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

With usual notation, given that total surface area $S=\pi rl+\pi r^2$
$\Rightarrow S=\pi r\sqrt{r^2+h^2}+\pi r^2(\because I=\sqrt{r^2+h^2})$
$\Rightarrow \frac{S}{\pi r}-r=\sqrt{r^2+h^2}\Rightarrow \frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }=h^2\Rightarrow h=\sqrt{\frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }}(\because \frac{S^2}{{\pi }^2{r}^2}>\frac{2S}{\pi })\dots \left(i\right)$
and volume $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi r^2\sqrt{\frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }}$
$\Rightarrow V=\frac{r}{3}\sqrt{S^2-2S\pi r^2},r^2<\frac{S}{2\pi }$ i.e., $0<r<\sqrt{\frac{S}{2\pi }}$
Since, V is maximum, then V^2 is maximum\\
Now, $V^2=\frac{S^2{r}^2}{9}-\frac{2S\pi r^4}{9},0<r<\sqrt{\frac{S}{2\pi }}$
$\therefore \frac{d}{dr}\left(V^2\right)=\frac{2r{S}^2}{9}=\frac{8S\pi r^3}{9}$
and $\frac{d}{d{r}^2}\left(V^2\right)=\frac{2S^2}{9}-\frac{24S\pi r^2}{9}$
For maxima put $\frac{dV}{dr}=0$
$\Rightarrow \frac{2r{S}^2}{9}-\frac{8S\pi r^3}{9}=0\Rightarrow r^2=\frac{S}{4\pi }\Rightarrow \frac{d^2\left(V^2\right)}{d{r}^2}<0forr=\sqrt{\frac{S}{4\pi }}$
From Eq. (i) $h=\sqrt{\frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }}=\sqrt{\frac{S^2\left(4\pi \right)}{{\pi }^{2}S}-\frac{2S}{\pi }}=\sqrt{\frac{2S}{\pi }}$
If $\theta$ is semi-vertical angle of the cone when the volume is maximum,
then in right triangle AOC,
$sin\theta =\frac{r}{r^2+h^2}=-\frac{\sqrt{\frac{S}{4\pi }}}{\sqrt{\frac{S}{4\pi }+\frac{}{2}S\pi }}=\frac{1}{\sqrt{1+8}}i.e.,\theta =si{n}^{-1}\left(\frac{1}{3}\right)$