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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.2012-04-03
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    I will edit in my work.2012-04-03

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