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)

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, $\frac{d^{2}A}{d{\theta }^2}=\frac{k^2}{2}\left[\frac{2\times (-3)}{{(\theta +2)}^4}-\frac{{(\theta +2)}^3\times 1-\theta \times 3{(\theta +2)}^2}{{[{(\theta +2)}^3]}^2}\right]$
$=\frac{k^2}{2}\left[\frac{-6}{{(\theta +2)}^4}-\frac{\theta +2-3\theta }{{(\theta +2)}^4}\right]$
$=\frac{-k^2}{2}\left[\frac{6}{{(\theta +2)}^4}+\frac{2-\theta }{{(\theta +2)}^4}\right]$ 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 $\theta =2^o$