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Math Help - Multivariable Maximum problem

  1. #1
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    Multivariable Maximum problem

    Find all the points (x, y, z) with x, y, z => 0 which maximizes the function
    f(x, y, z) = xy + xz + yz - 4xyz subject to the constraint x + y + z = 1.
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  2. #2
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    Question its urgent

    hi!!!
    please help me n answer my questions???? what are the effects of rearranging and regrouping series? their difference and some examples. what about convergence after regrouping and rearranging????
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  3. #3
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    Please do NOT "hijack" other peoples threads by asking completely unrelated questions in them! It is extremely rude.

    guess, use the "Lagrange Multiplier Method"- the max or min of f(x,y,z) subject to the constraint g(x,y,z)= constant must satisfy \nabla f= \lambda \nabla g for some number \lambda.

    For this problem, that is <y+ z- 4yz, x+ z- 4xz, x+ y- 4xy>= \lambda <1, 1, 1>. That is, you have y+ z- 4yz= \lambda, x+ z- 4xz= \lambda and x+ y- 4xy= \lambda.

    Those fairly easily reduce to y+ z- 4yz= x+ z- 4xz and y+ z- 4yz= x+ y- 4xy. Together with x+ y+ z= 1, you have three equations to solve for x, y, and z.
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  4. #4
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    Thanks

    Hi hallofivy,

    I have tried your method too. However, it does not return the absolute extremum value.

    when i re try it, the partial derviative of x and y is not equal to zero
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  5. #5
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    well , i am not keen on inequalities but i think it is a great attempt !

     xy + yz + zx - 4xyz

     = \frac{ (1-4x)(1-4y)(1-4z) - 1 + 4(x+y+z) }{16}

     =   \frac{ (1-4x)(1-4y)(1-4z) + 3  }{16}


     = \frac{1}{16} \left[ 3 + (1-4x)(1-4y)(1-4z) \right]

    use  GM \leq AM

     \leq   \frac{1}{16} \left[ 3 + \left( \frac{ (1-4x ) +  (1-4y ) +  (1-4z )   }{3} \right )^3\right]

     = \frac{1}{16} ( 3 - \frac{1}{27} ) = \frac{5}{27}

    the equality holds when  x =y =z = 1/3

    but it may be wrong because we can find that :

     GM \leq AM \leq 0 !!

    I hope it will be perfectly corrected by some experts .
    Last edited by simplependulum; March 29th 2010 at 07:46 PM.
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  6. #6
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    Quote Originally Posted by guess View Post
    Hi hallofivy,

    I have tried your method too. However, it does not return the absolute extremum value.

    when i re try it, the partial derviative of x and y is not equal to zero
    The partial derivatives wrt x and y, for a constrained max or min do not have to be 0!
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