The semivertical angle of a cone of maximum volume and of given total surface area is

given surface area=k

$$\begin{gathered} {\mathrm{\pi r}}^2+\mathrm{\pi rl}=\mathrm{k} \\ \mathrm{\pi rl}=\mathrm{k}-{\mathrm{\pi r}}^2 \\ {\mathrm{\pi }}^2{\mathrm{r}}^2{\mathrm{l}}^2={(\mathrm{k}-{\mathrm{\pi r}}^2)}^2 {\mathrm{l}}^2 \end{gathered}$$

volume $\mathrm{V}=\frac{\mathrm{\pi }}{3}{\mathrm{r}}^2 \mathrm{h}$

$$\begin{gathered} {\mathrm{V}}^2=\frac{{\mathrm{\pi }}^2}{9} {\mathrm{r}}^4 {\mathrm{h}}^2 \\ =\frac{{\mathrm{\pi }}^2}{9}{\mathrm{r}}^4\left(\frac{\left(\mathrm{k}-{\mathrm{\pi r}}^2\right)-{\mathrm{\pi }}^2{\mathrm{r}}^4}{{\mathrm{\pi }}^2{\mathrm{r}}^2}\right) \\ =\frac{1}{9} {\mathrm{r}}^2\left({\mathrm{k}}^2+{\mathrm{\pi }}^2{\mathrm{r}}^4-2\mathrm{k\pi } {\mathrm{r}}^2-{\mathrm{\pi }}^2 {\mathrm{r}}^4\right) \\ =\frac{1}{9}\left({\mathrm{k}}^2 {\mathrm{r}}^2-2{\mathrm{k\pi r}}^4\right) \\ {\mathrm{V}}^1=\frac{1}{9}\left(2{\mathrm{k}}^2\mathrm{r}-8{\mathrm{\pi r}}^3\right) \end{gathered}$$

V is min or max

$$\begin{gathered} {\mathrm{V}}^1=0 \\ 2{\mathrm{k}}^2\mathrm{r}=8\mathrm{k} {\mathrm{\pi r}}^3 \\ \mathrm{k}=4{\mathrm{\pi r}}^2 \\ {\mathrm{\pi r}}^2+\mathrm{\pi rl}=4{\mathrm{\pi r}}^2 \\ \mathrm{\pi rl}=3{\mathrm{\pi r}}^2 \\ \mathrm{l}=3\mathrm{r} \\ \frac{\mathrm{r}}{\mathrm{l}}=\frac{1}{3} \\ \mathrm{sin\theta }=\frac{\mathrm{r}}{\mathrm{l}}=\frac{1}{3} \\ \Rightarrow \mathrm{\theta }={\sin }^{-1} \frac{1}{3} \end{gathered}$$