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Homework Help: Optimization Problems

  1. Apr 28, 2006 #1
    Code (Text):

    A 51 meter length of wire is cut into two parts.  
    The first part is fashioned into a rectangle that is twice as long as it is wide.  
    The second part is fashioned into a square.  
    How much of the originial wire is used for each shape
    if the shapes' combined area is a minimum?  
    Use the second derivative test(S.D.T.) to determine the minimum.
    Couple of questions/comments on this problem. Why the SDT? Out of all the examples in the bookn either max or min optimization problems don't do the SDT. They do the first derivative test to determine critical values to solve the problem. Second question is my primary and secondary equations set up right? This the primary equation:
    [tex]A = xy + x^2[/tex]
    [tex]51 = 2(2x + y) + 4x[/tex]
  2. jcsd
  3. Apr 28, 2006 #2
    Why are you assuming the rectangle has a short side equal to a side of the square? That's not necessarily the case.
  4. Apr 28, 2006 #3
    So I need a third variable? I'm starting to understand this less and less.
  5. Apr 28, 2006 #4

    Tom Mattson

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    No you don't need a third variable, because the length of the wire was given. You've made two mistakes here.

    1.) The dimensions of the rectangle aren't 2x by y, they're 2x by x. They told you that one dimension is twice the other.

    Then there's the mistake that daveb pointed out.

    2.) You're assuming that each side of the square has the same length as the width of the rectangle. That isn't necessarily the case, so call it something else (like y).
  6. Apr 28, 2006 #5


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    No, you only need two variables because you haven't used "a rectangle that is twice as long as it is wide".

    You are cutting the wire into two pieces. One is used to form a rectangle. Since the "rectangle is twice as long as it is wide", you can call the width of the rectangle x and the length will be 2x. Then the area is 2x2 and the perimeter (the length of that piece of wire) is 2x+ 2(2x)= 6x. The other piece is used to form a square. Call the length of a side of the square y. Then the area is y2 and the perimeter (the length of that piece of wire) is 4y. Since the entire wire has length 51 m, you have 6x+ 4y= 51 and want to minimize 2x2+ y2 subject to that constraint.

    The second derivative test can be used determine if your critical point is a minimum.
  7. Apr 28, 2006 #6
    Thanks for the help! Times like these make me wonder about my math career.:frown:
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