Question

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

Answer 1

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

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

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

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

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

$=\frac{2V}{r}+2\pi r^2$

For maximum $\frac{dS}{dr}=0$

$\frac{-2V}{r^2}+4\pi r=0$

$r^3=\frac{V}{2\pi }$

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

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

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

Hence surface area minimum.