Calculus question with optimization homework

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.

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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

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 $x=30$.

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Since this is a fairly "standard" first-semester calculus textbook problem (if it isn't given for homework, it turns up on an exam), I thought I'd write something about it generally.

The problem usually asks for a length of wire $\ L \$ to be cut into two pieces, each one to be formed into a specified geometric figure; the typical choices are a square and a circle, or an equilateral triangle and a circle. (As we'll see, the actual figures used don't affect the character of the solution.) For these particular shapes, we have the relations

square : $$l_s \ = \ 4S \ \ \Rightarrow \ \ A_s \ = \ S^2 \ = \ \left( \frac{l_s}{4} \right)^2 \ = \ \frac{1}{16} \ l_s^2 \ \ ;$$

equilateral triangle : $$l_t \ = \ 3s \ \ \Rightarrow \ \ A_t \ = \ \frac{\sqrt{3}}{4}s^2 \ = \ \frac{\sqrt{3}}{4}\left( \frac{l_t}{3} \right)^2 \ = \ \frac{\sqrt{3}}{36} \ l_t^2 \ \ ;$$

square : $$l_c \ = \ 2 \pi \ r \ \ \Rightarrow \ \ A_c \ = \ \pi \ r^2 \ = \ \pi \left( \frac{l_c}{2 \pi} \right)^2 \ = \ \frac{1}{4 \pi} \ l_c^2\ \ .$$

The significant (and familiar) feature of these area functions is that they are always increasing with the length of the wire used to produce them and are always "concave upward"; they really only differ as to the constant of proportionality (what physicists call the "dimensionless geometric factor").

area functions of circular (blue), square (green), and equilateral-triangular (red) lengths of wire, with $\ 0 \ \le \ L \ \le \ 10 \$

Upon cutting the length of wire $\ L \$ into two pieces, and, say, allocating the length $\ x \$ to the triangle or square and the remainder, $\ L \ - \ x \$ to the circle, we produce the area-sum functions

square + circle : $$\ \ A \ = \ \frac{1}{16} \ x^2 \ + \ \frac{1}{4 \pi} \ (L - x)^2 \ \ ;$$

equilateral triangle + circle : $$\ \ A \ = \ \frac{\sqrt{3}}{36} \ x^2 \ + \ \frac{1}{4 \pi} \ (L - x)^2 \ \ .$$

What we are interested in noting here is the properties of such area-sum functions. For two wire shapes (they need not even be regular figures, but just simple closed curves) with "geometric factors" $\ F_1 \ \ \text{and} \ \ F_2 \$ , the first two derivatives of the area-sum function are

$$\ \frac{dA}{dx} \ = \ 2 \ F_1 \ x \ - \ 2 \ F_2 \ (L - x ) \ \ \text{and} \ \ \frac{d^2A}{dx^2} \ = \ 2 \ F_1 \ + \ 2 \ F_2 \ \ .$$

Since the geometric factors are necessarily positive, we will always have $\ \frac{d^2A}{dx^2} \ > \ 0 \$ in such problems, so the extremal value of the area-sum in the interval $\ 0 \ < \ x \ < \ L \$ is a minimum located as $\ x \ = \ \left( \frac{F_2}{F_1 + F_2} \right) \ L \$ . This will be true regardless of the choice for the two simple closed curves.

the red, green, and blue curves are those appearing in the graph above, but with the circle area function "reversed" on the interval $\ 0 \ \le \ x \ \le \ L \$ to represent the circle being the "second" piece of wire; the greenish and violet curves represent the "square-and-circle" and "triangle-and-circle" area-sum functions, respectively

What is often called the "extreme value theorem" tells us that a continuous function on a closed interval has an absolute maximum and an absolute minimum value. Since the open interval $\ 0 \ < \ x \ < \ L \$ doesn't contain a maximum, we must look for the absolute maximum at one of the endpoints of the closed interval, $\ x \ = \ 0 \ \ \text{or} \ \ x \ = \ L \$ . But here, we can use another familiar fact (not readily established using only first-semester calculus) that a circle of a specified "perimeter" encloses the greatest possible area of any simple closed curve. Hence, the solution for the maximal enclosed area is to use the entire length of wire (corresponding here to $\ x \ = \ 0 \$ ) to form a circle only. (The general solution for any two simple closed curves is to make only the figure with the larger geometric factor of the two.)

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