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Math Help - Lagrange Multiplier Fix..

  1. #1
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    Lagrange Multiplier Fix..

    Hi, I'm in the midst of revision for a calculus exam next week. I'm not sure if I tackled this problem correctly.

    Find the maximum and minimum distances from the origin for points on the curve described by
    3x^2 + 2y^2 + z^2 = 20 and z^2 = 2xy .

    I know the constraint eqns defined by g(x,y,z)= 3x^2 + 2y^2 + z^2 - 20 = 0 and h(x,y,z)= z^2 = 2xy
    Also I define the function f(x,y,z)= x^2 + y^2 + z^2, which I have to find the minimum and maximum values for.

    In short, what I did was to introduce a function F(x,y,z,k,j) = f(x,y,z) - k[g(x,y,z)] - j[h(x,y,z)] and solve for grad(F)=0 ( k and j are constants)

    The problem is when I realised 1 of the constants I had used turns out to be 0, is that correct?

    PLEASE, would really appreciate some guidance on this: my ans is minimum: (-sqrt(13/8), 0, 0) and max: (sqrt(13/8), 0, 0)
    Will be really glad if I can countercheck with some of your answers
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  2. #2
    Moo
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    Hello,

    Here is what I've done.
    I'm finding two extreme values : 10 and \frac{20}{3} (so the first one is a maximum and the second one a minimum), each of them attained at two points.

    Sorry for the writing, I've got a problem with writing on a straight line
    I hope it looks correct and helps you.

    ----------------------------------
    Note that your answers for min and max yield the same result for f.
    Since it is a symmetric function, f((-sqrt(13/8),0,0))=f((sqrt(13/8),0,0))
    Attached Thumbnails Attached Thumbnails Lagrange Multiplier Fix..-lagrange.jpg  
    Last edited by Moo; November 17th 2008 at 10:48 AM. Reason: added a remark
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  3. #3
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    k+j=1?

    Thanks! I could follow until the part when you consider z is not zero.

    Shouldnt k+j=1?

    If it's not too much trouble could you please explain more on this part?

    Many thanks
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  4. #4
    Moo
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    Yo,

    That's basic algebra !

    If ab=0, then either a=0, either b=0.

    Here, a=z and b=k+j-1


    funny enough, my friend did the exact same mistake ^^
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  5. #5
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    THANKS:)

    Thanks for your time spent on this. I misread one for lambda, lol..

    The factorisation part was really neat, I would have been defeated by tt equation.

    Glad tt I've finally understood this qn.

    Cheers.
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