Perimeter of a sector is p. The area of the sector is maximum when its radius is

Let r be the radius and $\theta$ be the sector angle. Then

Perimeter $=p\Rightarrow 2r+r\theta =p\Rightarrow \theta =\frac{p-2r}{r}$

Let A denote the area of the sector. Then,

$\begin{array}{l}A=\frac{1}{2}{r}^2\theta \\ \Rightarrow A=\frac{1}{2}{r}^{ }(p-2r)\\ \Rightarrow A=\frac{1}{2}\left(pr-2r^2\right)\\ \Rightarrow \frac{dA}{dr}=\frac{1}{2}(p-4r)\text{ and }\frac{d^{2}A}{d{r}^2}=-2\end{array}$

Now, $\frac{dA}{dr}=0\Rightarrow p=4r\Rightarrow r=\frac{p}{4}$

Clearly, $\frac{d^{2}A}{d{r}^2}<0\text{ for all }r$

Hence, A is maximum when $r=\frac{p}{4}$