Optimization problem (minimization)

[togame]

Homework Statement

I have a sector of a circle with area 12 square meters. If radius r and angle $\theta$ are chosen so that the that perimeter of the sector is the smallest possible, then what is the radius?

Homework Equations

I have area of sector as $A=\frac{\theta r^2}{2}$ which is 12
and the length of the arc as $L=\theta r$

The Attempt at a Solution

my attempt is as follows:
since i am attempting to minimize the circumference, i will need to minimize that function.
$$L=\theta r$$
and since i only want one variable, i use the area function and solve in terms of r and replace it in the circumference function.
$$12=\frac{\theta r^2}{2}$$
$$\theta=\frac{24}{r^2}$$
$$C=r\frac{24}{r^2}$$
$$C=\frac{24}{r}$$
now i need to take the derivative of this function and solve set equal to 0 to get r
$$C\prime=-\frac{24}{r^2}$$
now, this is where i am stuck. since r is squared, this derivative can never be 0, so i believe i am missing a step somewhere or am just confused about how to set it up. any help would be greatly appreciated.

 

[scurty]

The only thing I can think of is that what they mean by perimeter is the length of the arc + the length of the two radii.

Think about it for a moment, I believe your answer for the derivative makes sense. If the area of the sector remains constant, the larger your radius gets, while keeping the area constant, the smaller the length of the arc gets. Take r to infinity and you get, basically, 0 for the length of the arc. That's why I believe they mean the length AROUND the sector, not just the arc.

Try setting $L = 2r + \theta r$ and see what answer you get!

 

[togame]

The only thing I can think of is that what they mean by perimeter is the length of the arc + the length of the two radii.

Oh man, I can't believe I didn't see that! That got me to where I needed to be. Thanks for your help!