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Maximize the product xyz

  1. Jul 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Find three positive numbers x, y, and z whose sum is 100 such that (x^a)(y^b)(z^c) is a maximum.


    2. Relevant equations
    constraint: x+y+z=100
    maximize: (x^a)(y^b)(z^c)


    3. The attempt at a solution
    First I replaced the z in the maximization problem with 100-y-z. Then I took the partial derivatives of the maximization function with respect to x and to y. Solving these, I got x=100. This implies that y=-z. But the question asks for all positive numbers. I don't know what else to do...any tips?
     
  2. jcsd
  3. Jul 9, 2008 #2

    arildno

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    You most certainly haven't done this correctly!

    I advise you to use Lagranfe multipliers.
     
  4. Jul 9, 2008 #3
    We haven't learned that yet.
     
  5. Jul 9, 2008 #4

    arildno

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    All right, then!

    Post the equations you got after taking the partial derivatives and setting them to zero.
     
  6. Jul 9, 2008 #5
    x=(100a-100ay)/(c+a)
    y=100c/a
    Plugging this into x+y+z=100, I get z=-100c/a
     
  7. Jul 10, 2008 #6

    arildno

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    This is wrong!

    You are to differentiate:
    [tex]g(x,y)=x^{a}y^{b}(100-x-y)^{c}[/tex]
    The partial derivative of g with respect to x becomes, using the product&chain rules:
    [tex]\frac{\partial{g}}{\partial{x}}=ax^{a-1}y^{b}(100-x-y)^{c}-cx^{a}y^{b}(100-x-y)^{c-1}[/tex]

    Now, if this is set to zero, it means:
    [tex]ax^{a-1}y^{b}(100-x-y)^{c}=cx^{a}y^{b}(100-x-y)^{c-1}[/tex]
    Assuming that all factors are non-zero, we may divide through, say in this manner:
    [tex]a(100-x-y)=cx[/tex]
    Make a similar manipulation of the equation you gain from dg/dy!
     
  8. Jul 10, 2008 #7
    If you simplify the last equality you made, you get the same answer as I did for x.
     
  9. Jul 10, 2008 #8

    arildno

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    No, you don't. For one thing, 100y will not appear anywhere.
     
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