Question

The perimeter of a sector is given. The area is maximum when the angle of the sector is

• A. $1$ radian
• B. $2$ radians
• C. $3$ radians
• D. $4$ radians

Answer 1

Answer: (B)

$2$ radians

Perimeter of sector $=2r+l=2r+r\theta$
$\Rightarrow k=2r+r\theta$
$\Rightarrow \theta =\frac{k-2r}{r}$
$AreaA=\frac{1}{2}{r}^2\theta$
$\Rightarrow A=f\left(r\right)=\frac{1}{2}r\left(k-2r\right)$
$f^{\prime }\left(r\right)=\frac{1}{2}(k-4r)$
For maxima or minima,
$f^{\prime }\left(r\right)=0$
$\Rightarrow r=\frac{k}{4}$
$f^{\prime \prime }\left(r\right)=\frac{1}{2}(-4)=-2$
$f^{\prime \prime }\left(r\right)<0$ at $r=\frac{k}{4}$
Hence, f(x) has a maximum at $r=\frac{k}{4}$
So, $\theta =2$ radians