An open metallic tank is to be constructed, with a square base and vertical sides, having volume 500 cubic…
An open metallic tank is to be constructed, with a square base and vertical sides, having volume 500 cubic meter. Then the dimensions of the tank, for minimum area of the sheet metal used in its construction, are
$5 \mathrm{~m}, 5 \mathrm{~m}, 10 \mathrm{~m}$
$10 \mathrm{~m}, 10 \mathrm{~m}, 5 \mathrm{~m}$
$2 \mathrm{~m}, 2 \mathrm{~m}, 8 \mathrm{~m}$
$15 \mathrm{~m}, 15 \mathrm{~m}, 5 \mathrm{~m}$
Solution
Let the length, breadth and depth of open tank be $x, x$ and $y$ respectively.
Volume $(\mathrm{V})=x^2 y$
$\therefore \quad 500=x^2 y$... (i)
Total surface area of open tank is given by
$\mathrm{S}=x^2+4 x y$... (ii)
From (i), $y=\frac{500}{x^2}$
From (ii), $\mathrm{S}=x^2+4 x \times \frac{500}{x^2}$
$=x^2+\frac{2000}{x}$
Differentiating w.r.t. $x$, we get
$\frac{\mathrm{dS}}{\mathrm{d} x}=2 x-\frac{2000}{x^2}$
For minimum area, $\frac{\mathrm{dS}}{\mathrm{d} x}=0$
$\begin{aligned}
\therefore \quad & 2 x-\frac{2000}{x^2}=0 \\
& \Rightarrow 2000=2 x^3 \\
& \Rightarrow x^3=1000
\end{aligned}$
$\begin{aligned}
& \Rightarrow x=10 \mathrm{~m} \\
& \frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{~d} x^2}=2+\frac{4000}{x^3} \\
& \Rightarrow\left(\frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{~d} x^2}\right)_{x=10}>0
\end{aligned}$
$\mathrm{S}$ is minimum when $x=10 \mathrm{~m}$ and $y=5 \mathrm{~m}$
...[From (i)]