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

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)]