A wire of length 12" can be bent into a circle, a square or cut into 2 pieces and make both a circle and a square. How much wire should be used for the circle if the total area enclosed by the figure(s) is to be:

a) a Maximum

b) a Minimum

What I've got so far is that the formula for the square is $A_s=\frac{1}{16}s^2$ and the circumfrance of the circle to be $P=12-c$ and area to be $A_c = \pi(\frac{P}{2\pi})^2$ where $c$ is the length of the wire for the circle and $s$ is the length of the wire for the square.

Now I know I need to differentiate these formulas to then find the max and min they both can be, but what am I differentiating with respect to? The missing variable in each of the formulas?

Also, once, I find the derivitives, what would my next steps be to minimizing and maximizing these?

And did I set the problem up correctly?

Thanks for any help

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    $\begingroup$ You're confusing $s$ and $c$. You have $s \in [0,12]$ and $c = 12 - s$, so $P = c = 12-s$ in your notation. Having settled this, differentiate the sum of areas of circle and square with respect to $s$ (the variable). Don't forget to consider the values at $s = 0$ and $s = 12$! $\endgroup$ – t.b. Jul 18 '11 at 17:04
  • $\begingroup$ There is a need to be careful. For example, it looks as if you are exploring the consequences of using a length $s$ for making the square and the rest for making the circle. If so, the side of the square is $s/4$, and therefore the area of the square is $s^2/16$. $\endgroup$ – André Nicolas Jul 18 '11 at 17:14
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    $\begingroup$ And please note that $P=12-s$. You have basically done all the work. What is the combined area in terms of $s$? $\endgroup$ – André Nicolas Jul 18 '11 at 17:20
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    $\begingroup$ @RHP: Are you sure that two squares or two circles are allowed? The way I read it is that you can make either a circle, a square or a circle and a square. Please correct me if I'm wrong, my English (not my native language) may be playing tricks on me. $\endgroup$ – t.b. Jul 18 '11 at 18:30
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    $\begingroup$ On second reading I believe you are absolutely correct. Your English seems perfect, my mistake. $\endgroup$ – RHP Jul 18 '11 at 18:35

Let $s$ be the circumference of the square. Then the circumference of the circle is $12-s$ (because that's what is left from the wire). Now you already computed the formulas $A_{\mathrm{square}}(s) = \frac{1}{16} s^2$ and $A_{\mathrm{circle}}(s) = \frac{1}{4\pi}(12 - s)^2$. The total area is $A(s) = A_{\mathrm{square}}(s) + A_{\mathrm{circle}}(s)$, where $s \in [0,12]$ is the variable. To find the extrema (maximum/minimum) of this function, a necessary condition is $A'(s) = 0$ (differentiate with respect to $s$) when $0 \lt s \lt 12$ and you need also consider $A(0)$ and $A(12)$.

So the task you need to do is to differentiate $A(s)$ with respect to $s$, solve $A'(s) = 0$ for $s$ (there will be only one solution $s_0$). Now the maximum among $A(0)$, $A(12)$ and $A(s_0)$ will be the maximum and the minimum among them will be the minimum of $A(s)$. It may also help if you sketch the graph to convince yourself of the solution.

Here's a small sanity check: The circle is the geometric figure that encloses the largest area among all figures with the same circumference, so the maximum should be achieved for $s = 0$. Since enclosing two figures needs more wire than enclosing a single one, the minimum should be achieved at $s_0$.


Since the results you mention are a bit off, let me show you what I get:

First $$A(s) = \frac{1}{16}s^2 + \frac{1}{4\pi}(12-s)^2.$$ Differentiating this with respect to $s$ I get $$A'(s) = \frac{1}{8}s - \frac{1}{2\pi}(12-s)$$ Now solve $A'(s) = 0$ to find $$s_0 = \frac{12}{1+\frac{\pi}{4}} \approx 6.72$$

Plugging this in gives me $A(s_0) \approx 5.04$. (No warranty, I hope I haven't goofed)

  • $\begingroup$ I didn't take RHP's suggestion into account because that's not how I understand the question. If this is indeed allowed, I can elaborate on that. $\endgroup$ – t.b. Jul 18 '11 at 18:33
  • $\begingroup$ When I differentiate $A_{square}(s) + A_{circle}(s)$ I get that A(0) is a negative number which isn't in tho domain of the problem. Would I just ignore it? $\endgroup$ – OghmaOsiris Jul 18 '11 at 18:41
  • $\begingroup$ @OghmaOsiris: Wait, $A'(0)$ will be negative, that's right. But you want to know $A(0) = \frac{1}{4\pi}(12-0)^2 \gt 0$. $\endgroup$ – t.b. Jul 18 '11 at 18:47
  • $\begingroup$ By the way $A(s)$ will be positive for all $s$. $\endgroup$ – t.b. Jul 18 '11 at 18:53
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    $\begingroup$ I got the max to be A(0)=$36\pi$ and min to be A($s_0$)=1.60...... Which goes with what you said. Thanks so much! $\endgroup$ – OghmaOsiris Jul 18 '11 at 19:34

Every so often, one might mention the following sort of approach.

Let $x$ be the length of wire we will devote to the circle, and $y$ the length we will devote to the square.

Let $A$ be the combined area of the circle and square. A calculation identical to the one done by the OP shows that $$A=\frac{x^2}{4\pi}+\frac{y^2}{16}.$$

We want to find the values of $x$ that give maximum and minimum area, given that $x$ and $y$ are non-negative, and $x+y=12$.

Maximum and/or minimum values may be reached at an endpoint. So we compute $A$ when $x=0$, $y=12$, and also when $x=12$, $y=0$.

The remaining candidates for maximum/minimum are with $0<x<12$. At such a candidate $x$, we will have $\dfrac{dA}{dx}=0$. (We are doing one-variable calculus.)

It is easy to see that $$\frac{dA}{dx}=\frac{2x}{4\pi} +\frac{2y}{16}\frac{dy}{dx}.$$

But from $x+y=12$, we can see that $\dfrac{dy}{dx}=-1$. Now we have two equations in the two unknowns $x$ and $y$. Solve for $x$, compute $A(x)$, and compare with the endpoint values.

The above procedure carries no advantage in this case, and may increase the probability of mechanical error. However, when the "constraint" is non-linear, there can be real computational advantages to working with implicit functions, particularly if the constraint has symmetries.


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