Show that the maximum volume of the cylinder

Solution:

Let the height, radius of base and volume of a cylinder be $h, r$ and $V$, respectively. Then,

$\frac{h^{2}}{4}+r^{2}=R^{2}$

$\Rightarrow h^{2}=4\left(R^{2}-r^{2}\right)$

$\Rightarrow r^{2}=R^{2}-\frac{h^{2}}{4}$              $\cdots(1)$

Now,

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

$\Rightarrow V=\pi\left(h R^{2}-\frac{h^{3}}{4}\right)$                        $[$ From eq. $(1)]$

$\Rightarrow \frac{d V}{d h}=\pi\left(R^{2}-\frac{3 h^{2}}{4}\right)$            

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

$\frac{d V}{d h}=0$

$\Rightarrow \pi\left(R^{2}-\frac{3 h^{2}}{4}\right)=0$

$\Rightarrow R^{2}-\frac{3 h^{2}}{4}=0$

$\Rightarrow R^{2}=\frac{3 h^{2}}{4}$

$\Rightarrow h=\frac{2 R}{\sqrt{3}}$

$\frac{d^{2} V}{d h^{2}}=\frac{-3 \pi h}{2}$

$\frac{d^{2} V}{d h^{2}}=\frac{-3 \pi}{2} \times \frac{2 R}{\sqrt{3}}$

$\Rightarrow \frac{d^{2} V}{d h^{2}}=\frac{-3 \pi R}{\sqrt{3}}<0$

So, the volume is maximum when $\mathrm{h}=\frac{2 \mathrm{R}}{\sqrt{3}}$.

Maximum volume $=\pi h\left(R^{2}-\frac{h^{2}}{4}\right)$

$=\pi \times \frac{2 R}{\sqrt{3}}\left(R^{2}-\frac{4 R^{2}}{12}\right)$

$=\frac{2 \pi R}{\sqrt{3}} \frac{8 R^{2}}{12}$

$=\frac{4 \pi R^{3}}{3 \sqrt{3}}$

$=\frac{4 \pi(5 \sqrt{3})^{3}}{3 \sqrt{3}}$

$=500 \pi \mathrm{cm}^{3}$