Consider a closed cylinder of radius r with a fixed surface area. The…

The correct answer is Option (4) → $2r$

Surface area of a closed cylinder: $S=2\pi r^{2}+2\pi r h$ (fixed).

Volume: $V=\pi r^{2}h$.

From $S$ express $h$: $h=\frac{S-2\pi r^{2}}{2\pi r}=\frac{S}{2\pi r}-r$.

Substitute into $V$:

$V=\pi r^{2}\!\left(\frac{S}{2\pi r}-r\right)=\frac{S}{2}r-\pi r^{3}$.

Differentiate w.r.t. $r$ and set to zero for extremum:

$\frac{dV}{dr}=\frac{S}{2}-3\pi r^{2}=0 \;\Rightarrow\; r^{2}=\frac{S}{6\pi}$.

Using $S=2\pi r^{2}+2\pi r h$ gives $\frac{S}{2\pi r}=r+h$. With $S=6\pi r^{2}$ one gets $\frac{S}{2\pi r}=3r$, hence

$h=3r-r=2r$.

The volume is maximum when the height $h=2r$.