Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, of the dimensions of the can which has the minimum surface area.

Let $r$ cm be the radius, $h$ cm be the height, $Sc{m^2}$ be the total surface area and $Vc{m^3}$ be the volume, then

$V = \pi {r^2}h = 100 \Rightarrow h = \cfrac{{100}}{{\pi {r^2}}}$ …(i)
and $S = 2\pi {r^2} + 2\pi rh$ …(ii)

$\Rightarrow S = 2\pi {r^2} + 2\pi r\left( {\cfrac{{100}}{{\pi {r^2}}}} \right) = 2\pi {r^2} + \cfrac{{200}}{r}$ (using (i))

Differentiating $S = 2\pi {r^2} + \cfrac{{200}}{r}$ w.r.t. $r$, we get
$\cfrac{{dS}}{{dr}} = 4\pi r - \cfrac{{200}}{{{r^2}}}$

To find the maximum or minimum surface area,
$\cfrac{{dS}}{{dr}} = 0 \Rightarrow 4\pi r - \cfrac{{200}}{{{r^2}}} = 0 \Rightarrow {r^3} = \cfrac{{200}}{{4\pi }} \Rightarrow r = {\left( {\cfrac{{50}}{\pi }} \right)^{1/3}}$

$\cfrac{{{d^2}S}}{{d{r^2}}} = 4\pi + \cfrac{{200 \times 2}}{{{r^3}}} = 4\pi + \cfrac{{400}}{{{r^3}}}$
and ${\left( {\cfrac{{{d^2}S}}{{d{r^2}}}} \right)_{r = {{(50/\pi )}^{1/3}}}} = 4\pi + \cfrac{{400}}{{(50/\pi )}} > 0$

Therefore $S$ has a minimum value , when $r = {\left( {\cfrac{{50}}{\pi }} \right)^{1/3}}$

When $r = {\left( {\cfrac{{50}}{\pi }} \right)^{1/3}}$
$h = \cfrac{{100}}{{\pi {r^2}}} = \cfrac{{100}}{{\pi {{\left( {\cfrac{{50}}{\pi }} \right)}^{2/3}}}}$

$= \cfrac{{100}}{{{{(50)}^{2,3}}{\pi ^{1/3}}}} = \cfrac{{50 \times 2}}{{{{(50)}^{2/3}}{\pi ^{1/3}}}} = 2{\left( {\cfrac{{50}}{\pi }} \right)^{1/3}}$

Therefore, radius $= {\left( {\cfrac{{50}}{\pi }} \right)^{\cfrac{1}{3}}}$ cm and height $= 2{\left( {\cfrac{{50}}{\pi }} \right)^{\cfrac{1}{3}}}$ cm.