# Homework Help: Optimization math help

1. Oct 8, 2009

### fghtffyrdmns

1. The problem statement, all variables and given/known data

A piece of wire 8 cm long is cut into two pieces. One piece is bent to form a circle, and the other is bent to form a square. How should the wire be cut if the total enclosed area is to be small as possible? Keep $$\pi$$ in your answer.

2. Relevant equations
$$A= \pi r^{2}$$
$$A= lw$$

3. The attempt at a solution

this is the part I am confused about. I cannot seem to set up the question. I do now that I have to find the function, take the derivative and equal it to zero (use 2nd derivative test to check that it's a min), and solve.

2. Oct 8, 2009

### Donaldos

Re: Optimization

Let $$l$$ be the length of the piece that is used to form a square.

Prove that :

$$A(l)=\frac{l^2} {16} +\frac{\left(8-l\right)^2}{4\pi}$$

Then you can derivate that expression.

3. Oct 8, 2009

### fghtffyrdmns

Re: Optimization

Whoah, where did that come from?

I was going to say, can't you just treat the square and circle separately?

4. Oct 8, 2009

### Staff: Mentor

Re: Optimization

No, since you have to find the location to cut the wire so that you get the maximum area for both figures.

5. Oct 8, 2009

### fghtffyrdmns

Re: Optimization

What if you make x be the length of the wire?

Then you get $$A= (8-x)^{2}$$ for the area of the square

6. Oct 8, 2009

### Staff: Mentor

Re: Optimization

That won't work, since you're assuming that all four sides are 8 - x cm long.

If your wire extends from 0 to 8 cm, and you cut it at x, that gives you two pieces of wire. One of them is x cm long and the other is 8 - x cm long. One piece will be used to form a square - all four sides. The other piece will be used to form a circle.

So which piece will you use to make the square, and which piece will you use to make the circle? Keep in mind that each piece will form the perimeter of whatever object you make out of it.

7. Oct 11, 2009

### fghtffyrdmns

Re: Optimization

I still cannot understand this question : (

Let x be the lentgh

thus should you not get 8-x?

then you can plug it into the area of a cirlce equation to get pi(8-x)^{2}?

8. Oct 11, 2009

### Staff: Mentor

Re: Optimization

No, not at all.

Try to be more specific. x defines the point at which the 8 cm wire is cut.

You will end up with one piece that is x cm long, and another that is 8 - x cm long. You're going to bend these pieces to form a square and a circle. Is it reasonable that the circle's radius is 8 - x cm? That's the formula that you seem eager to use.

9. Oct 11, 2009

### fghtffyrdmns

Re: Optimization

So 8 - x is the square while x is the square?

10. Oct 11, 2009

### Staff: Mentor

Re: Optimization

Huh???

11. Oct 11, 2009

### fghtffyrdmns

Re: Optimization

I mean x-8 is the square while x is the circle.

I can't think :(

12. Oct 12, 2009

### fghtffyrdmns

Re: Optimization

I got $$A(x) = x^{2}+\pi(8-x)^{2}$$

13. Oct 12, 2009

### Staff: Mentor

Re: Optimization

Not even close.

You have the two pieces of wire: x cm for the square and 8 - x cm for the circle. Describe to me in words exactly how you will form the two figures from these pieces of wire.

14. Oct 12, 2009

### Office_Shredder

Staff Emeritus
Re: Optimization

It might help if you actually take a piece of string or something and try to make a circle out of it. What does the length of the string represent here?

15. Oct 12, 2009

### fghtffyrdmns

Re: Optimization

Circumference.

16. Oct 12, 2009

### Staff: Mentor

Re: Optimization

Right - not the radius. Good.

And for the square, what does the length of that string represent?

So if you bend the x cm piece to make a square and the 8 - x piece to make a circle (BTW, it might be easier to use the x piece for the circle and the 8 - x piece for the square, what are the dimensions of each figure?

17. Oct 12, 2009

### fghtffyrdmns

Re: Optimization

the length is is each side. If I make (8-x) for the square, I would just get $$8x-x^{2}$$

I don't seem to remember how to get area with circumference.

18. Oct 12, 2009

### Staff: Mentor

Re: Optimization

How did you get that? What are the dimensions of the square?

19. Oct 12, 2009

### Staff: Mentor

Re: Optimization

I think it would be useful for you to take a wire coathanger and cut a piece out of it 8 inches long. Then cut your 8" long piece into two pieces, say 3" and 5". Take one of them and form it into a square and form the other into a circle. If you use the 3" piece to make a square, what are the lengths of the each side of the square? If you use the 5" piece for a circle, what is the radius of the circle.

Once you know the dimensions of each object, then you can calculate the area of each, and hence, the total area.

20. Oct 12, 2009

### fghtffyrdmns

Re: Optimization

You can't tell the dimensions though. It doesn't say it was cut in half.

21. Oct 12, 2009

### Office_Shredder

Staff Emeritus
Re: Optimization

You take a piece of wire that is of length 8-x inches. You bend it into a square. What is the length of one side of the square? This question is definitely answerable. Do two steps:
1) What is the perimeter of this square?
2) Based on 1), what is the length of a side of this square?

Once you have done that you can calculate the area

22. Oct 12, 2009

### fghtffyrdmns

Re: Optimization

The length of the square is 8-x? Then you get 32 - 4x for perimeter

I am not following this.

23. Oct 12, 2009

### Office_Shredder

Staff Emeritus
Re: Optimization

If I have a piece of wire that is 8-x inches long, and I bend it into a square, you DO NOT get that each side is 8-x inches long. The perimeter will be 8-x inches long

24. Oct 12, 2009

### fghtffyrdmns

Re: Optimization

Oh, I thought you said that's the length. I was wondering what's up. I know that the area of the square is going to be $$\frac{(8-x)^{2}}{16}$$. It's the circle that I cannot seem to get.

Edit: ahhh, I got it now. $$\frac{x^{2}}{4\pi}$$

I solved for the diameter using the circumference formula then divided by 2 get the radius.

Last edited: Oct 12, 2009
25. Oct 12, 2009

### fghtffyrdmns

Re: Optimization

Wait, if the perimeter is 8-x, wouldn't each side be 8-x/4?

$$\frac{x^{2}}{4\pi}$$

is the area of the circle.