A closed cylinder of given volume will have least surface area when the ratio of its height and base radius is

For a closed cylinder let the height be $h$ and base radius be $r$.

Volume, $V=\pi r^{2}h$

i.e. $h=\frac{V}{{\mathrm{\pi r}}^2}$

Now, surface area, $S=2\mathrm{\pi rh}+2{\mathrm{\pi r}}^2=2\mathrm{\pi r}\frac{\mathrm{V}}{{\mathrm{\pi r}}^2}+2{\mathrm{\pi r}}^2$

$=\frac{2V}{r}+2{\mathrm{\pi r}}^2$

For least surface area, $\frac{dS}{dr}=0$

$\Rightarrow -\frac{2V}{r^2}+4\mathrm{\pi r}=0\Rightarrow V=2{\mathrm{\pi r}}^3$

Hence, $h=2r\Rightarrow h:r=2:1$