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More on mechanics

  1. Feb 9, 2006 #1
    of the classical kind

    All i need here is an explanation as why F is the way it is
    A person wished to fence off a maximum area for hi8s dog by attachin a flexible fence of length L to the side of his house whose width is 2a. What should the slope of the fence be?
    Assume L > 2a
    well then the action
    [tex] A[y] = \int_{-a}^{a} y(x) dx [/tex] is a maximum
    subject to constant L where [tex] L = \int_{-a}^{a} \sqrt{1+y'^2} dx [/tex]
    If L[y,y'} = constant then [itex] \delta L = 0 [/itex]
    and A[y,y'] = constant at its maximum
    [tex] \delta A + \lambda \delta L = 0 [/tex] where
    [tex] \lambda = \frac{[A]}{[L]} [/tex]
    we also want that [tex] \delta \int_{-a}^{a} F(y,y',x) dx = 0 [/tex]
    where [tex] F = y + \lambda \sqrt{1+y'^2} [/tex]
    thats the problem, Why is F the way it is ? Maybe the attached diagram helps...

    Attached Files:

    Last edited: Feb 9, 2006
  2. jcsd
  3. Feb 10, 2006 #2
    can anyone help out?

    is it because f takes on the shape of half a unit circle?

    Or will any function do?
  4. Feb 11, 2006 #3


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    Homework Helper

    maximizing the Area provided the "y" term,
    using all the fence material provided the sqrt term.
    Whether you put the lambda multplier with the sqrt term
    or with the y term was your choice, back in step 4 ...
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