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LaGrange Multipliers Help!

  1. Nov 15, 2005 #1
    LaGrange Multipliers!! Help!

    Use the Lagrange multiplier method for 3 variables to find the points on the surface 3xy-z^2=1 that are closest to the origin.

    I tried using the gradient= lamda(granient) and ended up getting (-3/2,0,-1). but i think i did it way wrong. Can someone please help? Thanks!
     
  2. jcsd
  3. Nov 15, 2005 #2

    mathwonk

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    the gradient should be parallel to the radius vector, no?
     
  4. Nov 16, 2005 #3

    benorin

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    Minimize [tex]f(x,y,z)=\sqrt{x^2+y^2+z^2}[/tex] subject to the constraint [tex]g(x,y,z)=3xy-z^2=1[/tex]

    and note that the min of f is also the min of f^2, so the vector eqn is

    [tex] 2<x,y,z> = \lambda <3y,3x,-2z>[/tex] and [tex]3xy-z^2=1[/tex]

    the z-component gives [tex]z=-\lambda z[/tex] and hence [tex]\lambda=-1[/tex] iff [tex]z\neq 0[/tex] and plugging [tex]\lambda=-1[/tex] into the vector eqn quickly gives both 2y=-3x and 2x=-3y so x=y=0, however this is an extraneous solution since (0,0,z) is not on the given suface (for real z anyhow). Therefore z=0 by contradiction and now it is easy.

    Have fun,
    -Ben
     
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