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Global optimization subject to constraints

  1. Apr 26, 2008 #1
    1a) Determine the maximum value of f(x,y,z)=(xyz)1/3 given that x,y,z are nonnegative numbers and x+y+z=k, k a constant.

    1b) Use the result in (a) to show that if x,y,z are nonnegative numbers, then (xyz)1/3 < (x+y+z)/3


    Attempt:
    1a) Using the Lagrange Multiplier method, I get that the absolute maximum of f subject to the constraints x+y+z=k and x,y,z>0 is k/3

    1b) Here, it seems to me that one of the constraints, namely x+y+z=k, is removed. If so, then how can we still use the result of part (a) here?

    I need some help on part (b). Any help is appreciated!
     
  2. jcsd
  3. Apr 26, 2008 #2

    HallsofIvy

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    For any x, y, z, x+ y+ z is something isn't it? For any x, y, z, define k= x+ y+ z. Then you have shown by a that [itex](xyz)^{1/3}\le k/3= (x+ y+ z)/3[/itex].
     
  4. Apr 26, 2008 #3
    For simplicity, let's take k=5.

    In part b, x,y,z are only required to be nonnegative numbers. There is no restriction that x+y+z=5 as there is in part a.
    Take e.g. x=5, y=5, z=5 which are nonnegative
    But x+y+z=15, which is not equal to 5.

    It seems to me that (xyz)1/3 < (x+y+z)/3 is true only if x+y+z=k, but NOT true for ANY nonnegative numbers, and in part b we have to prove the latter.

    Can someone explain more, please?
     
  5. Apr 27, 2008 #4

    Dick

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    You really didn't listen to Halls, did you? You proved the max for ALL k. If k=5 then the max is at x=y=z=5/3. And (xyz)^(1/3)<=5=x+y+z. If you take x=y=z=5 you'd better set k=15. Then (xyz)^(1/3)<=15<=x+y+z. Any other values of k you'd like me to address individually? I think you should think about this a little more before posting another question.
     
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