Step 1:

Let $r$ be the radius and $h$ be the the height of a cylinder of given volume $V$

$V=\pi r^2h$

$h=\large\frac{v}{\pi r^2}$

Let $S$ be total surface area.Since the cylinder is open at its top.

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

Substituting for $h$ we get,

$S=2\pi r\big(\large\frac{v}{\pi r^2}\big)$$+\pi r^2$

$\;\;=\large\frac{2v}{r}$$+\pi r^2$

Step 2:

Differentiating with respect to $x$ we get,

$\large\frac{dS}{dr}=\frac{-2v}{r^2}$$+2\pi r$

For maximum or minimum

$\large\frac{dS}{dr}=$$0$

$\Rightarrow \large\frac{2v}{r^2}+$$2\pi r=0$

(i.e) $\large\frac{-v}{r^2}$$+\pi r=0$

$\therefore v=\pi r^3$

$\pi r^2h=\pi r^3$

$\Rightarrow h=r$

Step 3:

Differentiating again with respect to $r$ we get,

$\large\frac{d^2S}{dr^2}=\frac{4v}{r^3}$$+2\pi$

When $r=h$

$\large\frac{d^2S}{dr^2}=\frac{4v}{r^3}$$+2\pi$ >0

Hence $S$ is minimum when $h=r$ (i.e) when the height of the cylinder is equal to the radius of the base.