Let $r$ be radius, $l$ be the slant height and $h$ be the height of the cone of given surface area $s$. Then
$s=\pi r^2+\pi r l$
$\Rightarrow l=\frac{s-\pi r^2}{\pi r} \quad \ldots(i)$
Let $\mathrm{V}$ be the volume of cone,then
$\mathrm{V}=\frac{1}{3} \pi r^2 h$
$\begin{aligned}
&\therefore \mathrm{V}^2=\frac{1}{9} \pi^2 r^4 h^2=\frac{1}{9} \pi^2 r^4\left(l^2-r^2\right) \text { Using (i) } \\
&\Rightarrow \mathrm{V}^2=\frac{\pi^2 r^4}{9}\left[\left(\frac{s-\pi r^2}{\pi}\right)^2-r^2\right] \\
&=\frac{1}{9} s\left(s r^2-2 \pi r^4\right)
\end{aligned}$
Let $z=\mathrm{V}^2$. Then $\mathrm{V}$ is maximum or minimum according as $z$ ismaximum or minimum.
$\begin{aligned}
&z=\frac{1}{9} s\left(s r^2-2 \pi r^4\right) \\
&\Rightarrow \frac{d z}{d r}=\frac{1}{9} s\left(2 s r-8 \pi r^3\right) \quad \ldots(ii)
\end{aligned}$
For maximum or minimum, we must have $\frac{d z}{d r}=0$ $\Rightarrow 2 s r-8 \pi r^3=0 \Rightarrow s=4 \pi r^2 \quad \ldots(iii)$
Diff. (ii) w.r.t. $r$, we get $\frac{d^2 z}{d r^2}=\frac{s}{9}\left(2 s-24 \pi r^2\right)$
$\frac{d^2 z}{d r^2}=\frac{s}{9}\left[2 s-24 \pi \frac{s}{4 \pi}\right]=-\frac{4 s^2}{9} < 0$
So $z$ is maximum when $s=4 \pi r^2$
Now $s=4 \pi r^2 \Rightarrow s=\pi r l+\pi r^2$
$\begin{aligned}
&\Rightarrow 4 \pi r^2=\pi r l+\pi r^2 \\
&\Rightarrow 3 \pi r^2=\pi r l \Rightarrow l=3 r \therefore \sin \alpha=\frac{r}{l}=\frac{r}{3 r}=\frac{1}{3}
\end{aligned}$
Hence $\mathrm{V}$ is maximum when $\alpha=\sin ^{-1} \frac{1}{3}$.