300-150)=3750 \mathrm{~m}^{2}$
(iii)
$\displaystyle A=\frac{x}{3}(300-2 x)=\frac{1}{3}\left(300 x-2 x^{2}\right)$
$\displaystyle \Rightarrow \frac{d A}{d x}=\frac{1}{3}(300-4 x)$
For critical points, put $\displaystyle \frac{d A}{d x}=0 \Rightarrow x=75$
As $\displaystyle \frac{\boldsymbol{d} \boldsymbol{A}}{\boldsymbol{d} \boldsymbol{x}}$ changes its sign from positive to negative as $\displaystyle \mathbf{x}$ passes through
$\displaystyle x=75$ from left to right, whichmeans $\displaystyle x=75$ is the point of maximum.
Also, maximum area is $\displaystyle A=\frac{75}{3}(300-150)=3750 \mathrm{~m}^{2}$
$\displaystyle 2 x+2 y=300$ or $\displaystyle 2 x+4 y=300$ or $\displaystyle 4 x+4 y=300$ or $\displaystyle 4 x+3 y=300$
The solutions of sub-parts will differ and marks may be given accordingly.
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.