Q98 / 180 · KCET 2012 · Mathematics
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$4^{\mathrm{c}}$
- Correct answerD$2^{\mathrm{c}}$
Solution
Let $r$ be the radius and $\theta$ be the sectorial angle. $\therefore$ Perimeter of sector,
$\begin{aligned} & k=2 r+\frac{\theta}{180^{\circ}} \pi r \\ \Rightarrow \quad & r=\frac{k}{2+\frac{\theta \pi}{180^{\circ}}} \end{aligned}$
$\begin{aligned} \therefore \text { Area of sector, } \mathrm{A} &=\frac{\theta}{360^{\circ}} \pi \mathrm{r}^{2} \\ &=\frac{\pi}{360^{\circ}}\left[\theta \times\left(\frac{\mathrm{k}}{2+\frac{\pi \theta}{180^{\circ}}}\right)^{2}\right] \\ \Rightarrow \quad \mathrm{A} &=\frac{\mathrm{k}^{2} \pi}{360^{\circ}}\left[\theta \times\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-2}\right] \end{aligned}$
On differentiating w.r.t. ' $\theta$ ', we get
$$
\begin{aligned}
\frac{\mathrm{dA}}{\mathrm{d} \theta}=& \frac{\mathrm{k}^{2} \pi}{360^{\circ}}\left[1 \times\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-2}-2\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-3}\right.\\
&\left.=\frac{\mathrm{k}^{2} \pi}{360^{\circ}}\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-2}\left[1-\frac{\pi}{180^{\circ}}\right)\right] \\
&=\frac{\mathrm{k}^{2} \pi}{360^{\circ}}\left(2+\frac{\pi \theta}{180^{\circ}}\right)^{-3}\left[2-\frac{\pi \theta}{180^{\circ}}\right] \\
\text { Put } \left.\frac{\mathrm{dA}}{\mathrm{d} \theta}=0 \Rightarrow 2-\frac{\pi \theta}{180^{\circ}}\right] \\
\Rightarrow \quad \theta=\frac{2 \times 180^{\circ}}{\pi}=2^{\circ}
\end{aligned}
$$
At $\theta=2^{c}, \quad \frac{d^{2} \mathrm{~A}}{d \theta^{2}} < 0$ (maximum)
Hence, the correct answer is option: 'D'.