The height of a closed cylinder of given volume and the minimum surface area is :

Solution (By Examveda Team)

$$\eqalign{ & V = \pi {r^2}h{\text{ and }} \cr & S = 2\pi rh + 2\pi {r^2} \cr & \,\,\,\,\,\,\, = 2\pi r\left( {h + r} \right) \cr & {\text{Where, }}h = \frac{V}{{\pi {r^2}}} \cr & \Rightarrow S = 2\pi r\left( {\frac{V}{{\pi {r^2}}} + r} \right) \cr & \Rightarrow S = \frac{{2V}}{r} + 2\pi {r^2} \cr & \Rightarrow \frac{{dS}}{{dr}} = \frac{{ - 2V}}{{{r^2}}} + 4\pi r{\text{ and}} \cr & \frac{{{d^2}S}}{{d{r^2}}} = \left( {\frac{{4V}}{{{r^3}}} + 4\pi } \right){\text{ > 0}} \cr} $$
∴ S is minimum when :
$$\eqalign{ & \frac{{dS}}{{dr}} = 0 \cr & \Rightarrow \frac{{ - 2V}}{{{r^2}}} + 4\pi r = 0 \cr & \Rightarrow V = 2\pi {r^3} \cr & \Rightarrow \pi {r^2}h = 2\pi {r^3} \cr & \Rightarrow h = 2r \cr} $$