Question

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

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

Answer 1

Answer: (D)

$2^C$

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

Then, perimeter$=2r+r\theta =k$(constant) ....[given]
$\Rightarrow r=\frac{k}{2+\theta }$
Let $A$ be the area of the sector, then
$A=\frac{1}{2}{r}^2\theta =\frac{k^2}{2}\cdot \frac{\theta }{(\theta +2{)}^2}$
On differentiating both sides w.r.t. $\theta$, we get
$\frac{dA}{d\theta }=\frac{k^2}{2}\left\{\frac{(\theta +2{)}^2-2\theta (\theta +2)}{(\theta +2{)}^4}\right\}$
$=\frac{k^2}{2}\frac{(2-\theta )}{(\theta +2{)}^3}$
For maximum, put $\frac{dA}{d\theta }=0$
$\Rightarrow \theta =2$
Now, $\frac{d^{2}A}{d{\theta }^2}=\frac{k^2}{2}\left[\frac{2\times (-3)}{(\theta -2{)}^4}-\frac{(\theta +2{)}^3\times 1-\theta \times 3(\theta +2{)}^2}{\left[\right(\theta +2{)}^3{]}^2}\right]$
$=\frac{k^2}{2}\left[\frac{-6}{(\theta +2{)}^4}-\frac{\theta +2-3\theta }{(\theta +2{)}^4}\right]$
$=\frac{-k^2}{2}\left[\frac{6}{(\theta +2{)}^4}+\frac{2-\theta }{(\theta +2{)}^4}\right]$
At $\theta =2$,
$\frac{d^{2}A}{d{\theta }^2}=-\frac{k^2}{2}\left[\frac{6}{4^4}+0\right]$
$=\frac{-3k^2}{256}<0$
Hence, $A$ is maximum, when $\theta =2^C$.