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Optimization problem

  1. Nov 25, 2007 #1
    1. The problem statement, all variables and given/known data
    You're building a walkway from the corner of one building to the corner of another building. The diagram looks like this.
    [​IMG]
    The street is 100 ft wide, and 50 ft long.
    The walkway will weigh 40 pounds per feet when it is parallel to the street and 30 pounds per feet when it is crossing the street.

    How should the walkway be laid out to have a minimal weight?

    2. Relevant equations

    I'm going to call the horizontal portion of the walkway x ft of length, and the portion that crosses the street y ft of length. So the total length of the walkway would be x+y.
    Then the equation for the total weight of the walkway would be 40x + 30y, right?
    Also I'm not sure if the pythagorean theorem should come into play here.

    3. The attempt at a solution

    This is where I'm really, really confused. I know that when you do an optimization problem , the hard part is finding the right equation. I have no idea what that is. Can someone just give me a little prod in the right direction? Thanks.
     
    Last edited: Nov 25, 2007
  2. jcsd
  3. Nov 25, 2007 #2

    Dick

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    x and y are not independent. You can write y in terms of x if you finally decide you do need the pythagorean theorem. The right triangle containing y has one leg of length 100-x and another of 50.
     
  4. Nov 25, 2007 #3
    So y would be:

    [tex]\sqrt{100^2 + (50-x)^2}[/tex]

    Now do I take the derivative of that?
     
  5. Nov 25, 2007 #4

    Dick

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    Use the chain rule. Besides, you actually want to take the derivative of 40x+30y, right?
     
  6. Nov 25, 2007 #5
    Would the derivative be

    [tex]-(50-x)^{2}[/tex]

    But how would taking the derivative help me? Is it related to the 40 lb/ft and the 30 lb/ft?
     
  7. Nov 25, 2007 #6

    Dick

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    You take the derivative and set it equal to zero. But your guess for the derivative is way off. The derivative of sqrt(f(x))=f'(x)/(2*sqrt(f(x)). How did I know that?
     
  8. Nov 25, 2007 #7
    Umm, all I can tell is that you must have gotten the derivative out of

    (1/2) x (y') x (y^(-1/2))

    But I don't really understand how you got that. See, I'm not very good at using the Chain Rule, and what I did was just

    [tex]\frac{1}{2}[/tex] x [tex]\sqrt{0 + (50-2)^{2}}^{-\frac{1}{2}}[/tex] x (2 x (-1))
     
  9. Nov 25, 2007 #8

    Dick

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    You've got a good start there but in this case y=100^2+(50-x)^2. y' isn't 2*(-1). What is it? And the denominator is sqrt(100^2+(50-x)^2), why did the the 100^2 turn to zero and the x disappear?
     
    Last edited: Nov 25, 2007
  10. Nov 25, 2007 #9
    I got the ( 2 * -1) part from differentiating (50-x)^2.
     
  11. Nov 25, 2007 #10

    Dick

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    The derivative of (50-x)^2 is 2*(50-x)*(-1).
     
  12. Nov 25, 2007 #11
    Oh right!
    The 100^2 turned to 0 because it is an integer and you can't differentiate it anymore. The x disappeared because it is only to the one-power.
     
  13. Nov 26, 2007 #12

    Dick

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    The sqrt comes from y^(-1/2). That's NOT (y')^(-1/2). You might want to review differentiation in general.
     
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