Question

If the perimeter of a sector is $40cm$, then the angle of sector (in radians) when its area is maximum, is equal to

Answer 1

Let r be the radius, $\theta$ be the angle of Sector

Given, $2r+r\theta =40\cdots \left(i\right)$

Area, $A=\frac{1}{2}{r}^2\theta$

$=\frac{1}{2}{r}^2\left(\frac{40-2r}{r}\right)$

$=\frac{1}{2}(40r-2r^2)$

$A=20r-r^2$

Now, $A^{\prime }=20-2r$ and $A^{\prime \prime }=-2<0$

$\therefore A\text{is maximum when}{A}^{\prime }=0$ $\Rightarrow r=10\therefore 20+10\theta =40\left(\text{from}\right(i\left)\right)\Rightarrow \theta =2$