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The perimeter of a sector is constant. If its area is to be maximum, the sectorical angle should be
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Verified Answer
The correct answer is:
$2^c$
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 $\theta=2^c$
$P=2 r+r \theta=r(2+\theta)$
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 $\theta=2^c$
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