Show that the semi-vertical angle of right circular cone of given surface area and maximum volume is ${\sin ^{ - 1}}\left( {\cfrac{1}{3}} \right)$

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Let $r$ be the radius, $l$ be the slant height and $h$ be the vertical height and $\alpha$ be the semi-vertical angle of cone of given surface area $S$. Then,

$S = \pi rl + \pi {r^2} \Rightarrow l = \cfrac{{S - \pi {r^2}}}{{\pi r}}$

Let $V$ be the volume of the cone.

Therefore, $V = \cfrac{1}{3}\pi {r^2}h$
$\Rightarrow {V^2} = \cfrac{1}{9}{\pi ^2}{r^4}{h^2} = \cfrac{1}{9}{\pi ^2}{r^4}\left( {{l^2} - {r^2}} \right)$

$\Rightarrow {V^2} = \cfrac{{{\pi ^2}}}{9}{r^4}\left[ {{{\left( {\cfrac{{S - \pi {r^2}}}{{\pi r}}} \right)}^2} - {r^2}} \right] = \cfrac{{{\pi ^2}{r^4}}}{9}\left[ {\cfrac{{{{(S - \pi {r^2})}^2} - {\pi ^2}{r^4}}}{{{\pi ^2}{r^2}}}} \right]$

$\Rightarrow {V^2} = \cfrac{1}{9}S(S{r^2} - 2\pi {r^4})$

Let $Z = {V^2}$, then $y$ is maximum or minimum according as $Z$ is maximum or minimum.
$\therefore Z = \cfrac{1}{9}S\left( {S{r^2} - 2\pi {r^4}} \right) \Rightarrow \;\cfrac{{dZ}}{{dr}} = \cfrac{1}{9}S\left( {2Sr - 8\pi {r^3}} \right)$

To find maximum or minimum $Z$, we have $\cfrac{{dZ}}{{dr}} = 0$

$\Rightarrow 2Sr - 8\pi {r^3} = 0 \Rightarrow S = 4\pi {r^2} \Rightarrow r = \sqrt {\cfrac{S}{{4\pi }}}$

Now, $\cfrac{{{d^2}Z}}{{d{r^2}}} = \cfrac{S}{9}(2S - 24\pi {r^2})$

$\Rightarrow {\left( {\cfrac{{{d^2}Z}}{{d{r^2}}}} \right)_{r = \sqrt {\cfrac{S}{{4\pi }}} }} = \cfrac{S}{9}\left( {2S - 24\pi \cfrac{S}{{4\pi }}} \right)$

$\Rightarrow$ $\cfrac{{{d^2}Z}}{{d{r^2}}} = - \cfrac{{4{S^2}}}{9} < 0$

So, $Z$ is maximum when $S = 4\pi {r^2}$.

Hence, $V$ is maximum when $S = 4\pi {r^2}$

Now, $S = 4\pi {r^2} \Rightarrow \pi rl + \pi {r^2} = 4\pi {r^2} \Rightarrow l = 3r$
In $\Delta AOC$

$\sin \alpha = \cfrac{r}{l} = \cfrac{r}{{3r}} = \cfrac{1}{3} \Rightarrow \alpha = {\sin ^{ - 1}}\cfrac{1}{3}$

Hence $V$ is maximum when $\alpha = {\sin ^{ - 1}}\cfrac{1}{3}$.