Question

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

• A. $\frac {\pi ^c}{6}$
• B. $\frac {\pi ^c}{4}$
• C. ${4}^c$
• D. ${2}^c$

Answer 1

The Correct Option is D

Let $r$ be the radius and $\theta$ be the sectorial angle.
$\therefore$ Perimeter of sector,
$k =2 r+\frac{0}{180^{\circ}} \pi r $
$\Rightarrow r=\frac{k}{2+\frac{\theta \pi}{180^{\circ}}}$
$\therefore$ Area of sector, $A=\frac{\theta}{360^{\circ}} \pi \,r^{2}$
$=\frac{\pi}{360^{\circ}}\left[\theta \times\left(\frac{k}{2+\frac{\pi \theta}{180^{\circ}}}\right)^{2}\right] $
$\Rightarrow A =\frac{k^{2} \pi}{360^{\circ}}\left[\theta \times\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-2}\right]$
On differentiating w.r.t. ' $\theta$, we get
$\frac{ dA}{d \theta}= \frac{k^{2} \pi}{360^{\circ}}\left[1 \times\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-2}-2\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-3}\right.$
$\left.\theta\left(\frac{\pi}{180^{\circ}}\right)\right] $
$= \frac{k^{2} \pi}{360^{\circ}}\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-2}\left[1-\frac{2 \pi \theta / 180^{\circ}}{2+\pi \theta / 180^{\circ}}\right] $
$= \frac{k^{2} \pi}{360^{\circ}}\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-3}\left[2-\frac{\pi \theta}{180^{\circ}}\right] $
Put $ \frac{d A}{d \theta}=0$
$ \Rightarrow 2-\frac{\pi \theta}{180^{\circ}} $
$\Rightarrow \theta=\frac{2 \times 180^{\circ}}{\pi}=2^{c}$
At $\theta=2^{c}, \frac{d^{2} A}{d \theta^{2}}