A given quantity of metal is to be cast into a half cylinder

Solution:

Volume, $V=\frac{1}{2} \pi l\left(\frac{D}{2}\right)^{2}$

$\Rightarrow V=\frac{\pi D^{2} l}{8}$

$\Rightarrow l=\frac{8 V}{\pi D^{2}}$                 ....(1)

Total surface area $=\frac{\pi D^{2}}{4}+l D+\frac{\pi D l}{2}$

$\Rightarrow S=\frac{\pi D^{2}}{4}+\frac{8 V}{\pi D}+\frac{8 V}{2 D}$         $[$ From eq. $(1)]$

$\Rightarrow \frac{d S}{d D}=\frac{\pi D}{2}-\frac{8 V}{\pi D^{2}}-\frac{8 V}{2 D^{2}}$

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

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

$\Rightarrow \frac{\pi D}{2}-\frac{8 V}{\pi D^{2}}-\frac{8 V}{2 D^{2}}=0$

$\Rightarrow \frac{\pi D}{2}=\frac{8 V}{D^{2}}\left(\frac{1}{\pi}+\frac{1}{2}\right)$

$\Rightarrow D^{3}=\frac{16 V}{\pi}\left(\frac{1}{\pi}+\frac{1}{2}\right)$

Now,

$\frac{d^{2} S}{d D^{2}}=\frac{\pi}{2}+\frac{16 V}{D^{3}}\left(\frac{1}{\pi}+\frac{1}{2}\right)$

$\Rightarrow \frac{d^{2} S}{d D^{2}}=\frac{\pi}{2}+\pi>0$

$l=\frac{8 \mathrm{~V}}{\pi D^{2}}$

$\Rightarrow l=\frac{8}{\pi D^{2}}\left[\frac{\pi D^{3}}{16}\left[\frac{2 \pi}{\pi+2}\right]\right]$

$\Rightarrow l=D\left(\frac{\pi}{\pi+2}\right)$

$\Rightarrow \frac{l}{D}=\frac{\pi}{\pi+2}$

Hence proved.