Explanation

(d)
Surface area, S=$2\pi rh + \pi r^{2}\$ …(i)
And $V = \pi r^{2}h\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eq. (i), $h = \frac{S - \pi r^{2}}{2\pi r}$
$\therefore\ \ \ From\ Eq.(ii),\ V = \frac{r}{2}\left( S - \pi r^{2} \right)$
$\Rightarrow \ \ \ \ \ \frac{dV}{dr} = \frac{1}{2}\left( S - 3\pi r^{2} \right) = 0$ [say]
$\Rightarrow \ \ \ \ S - 3\pi r^{2} = 0 \Rightarrow S = 3\pi r^{2}$
On putting the value of S in Eq.(i),we get
$3\pi r^{2} = 2\pi rh + + \pi r^{2} \Rightarrow r = h$