The perimeter of a sector is given. The area is maximum when the angle of the sector is

Given perimeter is constant

$$\begin{gathered} l+2\mathrm{r}=\mathrm{k}\Rightarrow l=\mathrm{k}-2\mathrm{r} \\ \mathrm{Area}=\mathrm{A}=\frac{1}{2} lr \\ =\frac{1}{2}\left(\mathrm{k}-2\mathrm{r}\right)\mathrm{r} \\ =\frac{1}{2}\left(\mathrm{kr}-2{\mathrm{r}}^2\right) \\ {\mathrm{A}}^{\mathrm{I}}=\frac{1}{2}\left(\mathrm{k}-4\mathrm{r}\right) \end{gathered}$$

A is Min or Max

$$\begin{gathered} \Rightarrow {\mathrm{A}}^{\mathrm{I}}=0 \\ \mathrm{k}-4\mathrm{r}=0 \\ \Rightarrow \mathrm{r}=\frac{\mathrm{k}}{4} \\ l=\mathrm{k}-2\mathrm{r}=\mathrm{k}-\frac{\mathrm{k}}{2} \\ l=\frac{\mathrm{k}}{2} \\ \mathrm{\theta }=\frac{l}{r}=\frac{{\displaystyle \frac{\mathrm{k}}{2}}}{\frac{\mathrm{k}}{4}}=2 \end{gathered}$$