# Another optimization problem

Asphyxiated

## 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

$$A_{square}=ab$$

$$A_{circle} = \pi r^{2}$$

$$r= \pi *circumference$$

## 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:

$$r = \pi C,\;\; C=12-x, \;\; then \;\; r= \pi (12-x)= 12 \pi - \pi x$$

then substitute into area:

$$A_{circle}= \pi r^{2} = \pi (12 \pi - \pi x)^{2}$$

so once you foil that out you get:

$$A_{circle} = 144 \pi^{3}-24 \pi^{3}x + \pi^{3}x^{2}$$

but we want the total area which is:

$$A_{total}=A_{square}+A_{circle}$$

$$A_{total}=\frac {x^{2}}{16} + 144 \pi^{3}-24 \pi^{3}x + \pi^{3}x^{2}$$

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

$$A_{total}'= \frac{x}{8} -24 \pi^{3} + 2 \pi^{3} x$$

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!

## Answers and Replies

Homework Helper
Gold Member
$$r= \pi *circumference$$

This should be $$C =2 \pi r$$. This should change the critical point, namely the value of $$x$$ where $$A'=0$$. You will need to check whether that's a max or min. The other extremum should be at one of the boundary points of the interval on which $$x$$ is defined, $$x\in [0,12]$$.

Asphyxiated
ok using C=2 pi r

$$r = \frac {12-x}{2 \pi}$$

so area:

$$A_{circle} = \pi (\frac{12-x}{2 \pi})^{2} = \pi (\frac {144-24x+x^{2}}{4 \pi^{2}}) = \frac {144-24x+x^{2}}{4 \pi}$$

total area:

$$A_{total} = \frac {x^{2}}{16} + \frac {1}{4 \pi}(144-24x+x^{2})$$

take the derivative:

$$A_{total}'=\frac {x}{8} - \frac {24}{4 \pi}+ \frac {2x}{4 \pi} = \frac {x}{8} - \frac {6}{ \pi} + \frac {x}{2 \pi}$$

set to zero and solve and you will get an answer of 1.126 or 1.13 which is not the answer either.

did i do something wrong again?

Homework Helper
$$A_{total}'=\frac {x}{8} - \frac {24}{4 \pi}+ \frac {2x}{4 \pi} = \frac {x}{8} - \frac {6}{ \pi} + \frac {x}{2 \pi}$$

set to zero and solve and you will get an answer of 1.126 or 1.13 which is not the answer either.

did i do something wrong again?
I'm getting x ≈ 6.72 here. Show what you did to solve
$$\frac {x}{8} - \frac {6}{ \pi} + \frac {x}{2 \pi} = 0$$
.

Asphyxiated
I must have messed up with the calculator because i didnt want to simplify it anymore, so i redid it and got 6.72 but, alas, that is not the answer either.

Homework Helper
Gold Member
I must have messed up with the calculator because i didnt want to simplify it anymore, so i redid it and got 6.72 but, alas, that is not the answer either.

It is one of the answers. You need to do more work to determine whether it is a maximum or minimum.

Asphyxiated
I already know that 12 is the correct answer for the maximum, and 0,12, 6.72 and 1.13 are not answers for the minimum. I have checked both endpoints as well as that critical number.