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

Answer: Option A

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