Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $ta{n}^{-1}\sqrt{2}$.

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

Let $\theta$ be the semi-vertical angle of the cone.
It is clear that $\theta ϵ(0,\frac{\pi }{2})$.

Let r, h and I be the radius, height and the slant height of the cone respectively.
Tge slant height of the cone is given i.e., consider as constant.
Let V be the volume of the cone; $V=\frac{\pi }{3}{r}^{2}h$
$\Rightarrow V=\frac{1}{3}\pi \left(l^{2}si{n}^2\theta \right)\left(lcos\theta \right)=\frac{1}{3}\pi l^{3}si{n}^2\theta cos\theta$
On differentiating w.r.t. $\theta$, we get
$\frac{dV}{d\theta }=\frac{l^3\pi }{3}\left[si{n}^2\theta \right(-sin\theta \left)\right]+cos\theta \left(2sin\theta cos\theta \right)$
$=\frac{l^3\pi }{3}(-si{n}^3\theta +2sin\theta co{s}^2\theta )$
and $\frac{d^{2}V}{d{\theta }^2}=\frac{l^3\pi }{3}(-3si{n}^2\theta cos\theta +2co{s}^3\theta -4si{n}^2\theta cos\theta )$
$=\frac{l^3\pi }{3}(2co{s}^3\theta -7si{n}^2\theta cos\theta )$
For maxima put $\frac{dV}{d\theta }=0$
$\Rightarrow si{n}^3\theta =2sin\theta co{s}^2\theta \Rightarrow ta{n}^2\theta =2\Rightarrow tan\theta =\sqrt{2}\Rightarrow \theta =ta{n}^{-1}\sqrt{2}$
Now, when $\theta =ta{n}^{-1}\sqrt{2}$, then $ta{n}^2\theta =2orsi{n}^2\theta =2co{s}^2\theta$
Then, we have
$\frac{d^{2}V}{d{\theta }^2}=\frac{I^3\pi }{3}(2co{s}^3\theta -14co{s}^3\theta )=-4\pi l^{3}co{s}^3\theta <0$ for $\theta ϵ(0,\frac{\pi }{2})$
$\therefore$ By second derivative test, the volume V is maximum when $\theta =ta{n}^{-1}\sqrt{2}$
Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is $ta{n}^{-1}\sqrt{2}$

With usual notation, given that total surface area $S=\pi rl+\pi r^2$
$\Rightarrow S=\pi r\sqrt{r^2+h^2}+\pi r^2(\because l=\sqrt{r^2+h^2})$
$\Rightarrow \frac{S}{\pi r}-r=\sqrt{r^2+h^2}\Rightarrow \frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }=h^2\Rightarrow h=\sqrt{\frac{S^2}{{\pi }^2{r}^2}-\frac{2S}{\pi }}(\because \frac{S^2}{{\pi }^2{r}^2}>\frac{2S}{\pi })\dots \left(i\right)$
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,
$sin\theta =\frac{r}{\sqrt{r^2+h^2}}=-\frac{\sqrt{\frac{S}{4\pi }}}{\sqrt{\frac{S}{4\pi }+\frac{2S}{\pi }}}=\frac{1}{\sqrt{1+8}}i.e.,\theta =si{n}^{-1}\left(\frac{1}{3}\right)$