6
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ On second reading I believe you are absolutely correct. Your English seems perfect, my mistake. $\endgroup$ – RHP Jul 18 '11 at 18:35
6
$\begingroup$

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 $s_0$.


Added:

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)

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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