Application of Derivatives — Class 12 Maths Solution

Let $r$ be the radius of the circular base, $h$ be the height and $S$ be the total surface area of a right circular cylinder, then
$S = 2\pi {r^2} + 2\pi rh$

Let $V$ be the volume of cylinder with the above dimensions, then $V = \pi {r^2}h = \pi {r^2}\left( {\cfrac{{S - 2\pi {r^2}}}{{2\pi r}}} \right) = \cfrac{r}{2}(S - 2\pi {r^2})$

$\Rightarrow V = \cfrac{{Sr}}{{\dot 2}} - \pi {r^3}$ … (i)

Differentiating (i) w.r.t. $r$, we get $\cfrac{{dV}}{{dr}} = \cfrac{S}{2} - 3\pi {r^2}$

To find the maximum or minimum volume,
$\cfrac{{dV}}{{dr}} = 0 \Rightarrow \cfrac{S}{2} - 3\pi {r^2} = 0 \Rightarrow {r^2} = \cfrac{S}{{6\pi }} \Rightarrow r = \sqrt {\cfrac{S}{{6\pi }}}$

$\cfrac{{{d^2}V}}{{d{r^2}}} = - 6\pi r\,\,{\rm{and }}{\left( {\cfrac{{{d^2}V}}{{d{r^2}}}} \right)_{r = \sqrt {S/(6\pi )} }} = - 6\pi \sqrt {\cfrac{S}{{6\pi }}} < 0$

Therefore we can say that Volume has a maximum value at $r = \sqrt {\cfrac{S}{{6\pi }}}$.

When $r = \sqrt {\cfrac{S}{{6\pi }}}$, then $h = \cfrac{{S - 2\pi {r^2}}}{{2\pi r}} = \cfrac{{S - 2\pi \left( {\cfrac{S}{{6\pi }}} \right)}}{{2\pi \sqrt {\cfrac{S}{{6\pi }}} }}$

$= \cfrac{{4\pi S/6\pi }}{{2\pi \sqrt {\cfrac{S}{{6\pi }}} }}$

$\Rightarrow h = 2\sqrt {\cfrac{S}{{6\pi }}} = 2$ (radius)=diameter

So, volume is maximum when the height is equal to the diameter of the base.