Show that semi vertical angle of …

{tex}s = \pi {r^2} + \pi rl{/tex} (given)
{tex}\Rightarrow{/tex}{tex}l = \frac{{s - \pi {r^2}}}{{\pi r}}{/tex} 
Let v be the volume
{tex}v = \frac{1}{3}\pi {r^2}h{/tex} 
{tex}\Rightarrow{/tex}{tex}{v^2} = \frac{1}{9}{\pi ^2}{r^4}{h^2}\,\,\left[ {{h^2} = {l^2} - {r^2}} \right]{/tex} 
{tex}\Rightarrow{/tex}{tex}{v^2} = \frac{1}{9}{\pi ^2}{r^4}\,\left( {{l^2} - {r^2}} \right){/tex} 
{tex}\Rightarrow{/tex}{tex}{v^2} = \frac{1}{9}{\pi ^2}{r^4}\left[ {\,{{\left( {\frac{{s - \pi {r^2}}}{{\pi r}}} \right)}^2} - {r^2}} \right]{/tex} 
{tex}= \frac{1}{9}{\pi ^2}{r^4}\left[ {\frac{{\left( {s - \pi {r^2}} \right)}^2}{{{\pi ^2}{r^2}}} - \frac{{{r^2}}}{1}} \right]{/tex} 
{tex}= \frac{1}{9}{r^2}\left[ {{{\left( {s - \pi {r^2}} \right)}^2} - {\pi ^2}{r^4}} \right]{/tex} 
{tex}= \frac{1}{9}{r^2}\left[ {{s^2} + {\pi ^2}{r^4} - 2s\pi {r^2} - {\pi ^2}{r^4}} \right]{/tex} 
{tex}= \frac{1}{9}{r^2}\left[ {{s^2} - 2s\pi {r^2}} \right]{/tex} 
{tex}z = \frac{1}{9}\left[ {{s^2}{r^2} - 2s\pi {r^4}} \right]{/tex} 
{tex}\left[ {\because {v^2} = z} \right]{/tex} 
Now {tex}\frac{{dz}}{{dr}} = \frac{1}{9}\,\left[ {2r{s^2} - 8s\pi {r^3}} \right]{/tex} 
{tex}0 = \frac{1}{9}\,\left[ {2r{s^2} - 8s\pi {r^3}} \right]{/tex} 
{tex}8s\pi {r^2} = 2r{s^2}{/tex} 
 {tex}\implies{/tex} {tex}4\pi {r^2} = s{/tex} 
Now  {tex}\frac{{{d^2}z}}{{d{x^2}}} = \frac{1}{9}\,\left[ {2{s^2} - 24s\pi {r^2}} \right]{/tex} 
{tex}{\left. {\frac{{{d^2}z}}{{d{x^2}}}} \right]_{{r^2} = \frac{s}{{4\pi }}}} = \frac{1}{9}\,\left[ {{{25}^2} - 24\pi .\frac{5}{{4\pi }}} \right]{/tex} 
= + ve
Hence minimum
Now {tex}s = 4\pi {r^2}{/tex} 
We have {tex}s = \pi rl + \pi {r^2}{/tex} 
{tex}4\pi {r^2} = \pi rl + \pi {r^2}{/tex} 
{tex}\Rightarrow{/tex} {tex}3\pi {r^2} = \pi rl{/tex} 
{tex}\Rightarrow{/tex}3 r = l
{tex}\Rightarrow{/tex}{tex}\frac{r}{l} = \frac{1}{3}{/tex} 
{tex}\Rightarrow{/tex}{tex}\sin \alpha = \frac{1}{3}{/tex} 
{tex}\therefore{/tex}{tex}\alpha = {\sin ^{ - 1}}\left( {\frac{1}{3}} \right){/tex}