Explanation

(d)
Let length of sector is $l$ and radius of sector is$r$.
$\therefore\ \ \ \ \ \ l = \frac{2\pi r\theta}{360{^\circ}}$
Perimeter of sector $P = \frac{2\pi r\theta}{360{^\circ}} + 2r$
$\Rightarrow \ \ \ \ \ r = \frac{p}{\left( \frac{2\pi r\theta}{360{^\circ}} + 2 \right)}$
$\because\ \ \ \ \ A = \frac{\pi r^{2}\theta}{360{^\circ}} = \frac{\pi}{360{^\circ}}\left\lbrack \frac{P^{2}}{\left( \frac{2\pi\theta}{360{^\circ}} + 2 \right)^{2}} \right\rbrack\theta$
$\Rightarrow \ \ \ \ A = \frac{\pi P^{2}}{360{^\circ}}\left\lbrack \frac{\theta}{\left( \frac{2\pi\theta}{360{^\circ}} + 2 \right)^{2}} \right\rbrack$
$\frac{dA}{d\theta} = \frac{\pi P^{2}}{360{^\circ}}\ \left\lbrack \frac{\left( \frac{2\pi\theta}{360{^\circ}} + 2 \right)^{2} - \theta.2\left( \frac{2\pi\theta}{360{^\circ}} + 2 \right)\frac{2\pi}{360{^\circ}}}{\left( \frac{2\pi\theta}{360{^\circ}} + 2 \right)^{4}} \right\rbrack$
For maxima for minima, put $\frac{dA}{d\theta} = 0$
$\left( \frac{2\pi\theta}{360{^\circ}} + 2 \right) - \frac{4\theta\pi}{360{^\circ}} = 0$
$\Rightarrow \ \ \frac{2\pi\theta}{360{^\circ}} = 2\ \ \ \Rightarrow \ \ \ \ \theta = 2$rad
Thus, area of sector will be maximum, if sectorial angle is of 2 rad.