Show that the height of the cylinder of maximum volume

Solution:

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

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

$\Rightarrow h=2 \sqrt{R^{2}-r^{2}}$           ......(1)

Volume of cylinder, $V=\pi r^{2} h$

Squaring both sides, we get

$\Rightarrow V^{2}=\pi^{2} r^{4} h^{2}$

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

Now,

$Z=4 \pi^{2}\left(r^{4} R^{2}-r^{6}\right)$

$\Rightarrow \frac{d Z}{d r}=4 \pi^{2}\left(4 r^{3} R^{2}-6 r^{5}\right)$

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

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

$\Rightarrow 4 \pi^{2}\left(4 r^{3} R^{2}-6 r^{5}\right)=0$

$\Rightarrow 4 r^{3} R^{2}=6 r^{5}$

$\Rightarrow 6 r^{2}=4 R^{2}$

$\Rightarrow r^{2}=\frac{4 R^{2}}{6}$

$\Rightarrow r=\frac{2 R}{\sqrt{6}}$

Substituting the value of $r$ in eq. (1), we get

$\Rightarrow h=2 \sqrt{R^{2}-\left(\frac{2 R}{\sqrt{6}}\right)^{2}}$

$\Rightarrow h=2 \sqrt{\frac{6 R^{2}-4 R^{2}}{6}}$

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

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

Now,

$\frac{d^{2} Z}{d r^{2}}=4 \pi^{2}\left(12 r^{2} R^{2}-30 r^{4}\right)$

$\Rightarrow \frac{d^{2} Z}{d r^{2}}=4 \pi^{2}\left(12\left(\frac{2 R}{\sqrt{6}}\right)^{2} R^{2}-30\left(\frac{2 R}{\sqrt{6}}\right)^{4}\right)$

$\Rightarrow \frac{d^{2} Z}{d r^{2}}=4 \pi^{2}\left(8 R^{4}-\frac{80 R^{4}}{6}\right)$

$\Rightarrow \frac{d^{2} Z}{d r^{2}}=4 \pi^{2}\left(\frac{48 R^{4}-80 R^{4}}{6}\right)$

$\Rightarrow \frac{d^{2} Z}{d r^{2}}=4 \pi^{2}\left(-\frac{16 R^{4}}{3}\right)<0$

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

Hence proved.