Explanation

(c)
Perimeter of sector$= 2r + r\theta$
$\Rightarrow \ \ \ 60 = 2r + r\theta$ [given]
[image:6_image3.emf]
$\Rightarrow \ \ \ \ \theta = \frac{60 - 2r}{r}$
Area of sector, $A = \frac{\pi r^{2}\theta}{360{^\circ}} = \frac{\pi r^{2}(60 - 2r)}{r360{^\circ}}$
$= \frac{\pi r}{180{^\circ}}(30 - r)$
$\Rightarrow \ \ \frac{dA}{dr} = \frac{\pi}{180{^\circ}}(30 - 2r)\$
For maximum area, $\frac{dA}{dr} = 0\ \ \ \Rightarrow \ \ \ r = 15$
Now, $\frac{d^{2}A}{dr^{2}} = \frac{\pi}{180{^\circ}}\ (0 - 2) = - \frac{\pi}{90} < 0$
$\therefore$ It is maximum at $r = 15$ m