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 piece of wire 30 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle.

(a) How much of the wire should go to the square to maximize the total area enclosed by both figures? m

(b) How much of the wire should go to the square to minimize the total area enclosed by both figures? m

This is What I have done so far:

$\pi r^2 +(30-r)^2$

$\frac{dy}{dx}= 2\pi r-2(30-r)$

$60=2\pi r-2r$ $60=r(2{\pi}-2)$ $30=r(\pi-1)$

$r= \frac{30}{\pi-1}$

$r=14.01$ I do not understand What I am meant to do with this number now.. Is it the answer to a or b? and how do I find the other.

share|improve this question
Your first formula appears to be wrong. –  Michael Hoppe Oct 17 '13 at 7:43
Don't forget that there are always extrema on the boundary points. That's crucial for that problem. –  Michael Hoppe Oct 17 '13 at 7:46
As reference to future viewers, there is more discussion of this problem (differing only as to the length of wire) here: math.stackexchange.com/questions/127493/… –  RecklessReckoner Jul 1 '14 at 1:57

1 Answer 1

Hint: take $x$ meter of the 30 meters to form the circle, the radius of the circle is not $x$ but $\frac{x}{2\pi}$. Then how long is the side of the square? And don't forget the boundaries $x=0$ and $x=30$.

share|improve this answer

Your Answer


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.