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Math Help - Lagrange Multipliers, help with solving system of equations

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    Lagrange Multipliers, help with solving system of equations

    Use Lagrange multipliers to find the maximum and minumum values of the function subject to the given constraint

    f(x, y, z) = xyz
    g(x, y, z) = x^2+2y^2+3z^2 = 6



    \nabla f = [yz, xz, xy]
    \nabla g = [2x, 4y, 6z]

    yz - \lambda 2x = 0
    xz - \lambda 4y = 0
    xy - \lambda 6z = 0
    x^2+2y^2+3z^2 = 6

    I've gotten the system of equations. (shown above) I'm just not sure how to go about tackling this one. Any words of wisdom from someone more clever would be appreciated
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    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Lagrange Multipliers, help with solving system of equations

    Play around a bit with those multipliers to reach:

    Max(xyz) = \frac{2}{\sqrt{3}}

    for: {x,y,z} = \left\{x\to \sqrt{2},y\to -1,z\to -\sqrt{\frac{2}{3}}\right\}
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    Re: Lagrange Multipliers, help with solving system of equations

    Can you expand a bit on what you mean by "play around", please? Something more to go on? I've been playing around off and on this afternoon and so far am seeing little progress. That is the answer in back of book though.
    Last edited by sgcb; October 12th 2012 at 09:31 PM.
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    Re: Lagrange Multipliers, help with solving system of equations

    Ah, here we are:

    \lambda = \frac{yz}{2x} so y^2 = \frac{x^2}{2} and z^2 = \frac{x^2}{3}

    plugging into constraint gives:
    x = \pm \sqrt 2

    so
    f(\sqrt{2}, 1, \sqrt{\frac{2}{3}}) = \frac{2}{\sqrt{3}} max
    f(-\sqrt{2}, -1, -\sqrt{\frac{2}{3}}) = -\frac{2}{\sqrt{3}} min
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    Re: Lagrange Multipliers, help with solving system of equations

    Since a specific value of \lambda is not necessary for a solution, I find it is often easiest to eliminate \lambda first by dividing one equation by another. Here, I would first rewrite the equations as yz= \lambda 2x, xz= \lambda 4y, and xy= \lambda 6z.

    Dividing the first equation by the second, \frac{y}{x}= \frac{x}{2y} which is the same as 2y^2= x^2. Similarly, dividing the second equation by the third, \frac{z}{y}= \frac{2y}{3z} which is the same as 3z^2= 2y^2.

    The constraint that has to be satisied is x^2+ 2y^2+ 3z^2= 6 and, from the above, we can replace both x^2 and 3z^2 by 2y^2 making the equation 6y^2= 6 so that y^2= 1 and y= \pm 1. With y= \pm 1, x^2= 2y^2= 2 so that x= \pm\sqrt{2} and 3z^2= 2y^2= 2 so that z^2= \frac{2}{3} and z= \pm\sqrt{\frac{2}{3}}= \pm\frac{\sqrt{6}}{3}. That gives a total of eight points that should be checked for max and min values.
    Last edited by HallsofIvy; October 13th 2012 at 11:25 AM.
    Thanks from sgcb
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