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Math Help - LaGrange Multipliers

  1. #1
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    LaGrange Multipliers

    Consider the problem of finding the points on the surface xy+yz+zx=3 that are closest to the origin.

    1) Use the identity (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx) to prove that x+y+z is not equal to 0 for any point on the given surface.

    2) Use the method of Lagrange multipliers to find a system of four equations in x,y,z and \lambda whose solutions will give the closest points.

    3) Find the points on xy+yz+zx=3 that are closest to the origin.

    I'm clueless on what to do for the 1st part (although I imagine it's actually something simple), but I think I have the second part down. Problem is, I think I probably need to use the 1st part for the 3rd somehow.

    For the second part, I found the system of four equations to be:

    2x=\lambda(y+z)
    2y=\lambda(x+z)
    2z=\lambda(x+y)
    xy+yz+zx=3
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Part1

    since xz+yx+yz = 3

    x^2 +y^2 +z^2 +2(xz+yx+yz) >6 on the surface
    Last edited by Calculus26; April 16th 2009 at 12:47 PM.
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  3. #3
    MHF Contributor Calculus26's Avatar
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    solution

    eliminating lambda

    you should be able to show

    x^2+ xz = y^2 + yz

    (x-y)(x+y) = -z(x-y)

    if x doesn't equal y then x+y = -z but this yields x+y+z = 0
    but this can't happen so x = y

    similarly you can show y^2 +xy = z^2 + xz

    and use the same argument so x = y = z

    use this in your constraint xy + xz +yz =3 to solve
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  4. #4
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    Quote Originally Posted by Calculus26 View Post
    eliminating lambda

    you should be able to show

    x^2+ xz = y^2 + yz

    (x-y)(x+y) = -z(x-y)

    if x doesn't equal y then x+y = -z but this yields x+y+z = 0
    but this can't happen so x = y

    similarly you can show y^2 +xy = z^2 + xz

    and use the same argument so x = y = z

    use this in your constraint xy + xz +yz =3 to solve
    Thank you so much, I understand the identity now, but if you don't mind me asking:

    How would I go about eliminating lambda?
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  5. #5
    MHF Contributor Calculus26's Avatar
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    For example

    from 2x = L(y+z)

    Solve L =2x/(y+z)

    From 2y = L(x+z)

    Solve L = 2y/(x+z)

    Therefore 2x/(y+z) = 2y/(x+z)

    Similarly you can eliminate lambda using the 2d and 3d equations
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