# Calculus Question?

## The diagram shows a sector of a circle with radius r cm. The perimeter of the sector is 12 cm. Show that the area of the sector, $A c {m}^{2}$, in terms of r, is $A = \frac{72 \theta}{\theta + 2} ^ 2$. Hence determine the maximum value of A as r varies.

Jul 27, 2018

#### Explanation:

The area of the sector is

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

The perimeter of the sector is

$P = 2 r + r \theta$

As $P = 12$

$2 r + r \theta = 12$

$=$, $r \left(2 + \theta\right) = 12$

$r = \frac{12}{2 + \theta}$

Therefore,

$A = \frac{1}{2} \cdot {\left(\frac{12}{2 + \theta}\right)}^{2} \theta$

$= \frac{144}{2} \cdot \frac{\theta}{2 + \theta} ^ 2$

$A = \frac{72 \theta}{2 + \theta} ^ 2$

$A = f \left(\theta\right)$

$\frac{\mathrm{dA}}{d \theta} = \frac{72 {\left(2 + \theta\right)}^{2} - 2 \left(2 + \theta\right) 72 \theta}{2 + \theta} ^ 4$

$= \frac{144 + 72 \theta - 144 \theta}{2 + \theta} ^ 3$

$= \frac{144 - 72 \theta}{2 + \theta} ^ 3 = \frac{72 \left(2 - \theta\right)}{2 + \theta} ^ 3$

The maximum is when $\frac{\mathrm{dA}}{d \theta} = 0$

That is $\theta = 2$, $\implies$, $A = 9$

Also,

$\theta = \frac{12 - 2 r}{r}$

$A = \frac{72 \left(12 - 2 r\right)}{r \left(\frac{144}{r} ^ 2\right)} = \frac{r}{2} \left(12 - 2 r\right) = 6 r - {r}^{2}$

$A = f \left(r\right)$

Then,

$\frac{\mathrm{dA}}{\mathrm{dr}} = 6 - 2 r$

The maximum is when $\frac{\mathrm{dA}}{\mathrm{dr}} = 0$

That is

$6 - 2 r = 0$, $\implies$, $r = 3$

And $A = 9$