A closed cylinder has volume

Solution:

Let the height, radius of the base and surface area of the cylinder be $h, r$ and $S$, respectively. Then,

Volume $=\pi r^{2} h$

$\Rightarrow 2156=\pi r^{2} h$

$\Rightarrow 2156=\frac{22}{7} r^{2} h$

$\Rightarrow h=\frac{2156 \times 7}{22 r^{2}}$

$\Rightarrow h=\frac{686}{r^{2}}$             ......(1)

Surface area $=2 \pi r h+2 \pi r^{2}$

$\Rightarrow S=\frac{4312}{r}+\frac{44 r^{2}}{7}$              $[$ From eq. (1) $]$

$\Rightarrow \frac{d S}{d r}=\frac{4312}{-r^{2}}+\frac{88 r}{7}$

For maximum or minimum values of $S$, we must have

$\frac{d S}{d r}=0$

$\Rightarrow \frac{4312}{-r^{2}}+\frac{88 r}{7}=0$

$\Rightarrow \frac{4312}{r^{2}}=\frac{88 r}{7}$

$\Rightarrow r^{3}=\frac{4312 \times 7}{88}$

$\Rightarrow r^{3}=343$

$\Rightarrow r=7 \mathrm{~cm}$

Now,

$\frac{d^{2} s}{d r^{2}}=\frac{8624}{r^{3}}+\frac{88}{7}$

$\Rightarrow \frac{d^{2} s}{d r^{2}}=\frac{8624}{343}+\frac{88}{7}$

$\Rightarrow \frac{d^{2} s}{d r^{2}}=\frac{176}{7}>0$

So, the surface area is minimum when $r=7 \mathrm{~cm}$.