# Maximization of Two Areas — Calculus 1

"A 10-meter length of wire is available for making a circle and a square. How should the wire be distributed between the two shapes to maximize the sum of the enclosed areas?"

Here's what I have: $$Area_c = \pi r^2$$ $$Area_s = 4r^2$$

So, I'm thinking that I need to find the maximized radius size to figure everything else out. $$Area_c + Area_s = 10$$ $$(\pi r^2) + (4r^2) = 10$$

$${\operatorname{d}\over\operatorname{d}r} [(\pi r^2) + (4r^2) - 10] = 2\pi r + 8r$$

But here's my dilemma; if I take the derivative of that and solve for $r$, it comes out 0. So I'm not sure where I'm going wrong. Any advice?

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The length of the wire is $10$ meters. Your equation says that the sum of the areas is $10$. The length of the wire is the sum of the perimeters. – joriki Apr 20 '11 at 18:07
I'm a little unsure how to pull this off without another constraint. It seems like a circle is the most efficient shape in terms of containing the most area, so any expression you come up with would force the square's perimeter to zero and give you a circle with circumference = 10 meters. – Brian Vandenberg Apr 20 '11 at 18:10

You want to maximize $\pi x^2 + y^2$ subject to $2\pi x+4y=10$, $x\ge0$, $y\ge0$.
If $r$ is the radius of the circle and $x$ is the side of the square then you are given that $2 \pi r + 4x = 10$, or $x = \frac{10-2 \pi r}{4}$. You want to maximize $\pi r^2 + x^2 = \pi r^2 + (\frac{10-2 \pi r}{4})^2$. We know that $0 \leq r \leq 10$. So now you need to differentiate $\pi r^2 + (\frac{10-2 \pi r}{4})^2$, find all critical points, and then compare the values of this function at the end points ($r = 0$ and $r = 10$) and also at the critical points.