Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

"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?

share|improve this question
    
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

2 Answers 2

You want to maximize $\pi x^2 + y^2$ subject to $2\pi x+4y=10$, $x\ge0$, $y\ge0$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.