# Max/min problem, find max area of sector of circle.

A sector of a circle has fixed perimeter. For what central angle $θ$ (in radians) will the area be greatest? First I put together an equation for the perimiter of the sector being $p$ which I assume as a constant since it is fixed:

$$p = rθ + 2r$$

Since p is a contant and need to reduce the number of variables to one in the area equation before diffrenciating, I rearranged the equation in terms of $r$:

$$r = \dfrac{p}{θ + 2}$$

I noticed, since the perimeter is constant, $θ$ and $r$ must be inverse proportional to another, problem is I wasn't sure what the coefficient proportionality or constant ($p$) was, so I assumed it to be $1$:

$$r = \dfrac{1}{θ + 2}$$

I then plugged that back into the area of a sector forumla and diffrenciated with respect to $θ$:

$$\dfrac{d}{dθ}{(\dfrac{θ}{2(θ+2)^2})}=\dfrac{2(θ+2)^2-θ(4(θ+2))}{(2(θ+2)^2)^2}$$

I put that equal to zero and solved for theta and received 2 radians.

Did I do this correctly? if so, was I to assume the $p$ to be 1? or is there a way of calculating it? constructive criticism is welcome.

Thanks

## 2 Answers

This is, in fact a correct way to solve this problem. And if you did keep the perimeter as a general $p$ instead of assuming it to be 1, in your derivative, you would simply have an extra factor of $p$, which could be divided out, since the expression was set equal to zero in the end.

• You mean like this? $$\dfrac{d}{dθ}{(\dfrac{θp^2}{2(θ+2)^2})}=\dfrac{2p^2(θ+2)^2-θp^2(4(θ+2))}{(2(θ+2)^2)^2}$$ $$\dfrac{2p^2(θ+2)^2-θp^2(4(θ+2))}{(2(θ+2)^2)^2} =0$$ $$p^2(θ+2)(2(θ+2)-θ(4))=0$$ $$2θ+4-4θ=0$$ $$θ=2$$ – Modrisco Jun 20 '15 at 2:03
• Yes. That's it. – JoDraX Jun 20 '15 at 2:04

The area $A$ of a sector is $A = (1/2)(R^2)(\theta)$.

Substitute $R = {P \over {(\theta + 2)}}$ where $P$ is perimeter in above equation then differentiate w.r.t $\theta$. Then equate it to zero to find theta for which Area is Max.

Leave $P$ as it is. Don't put it equal to 1. Treat it as you treat constants.

Note: I am sorry I could not use latex as I am on mobile device.