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Q.
The perimeter of a sector is a constant. If its area is to be maximum, the sectorial angle is :
EAMCETEAMCET 2006
Solution:
Let length of sector is land radius of sector is $r$.
$\therefore l=\frac{2 \pi r \theta}{360^{\circ}}$
Perimeter of sector $P=\frac{2 \pi r \theta}{360^{\circ}}+2 r$
$\Rightarrow \quad P=\left(\frac{2 \pi \theta}{360^{\circ}}+2\right) r $
$\Rightarrow r=\frac{P}{\left(\frac{2 \pi \theta}{360^{\circ}}+2\right)}$
$\because A=\frac{\pi r^{2} \theta}{360^{\circ}}$
$A=\frac{\pi}{360^{\circ}}\left[\frac{P^{2}}{\left(\frac{2 \pi \theta}{360^{\circ}}+2\right)^{2}}\right] \theta$
$A=\frac{\pi P^{2}}{360^{\circ}}\left[\frac{\theta}{\left(\frac{2 \pi \theta}{360^{\circ}}+2\right)^{2}}\right]$
$\frac{d A}{d \theta}=\frac{\pi P^{2}}{360^{\circ}}\left[\frac{\left(\frac{2 \pi \theta}{360^{\circ}}+2\right)^{2}-\theta \cdot 2\left(\frac{2 \pi \theta}{360^{\circ}}+2\right) \frac{\pi}{360^{\circ}}}{\left(\frac{2 \pi \theta}{360^{\circ}}+2\right)^{4}}\right]$
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 \frac{\pi \theta}{180^{\circ}}=2 $
$\Rightarrow \theta =\frac{2 \times 180^{\circ}}{\pi} $
$= 2$ rad
Thus Area of sector will be maximum, if sectorial angle is of $2$ rad.