The perimeter of a sector is a constant. If its area is to be maximum, the sectorial angle is :
The perimeter of a sector is a constant. If its area is to be maximum, the sectorial angle is :
$\frac{\pi^c}{6}$
$\frac{\pi^c}{4}$
$4^c$
$2^c$
Solution
Let length of sector is land radius of sector is $r$.
$\therefore \quad l=\frac{2 \pi r \theta}{360^{\circ}}$
Perimeter of sector $P=\frac{2 \pi r \theta}{360^{\circ}}+2 r$
$\Rightarrow \quad P=\left(\frac{2 \pi \theta}{360^{\circ}}+2\right) r \Rightarrow r=\frac{P}{\left(\frac{2 \pi \theta}{360^{\circ}}+2\right)}$
$\because \quad A=\frac{\pi r^2 \theta}{360^{\circ}}$
$A=\frac{\pi}{360^{\circ}}\left[\frac{P^2}{\left(\frac{2 \pi \theta}{360^{\circ}}+2\right)^2}\right] \theta$
$A=\frac{\pi P^2}{360^{\circ}}\left[\frac{\theta}{\left(\frac{2 \pi \theta}{360^{\circ}}+2\right)^2}\right]$
$\frac{d A}{d \theta}=\frac{\pi P^2}{360^{\circ}}\left[\frac{\left(\frac{2 \pi \theta}{360^{\circ}}+2\right)^2-\theta \cdot 2\left(\frac{2 \pi \theta}{360^{\circ}}+2\right) \frac{\pi}{360^{\circ}}}{\left(\frac{2 \pi \theta}{360^{\circ}}+2\right)^4}\right]$
Put $\frac{d A}{d \theta}=0$, for maxima or minima
$\left(\frac{2 \pi \theta}{360^{\circ}}+2\right)-\frac{4 \theta \pi}{360^{\circ}}=0$
$\Rightarrow \quad \frac{\pi \theta}{180^{\circ}}=2 \Rightarrow \theta=\frac{2 \times 180^{\circ}}{\pi}$
$=2 \mathrm{rad}$
Thus Area of sector will be maximum, if sectorial angle is of $2 \mathrm{rad}$.