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I am really stuck on this, the instructor went over the problem in class but I couldn't follow what was happening or why.

A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is vent into a circle. How should the wire be cut so that the total area enclosed is a maximum and B) a minimum.

So here is where I get confused, this is how I set up the problem.

square + circle = 10m

$4s + 2\pi r = 10$ where s is side of a square and r is radius of the circle

and $s^2 + \pi r ^2 = area$

$(\frac{10-2\pi r}{4})^2 + \pi r^2 = 10$

then I take the derivative

$\frac {-10\pi + 2\pi ^2}{4} + 2 \pi$

then I attempt to find zeroes

now I realize that I really messed this up so I have to start all over. I will edit that back in in 20 or so minutes.

I see what I did wrong, the derivative should be

$\frac {-10\pi + 2\pi ^2}{4} + 2 \pi r$

which gives me zeroes of

$ x = \frac {5- \pi}{4}$

so what I did was solve for one variable and plug it into the area formula and then find a min or a max. This was very wrong and I don't know why. For some reason the teacher used 10-x and x for the lengths of wire but I do not see why that is necessary or why my set up is wrong.

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Your method is absolutely correct, you must just be making an error in the work that you aren't showing us. Personally, I think your method is much more natural than the "x, 10-x" one. Once you have the area expressed as a quadratic function of s (or r, your choice), you then use what you know about quadratic functions to answer the question. –  Matthew Conroy Apr 3 '12 at 3:21
    
I will edit in my work. –  user138246 Apr 3 '12 at 3:23

2 Answers 2

So basically this is how you tackle it: Let $x$ be the amount of wire that was cut that will be used to create the square. This implies that you have the rest ($10-x$) to do the circle.

Let us denote by $S(x)$ the area of the square. The reason why I write $S(x)$ is because it is a function of $x$. Similarly, let $C(x)$ denote the area of the circle.

Now let us calculate what the functions ought to be. For $S(x)$, given that the perimeter of a square is four times its side, and knowing that the area of the square is side to the power of two, you obtain $$S(x)=(x/4)^2$$For the circle you now you have to calculate the radius. You are given its circumference which is $10-x$, hence $$10-x=2\pi r$$so solving for $r$ you get $r=\frac{10-x}{2\pi}$, and this implies that the area is equal to $$C(x)=(\frac{10-x}{2\pi})^2\pi$$From here the total area is going to be given by $A(x)=S(x)+C(x)$, now to finish the problem you should find the min and max values for this function (remember that $x$ attains values in $[0,10]$ only!).

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I don't understand what is wrong with what I did and I have no idea how you got that for the perimiter of a square. –  user138246 Apr 3 '12 at 2:53
    
Basically you want to have only one variable. This variable is going to be the $x$ defined above. Then if you use $x$-meters to bend it into a square, it means that each side has length $\frac{x}{4}$, so the side $s$ is just that. The area of the square is $s^2$, but we use $x$ instead, and we have that area of square is equal to $(x/4)^2$. –  Daniel Montealegre Apr 3 '12 at 2:57
    
What is wrong with using the square + circle = 10? –  user138246 Apr 3 '12 at 3:00
    
@Jordan Your version has two variables (side of the square and radius of the circle), while Daniel's has only one. Having only one variable makes it much easier to optimize. –  Austin Mohr Apr 3 '12 at 3:04
2  
But I can easily make mine have one variable like I did in class, $r= \frac{5-2s}{\pi}$ –  user138246 Apr 3 '12 at 3:08

You may be shocked to discover that you are absolutely right. (The approach that uses $x$ and $10-x$ is also right.)

Using your notation, we have $4s + 2\pi r = 10$. We are trying to maximize and minimize
$s^2 + \pi r ^2$.

Using the calculation that you mention in a comment, we have $r=\frac{5-2s}{\pi}$. Substituting in the formula for area, we see that we want to maximize/minimize $A(s)$, where $$A(s)=s^2+\pi\left(\frac{5-2s}{\pi}\right)^2.$$ If we allow $0$ radius or side, as we should here, we have the bounds $0\le s\le 2.5$.

The expression for $A(s)$ can be simplified slightly to $s^2+(1/\pi)(5-2s)^2$. Now differentiate as usual. We get $A'(s)=2s-(4/\pi)(5-2s)$. Solve for $s$.

Calculate $A(s)$ for the value of $s$ that you find. Calculate also $A(0)$ and $A(2.5)$, and among the three values you get, pick the big winner and the big loser.

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