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Optimization math help

  1. Oct 8, 2009 #1
    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 [tex]\pi[/tex] in your answer.

    2. Relevant equations
    [tex]A= \pi r^{2}[/tex]
    [tex]A= lw[/tex]

    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. jcsd
  3. Oct 8, 2009 #2
    Re: Optimization

    Let [tex]l[/tex] be the length of the piece that is used to form a square.

    Prove that :

    [tex]A(l)=\frac{l^2} {16} +\frac{\left(8-l\right)^2}{4\pi}[/tex]

    Then you can derivate that expression.
     
  4. Oct 8, 2009 #3
    Re: Optimization

    Whoah, where did that come from?

    I was going to say, can't you just treat the square and circle separately?
     
  5. Oct 8, 2009 #4

    Mark44

    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.
     
  6. Oct 8, 2009 #5
    Re: Optimization

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

    Then you get [tex]A= (8-x)^{2}[/tex] for the area of the square
     
  7. Oct 8, 2009 #6

    Mark44

    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.
     
  8. Oct 11, 2009 #7
    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}?
     
  9. Oct 11, 2009 #8

    Mark44

    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.
     
  10. Oct 11, 2009 #9
    Re: Optimization

    So 8 - x is the square while x is the square?
     
  11. Oct 11, 2009 #10

    Mark44

    Staff: Mentor

    Re: Optimization

    Huh???
     
  12. Oct 11, 2009 #11
    Re: Optimization

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

    I can't think :(
     
  13. Oct 12, 2009 #12
    Re: Optimization

    I got [tex] A(x) = x^{2}+\pi(8-x)^{2}[/tex]
     
  14. Oct 12, 2009 #13

    Mark44

    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.
     
  15. Oct 12, 2009 #14

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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?
     
  16. Oct 12, 2009 #15
    Re: Optimization

    Circumference.
     
  17. Oct 12, 2009 #16

    Mark44

    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?
     
  18. Oct 12, 2009 #17
    Re: Optimization

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

    I don't seem to remember how to get area with circumference.
     
  19. Oct 12, 2009 #18

    Mark44

    Staff: Mentor

    Re: Optimization

    How did you get that? What are the dimensions of the square?
    I'm not asking for area. I'm asking for the dimensions of the circle, namely its radius.
     
  20. Oct 12, 2009 #19

    Mark44

    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.
     
  21. Oct 12, 2009 #20
    Re: Optimization

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