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Math Help - Lagrange Multipliers question

  1. #1
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    Lagrange Multipliers question

    Find the maximum of the function f(x,y)= -(x-a)^2 - (y-b)^2 subject to the constraint y=kx.


    So here's where I'm at:

    1. δf/δx = -2(x*-a) ; δf/δy = -2(y*-b) (* means for a particular x or y)

    2. g(x,y) = y-kx = 0

    3. δg/δx = -λ ; δg/δy = λ

    And then from here, I'm not really sure where to go. Any suggestions? Thanks,

    Kim
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  2. #2
    Junior Member
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    Jan 2009
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    Australia
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    The way to do this is to define a new function

    Your original function is

    f(x,y)=-(x-a)^2 -(y-b)^2

    Your constraint is

    g(x,y) = y - kx = 0

    So you form a new function

    L(x,y;\lambda) = f(x,y) - \lambda g(x,y)

    L(x,y;\lambda) = -(x-a)^2 -(y-b)^2 - \lambda(y-kx)

    Now find the partial derivatives w.r.t x and y and set them to zero

    \frac{\partial L}{\partial x}= -2(x-a) + \lambda k = 0

    \frac{\partial L}{\partial y}= -2(y-b) - \lambda = 0

    Now you want to eliminate lambda. So you multiply the second equation by k as add them to give

    -2(x-a) -2k(y-b) = 0

    And finally you substitute from your constraint y = kx to give

    -2(x-a) -2k(kx -b) = 0

    which simplifies to

    x = \frac{kb+a}{k^2+1}

    Lastly you need to find the y which can be found by substituting in the constraint (y = kx) therefore

    y = k\left(\frac{kb+a}{k^2+1}\right).

    Hope this helps. Lagrangian multipliers are very useful devices and come up in lots of interesting places.
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