CBSE 2025 · Region 5 · Set 1 · Q36 · 4 marks
A carpenter needs to make a wooden cuboidal box, closed from all sides, which has a square base and fixed volume. Since he is short of the paint required to paint the box on completion, he wants the surface area to be minimum. On the basis of the above information, answer the following questions :(i)Taking length = breadth = x m and height = y m, express the surface area ( S ) of the box in terms of x and its volume ( V ), which is constant.(ii)Find $\displaystyle \frac{\mathrm{dS}}{\mathrm{dx}}$.(iii)Find a relation between x and y such that the surface area (S) is minimum.If surface area (S) is constant, the volume (V) $\displaystyle =\frac{1}{4}\left(\mathrm{Sx}-2 \mathrm{x}^{3}\right)$, $\displaystyle \mathrm{x}$ being the edge of base. Show that volume (V) is maximum for $\displaystyle \mathrm{x}=\sqrt{\frac{\mathrm{S}}{6}}$. Case Study - $\displaystyle 2$
A carpenter needs to make a wooden cuboidal box, closed from all sides, which has a square base and fixed volume. Since he is short of the paint required to paint the box on completion, he wants the surface area to be minimum. On the basis of the above information, answer the following questions :
(i)
Taking length = breadth = x m and height = y m, express the surface area ( S ) of the box in terms of x and its volume ( V ), which is constant.
(ii)
Find $\displaystyle \frac{\mathrm{dS}}{\mathrm{dx}}$.
(iii)
Find a relation between x and y such that the surface area (S) is minimum.
If surface area (S) is constant, the volume (V) $\displaystyle =\frac{1}{4}\left(\mathrm{Sx}-2 \mathrm{x}^{3}\right)$, $\displaystyle \mathrm{x}$ being the edge of base. Show that volume (V) is maximum for $\displaystyle \mathrm{x}=\sqrt{\frac{\mathrm{S}}{6}}$. Case Study - $\displaystyle 2$
Marking-scheme solution
(i)
$\displaystyle \mathrm{V}=\mathrm{x}^{2} \mathrm{y} \Rightarrow \mathrm{y}=\frac{\mathrm{V}}{\mathrm{x}^{2}} \ldots \ldots \ldots \ldots$ (i)
Hence, $\displaystyle \mathrm{S}=2 \mathrm{x}^{2}+4 \mathrm{x} \mathrm{y}=2 \mathrm{x}^{2}+\frac{4 \mathrm{v}}{\mathrm{x}}$
(ii)
$\displaystyle \frac{\mathrm{dS}}{\mathrm{dx}}=4\left(\mathrm{x}-\frac{\mathrm{v}}{\mathrm{x}^{2}}\right)$
(iii)
$\displaystyle \frac{\mathrm{dS}}{\mathrm{dx}}=\mathrm{O} \Rightarrow \mathrm{V}=\mathrm{x}^{3} \Rightarrow \mathrm{x}^{2} \mathrm{y}=\mathrm{x}^{3} \Rightarrow \mathrm{y}=\mathrm{x} \frac{d^{2} \mathrm{S}}{d \mathrm{x}^{2}}=4\left(1+\frac{2 \mathrm{V}}{\mathrm{x}^{3}}\right)>0 \Rightarrow \mathrm{S}$ is minimum if $\displaystyle \mathrm{y}=\mathrm{x}$.
$\displaystyle \mathrm{V}=\frac{1}{4}\left(\mathrm{Sx}-2 \mathrm{x}^{3}\right) \Rightarrow \frac{\mathrm{dV}}{\mathrm{dx}}=\frac{1}{4}\left(\mathrm{~S}-6 \mathrm{x}^{2}\right)$
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