Question

A circular sector with radius r and angle Q has area A. Find r and Q so that the perimeter is smallest for a given area A. (Note: A=1/2r^2Q, and the length of the arc s=rQ, when Q is measured in radians)?

Answer 1

I believe its r = sqrt(A) and Q = 2
is this for a calculus class?

Answer 2

Yes. how did you get it?

Answer 3

ok just wanted to make sure because it involves taking a derivative to minimize the perimeter.
\[A = \frac{r^{2}Q}{2} , P = 2r + rQ\]
Now using the Area equation, solve for Q
\[Q = \frac{2A}{r^{2}}\]
Substitute into Perimeter equation
\[P = 2r + r(\frac{2A}{r^{2}}) = 2r + \frac{2A}{r}\]
Differentiate
\[dP/dr = 2 - \frac{2A}{r^{2}}\]
Set equal to zero and solve for r
\[2 - \frac{2A}{r^{2}} = 0\]
\[\rightarrow 2r^{2} = 2A\]
\[\rightarrow r = \sqrt{A}\]
Now substitute this value back in for r to solve for Q.
\[Q = \frac{2A}{r^{2}} = \frac{2A}{\sqrt{A}^{2}} = \frac{2A}{A} = 2\]

Answer 4

Thank you so much. But can you tell me how did you compute P=2r+rQ? Is this a formula?

Answer 5

no problem
its the sum of the arc length and the 2 radius.
a sector is like an isosceles triangle except the base is curved but its length is the arc length

Answer 6

Okay, thanks. Can you please go through more word problems. I have a exam tomorrow so I really need to figure these out!