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## Homework Statement

A piece of wire 12 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle. (Give your answers correct to two decimal places.)

Part A) how much of the wire should be used for the circle to maximize the area? (Solved this part, it is 12, the entire length)

Part B) how much of the wire should be used for the circle to minimize the area? (unsolved, it is not the number i get below, it is also not 0 or 12).

## Homework Equations

[tex] A_{square}=ab [/tex]

[tex] A_{circle} = \pi r^{2} [/tex]

[tex] r= \pi *circumference [/tex]

## The Attempt at a Solution

So cutting the wire of 12 m in length into two pieces, x and 12-x. The x side will be made into the square so that the square has sides x/4 and an area of (x/4)^2. This means the length of the wire 12-x will be the circumference of the circle. So if:

[tex] r = \pi C,\;\; C=12-x, \;\; then \;\; r= \pi (12-x)= 12 \pi - \pi x [/tex]

then substitute into area:

[tex] A_{circle}= \pi r^{2} = \pi (12 \pi - \pi x)^{2} [/tex]

so once you foil that out you get:

[tex] A_{circle} = 144 \pi^{3}-24 \pi^{3}x + \pi^{3}x^{2} [/tex]

but we want the total area which is:

[tex] A_{total}=A_{square}+A_{circle} [/tex]

[tex] A_{total}=\frac {x^{2}}{16} + 144 \pi^{3}-24 \pi^{3}x + \pi^{3}x^{2}[/tex]

now take the derivative to find the critical number(s):

[tex] A_{total}'= \frac{x}{8} -24 \pi^{3} + 2 \pi^{3} x [/tex]

solving that for x you get 11.97.

As I said about this number is not the answer to anything, as far as the program is concerned, so where did i go wrong? I have done this over a couple times so i dont think it is my math, perhaps i am just not fully understanding what it is that i need to do here.

Thanks for any help in advance!