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

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)