Question

Prove that a cylindrical vessel of given volume requires least surface area where its height is twice the radius.

Answer 1

Let $S$ and $V$ denote the surface and volume of right circular cylinder of height hard base radius $r$.

Then,

$S=2\pi rh+2\pi r^2$

$V=\pi r^{2}h\Rightarrow h=\frac{V}{\pi r^2}$

$\therefore$ $S=2\pi r.\left(\frac{V}{{\pi r}^2}\right)+2\pi r^2=\frac{2V}{r}+2\pi r^2$

For max or min $\frac{dS}{dr}=0$ $\frac{-2V}{r^2}+4\pi r=0$ $r^3=\frac{V}{2\pi }$

Also $\frac{d^{2}S}{{dr}^2}=\frac{4V}{r^3}+4V=\frac{4V\times 2\pi }{V}+4\pi =12\pi >0$

Thus $S$ is minimum when $r^3=\frac{V}{2\pi }=\frac{\pi r^{2}h}{2\pi }$

$\Rightarrow r=\frac{h}{2}$ $2r=h$

Hence Surface area is minimum.