An open tank with a square bottom, to contain 4000 cubic cm . of liquid, is to be constructed. The dimensions of the tank, so that the surface area of the tank is minimum, are

Let \(x\) be the length of the side of square bottom, h be the height, V be the volume and A be the surface area of open tank.
Then,
\(\begin{aligned}
& \mathrm{V}=x^2 \mathrm{~h}=4000...(i) \\
& \mathrm{~A}=x^2+4 x \mathrm{~h}...(ii)
\end{aligned}\)
From (i),
\(\mathrm{h}=\frac{4000}{x^2}\)
Substituting the value of \(h\) in (ii), we get
\(\begin{aligned}
& \mathrm{A}=x^2+\frac{16000}{x} \\
\therefore \quad & \frac{\mathrm{dA}}{\mathrm{~d} x}=2 x-\frac{16000}{x^2}
\end{aligned}\)
\(A\) is minimum, if \(\frac{d A}{d x}=0\)
\(\begin{aligned}
& \Rightarrow 2 x-\frac{16000}{x^2}=0 \\
& \Rightarrow x^3=8000 \\
& \Rightarrow x=20
\end{aligned}\)
Now, \(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{~d} x^2}=2+\frac{32000}{x^3}\)
\(\Rightarrow\left(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{~d} x^2}\right)_{x=20}=6 \gt 0\)
\(\therefore \quad \mathrm{A}\) is minimum when \(x=20 \mathrm{~cm}\).
\(\mathrm{h}=\frac{4000}{x^2}=\frac{4000}{400}=10 \mathrm{~cm}\)