Among all sectors of a fixed perimeter, choose the one with maximum area. Then the angle at the center of this sector (i.e., the angle between the bounding radii) is

• A

  $\frac{\pi}{3}$
• B

  $\frac{3}{2}$
• C

  $[\mathrm]\sqrt{3}$
• Correct answer
  D

  2

Solution

Perimeter $=2 \mathrm{r}+\mathrm{r} \theta=\mathrm{k}$ (const.) Area $=\frac{\theta}{2} \mathrm{r}^{2}=\frac{1}{2}(\mathrm{k}-2 \mathrm{r}) \cdot \mathrm{r}$ $\mathrm{A}=\frac{1}{2}\left(\mathrm{kr}-2 \mathrm{r}^{2}\right)$ $\frac{\mathrm{d} \mathrm{A}}{\mathrm{dr}}=\frac{1}{2}(\mathrm{k}-4 \mathrm{r})=0 \Rightarrow \mathrm{r}=\frac{\mathrm{k}}{4}$ $\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dr}^{2}} < 0$ So, $2 \mathrm{r}+2 \theta=\mathrm{k}$ $$ \begin{array}{l} \frac{\mathrm{k}}{2}+\frac{\mathrm{k} \theta}{4}=\mathrm{k} \\ \theta=2 \end{array} $$

Hence, the correct answer is option: 'D'.