Explanation:
(D) : Suppose \(\pi\) be the radius of the circle and \(\theta\) be the sectorial angle of a sector of it.
Then,
Perimeter, \(2 \pi+\pi \theta=\mathrm{k}\) (constant) [given]
\(\pi=\frac{\mathrm{k}}{2+\theta}\)
Let, A be the area of the sector,
Then,
\(\mathrm{A}=\frac{1}{2} \pi^{2} \theta=\frac{\mathrm{k}^{2}}{2} \cdot \frac{\theta}{(\theta+2)^{2}}\)
On differentiating both sides with respect to \(\theta\), we get-
\(\frac{\mathrm{dA}}{\mathrm{d} \theta}=\frac{\mathrm{k}^{2}}{2}\left\{\frac{(\theta+2)^{2}-2 \theta(\theta+2)}{(\theta+2)^{4}}\right\}=\frac{\mathrm{k}^{2}}{2}=\frac{(2-\theta)}{(\theta+2)^{3}}\)
For maximum Put-
\(\frac{d A}{d \theta} =0\)
\(\theta =2\)
Now,
\(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{~d} \theta^{2}}= \frac{\mathrm{k}^{2}}{2}\left[\frac{2 \times(-3)}{(\theta+2)^{4}}-\frac{(\theta+2)^{3} \times 1-\theta \times 3(\theta+2)^{2}}{\left[(\theta+2)^{3}\right]^{2}}\right]\)
\(=\frac{\mathrm{k}^{2}}{2}\left[\frac{-6}{(\theta+2)^{4}}-\frac{\theta+2-3 \theta}{(\theta+2)^{4}}\right]\)
\(=\frac{-\mathrm{k}^{2}}{2}\left[\frac{-6}{(\theta+2)^{4}}-\frac{2-\theta}{(\theta+2)^{4}}\right]\)
At, \(\theta=2\),
\(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{~d} \theta^{2}}=\frac{-\mathrm{k}^{2}}{2}\left[\frac{6}{4^{4}}+0\right]=\frac{-3 \mathrm{k}^{2}}{256}\lt 0\)
Hence, \(\mathrm{A}\) is maximum,
When, \(\theta=2^{\mathrm{C}}\)
