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

1. $\frac{\pi^c}{6}$
2. $\frac{\pi^c}{4}$
3. $4^C$
4. $2^c$

Solution

Perimeter, $P=2 r+r \theta=r(2+\theta)$ According to question,
$r(2+\theta)=k \Rightarrow r=\frac{k}{\theta+2}$ Let $A$ be the area to the sector $A=\frac{1}{2} r^2 \theta=\frac{k^2}{2} \cdot \frac{\theta}{(\theta+2)^2}$ $\Rightarrow \quad \frac{d A}{d \theta}=\frac{k^2}{2}\left[\frac{(\theta+2)^2 \cdot 1-\theta \cdot 2(\theta+2)}{(\theta+2)^4}\right]=\frac{k^2}{2} \cdot \frac{(2-\theta)}{(2+\theta)^3}$ For maximum area, $\frac{d A}{d \theta}=0$ $\Rightarrow \quad \theta=2 \Rightarrow\left[\frac{d^2 A}{d \theta^2}\right] < 0$ $\theta=2$ $\therefore A$ is maximum, when $\theta=2^c$

Asked in: AP EAMCET 2022 (07 Jul Shift 1)

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