Question

The length of the perimeter of a sector of a circle is 20cm.Give an expression for area of sector in terms of r(radius of the circle) and find maximum area of sector.

Answer 1

Let $OAB$ be the given sector of a circle of radius $r$ cm such that $AB=1$ cm and $\angle AOB=\theta$ radius then

Area of sector $OAB=20$ cm

$\Rightarrow 2r+l=20⟶\left(1\right)$

Let $S$ be the area of sector $OAB$. Then

$S=\frac{1}{2}{r}^2\theta$

$S=\frac{1}{2}{r}^2\times \frac{l}{r}=\frac{1}{2}lr$

$S=\frac{1}{20}(20-2r)r$

$S=10r-r^2$

This is the required expression for the area $S$ if the sector in term of $r$

$S=10r-r^2$

$\frac{dS}{dr}=10-2r$

$\frac{d^{2}S}{{dr}^2}=-2$

For maximum / minimum value of $S$ we must have

$\frac{dS}{dr}=0\Rightarrow 10-2r=0r=5cm$

Clearly $\frac{d^{2}S}{{dr}^2}<$ for all $r$. $S$ is maximum where $r=10$

Maximum value of $S=10\times 5-{\left(5\right)}^2=25{cm}^2$.