The perimeter of a sector is a constant. If its area is to be maximum, the sectorial angle is

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

$\therefore 2r+r\theta =k\Rightarrow r=\frac{k}{\theta +2}$

Area, $A=\frac{1}{2}{r}^2\theta =\frac{k^2\theta }{2(\theta +2{)}^2}$;

$\frac{dA}{d\theta }=0\Rightarrow 2-\theta =0\Rightarrow \theta =2$

$\begin{array}{r}\text{ Now, }\frac{{\mathrm{d}}^2\mathrm{A}}{\mathrm{d}{\theta }^2}=\frac{{\mathrm{k}}^2}{2}\left[\frac{2\times (-3)}{(\theta -2{)}^4}-\frac{(\theta +2{)}^3\times 1-\theta \times 3(\theta +2{)}^2}{{\left[(\theta +2{)}^3\right]}^2}\right]\\ =\frac{{\mathrm{k}}^2}{2}\left[\frac{-6}{(\theta +2{)}^4}-\frac{\theta +2-3\theta }{(\theta +2{)}^4}\right]\\ =\frac{-{\mathrm{k}}^2}{2}\left[\frac{6}{(\theta +2{)}^4}+\frac{2-\theta }{(\theta +2{)}^4}\right]\\ \text{ At }\theta =2\\ \frac{{\mathrm{d}}^2\mathrm{A}}{\mathrm{d}{\theta }^2}=-\frac{{\mathrm{k}}^2}{2}\left[\frac{6}{4^4}+0\right]\\ =\frac{-3{\mathrm{k}}^2}{256}<0\end{array}$

$\text{ Hence, }A\text{ is maximum, when }\theta =2^{\mathrm{C}}\text{. }$