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Verify that the sum of three quantities x, y, z

  1. Jun 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Verify that the sum of three quantities x, y, z, whose product is a constant k, is maximum when these three quantities are equal.

    2. Relevant equations
    w = x + y + z
    k = x * y * z

    3. The attempt at a solution
    Assuming that my understanding of the question is correct i.e. that we want the maximum of w and x*y*z has to be equal to a constant k then I'm pretty lost. If they mean the maximum as the largest value of w possible then it's definitely not when x = y = z.

    If k = 1 then if x = y = z would give w = 3. But if k = 1 and x = 100, y = 0.1, z = 0.1 then k is still 1 and w = 100.2.

    If they mean maximum as in where the rate of change of w is 0 then that doesn't make sense either because even though the rate of change of w is 0 when x = y = z, w is at it's lowest so that would make that point into a minimum.

    The only thing I can think of is that they wrote maximum instead of minimum.
     
  2. jcsd
  3. Jun 9, 2017 #2

    WWGD

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    Haven't you seen the method of Lagrange multipliers?
     
  4. Jun 9, 2017 #3
    Never heard of it. I'm working with the Calculus made easy book and I haven't encountered it there.
     
  5. Jun 9, 2017 #4

    WWGD

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    What kind of techniques have you seen/used to solve these types of problems?
     
  6. Jun 9, 2017 #5
    This chapter was the first and only chapter on partial differentation. They showed that to find the maxima and minima of a multivariable function, the derivative of all partial derivatives have to be zero.
     
  7. Jun 9, 2017 #6

    ehild

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    You are right. The sum should be minimum.
     
  8. Jun 9, 2017 #7

    WWGD

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    Ah, sorry, I missed that, I thought they were asking for a minimum.
     
  9. Jun 9, 2017 #8

    Ray Vickson

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    The result is true for a minimum sum. There is no maximum sum under those conditions, because we can write the sum as
    $$S = x + y + \frac{k}{xy},$$
    and so can have ##S \to +\infty## by taking ##x,y \to 0+##.
     
  10. Jun 9, 2017 #9
    In that case, k = xyz, w = x + y + z = k/(yz) + k/(xz) + k/(xy)
    ∂w/∂x = -k/(x2z) - k/(x2y) = -y/x - z/x = 0 ∴ y = -z
    ∂w/∂y = -k/(y2z) - k/(y2x) = -x/y - z/y = 0 ∴ z = -x
    ∂w/∂z = -k/(z2y) - k/(z2x) = -x/z - y/z = 0 ∴ x = -y

    Now this confuses me, because y = -z therefore x = -y = z. Now I have x = z and z = -x? Am I allowed to do this?

    I tried testing with values and w doesn't even seem to be a minimum when x = y = z considering that if x = 1, y = -1, z = -1 then w = -1
     
  11. Jun 9, 2017 #10

    WWGD

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    What
    constraints do you have on x,y,z ? Are they positive?
     
  12. Jun 9, 2017 #11

    Ray Vickson

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    Please use the appropriate reply button that identifies which post you are replying to. I cannot tell what you are commenting on in your response above.

    For non-negative values of ##x,y,z## the point ##(x,y,z) = (c,c,c)## (with ##c = k^{1/3}## and ##k > 0##) really is a provable minimum.
    However, if you allow for some negative values of the variables, the sum has no finite minimum for the same reason I outlined in #8 above: you can have
    $$S = x + y + \frac{k}{xy} \to -\infty$$
    by having ##x \to 0+## and ##y \to 0-## (again, when ##k > 0##).
     
    Last edited: Jun 9, 2017
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