# A sector of a circle whose radius is r and whose angle is theta has a fixed perimeter P. How do you find the values of r and theta so that the area of the sector is a maximum?

Dec 4, 2016

$r = \frac{P}{4}$ and $\theta = 2$

#### Explanation:

The perimeter of the sector is two radii and the arc cut off by $\theta$. So, the perimeter is given by

$P = 2 r + r \theta$

The area of a sector is $A = \frac{1}{2} {r}^{2} \theta$

Since $P$ is fixed, the only variables in $P = 2 r + r \theta$ are $r$ and $\theta$

$r = \frac{P}{2 + \theta}$ and $\theta = \frac{P - 2 r}{r} = \frac{P}{r} - 2$. Is we rewrite for $A$ using only $\theta$ we'll need the quotient rule to differentiate. So let's rewrite $A$ using only $r$.

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

We want to maximize $A$, so . . .

$A ' = \frac{P}{2} - 2 r = 0$ at $r = \frac{P}{4}$

Note that $A ' ' = - 2$ so $A \left(\frac{P}{4}\right)$ is a maximum, not a minimum.

Use $\theta = \frac{P}{r} - 2$ from above to get $\theta = 2$