1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook