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It makes logical sense but can't mathematical prove it

  1. Apr 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the minimum and maximum of the given equation, f(x,y,z)=x+y+z, with the constraint 1/x+1/y+1/z=1.

    2. Relevant equations

    f(x,y,z)=x+y+z, with the constraint 1/x+1/y+1/z=1.

    3. The attempt at a solution

    I know the solution, which is minimum {(1,1,-1), (1,-1,1), (-1,1,1)} and this part is easy to prove using lagrange multiplier.

    The hard part is proving the maximum {(6, 3, 2), (6, 2, 3), (3, 2, 6), (3, 6, 2), (2, 3, 6), (2, 6, 3)}
    I just can't prove it, but looking at it is makes perfect logical sense that it is a solution.
    The question is how to prove it.

    A side note using the lagrange multiplier method also yields a critical point of (3,3,3) but it is not a max or min.
     
    Last edited: Apr 5, 2013
  2. jcsd
  3. Apr 5, 2013 #2

    Dick

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    Why do you think that's a maximum? x=16, y=2, z=16/7 also satisfies your constraint. But x+y+z is definitely greater than 6+3+2.
     
  4. Apr 5, 2013 #3
    I agree that your solution works and it satisfies the constraint. Following your logic there should be infinitely many solutions. What I think the problem writer left out was that the solution needs to be a integer.

    With that being said and going back to solution that was given for the maximum. How would you find/solve for that solution without just plugging in numbers and trying to see if it works?
     
  5. Apr 5, 2013 #4

    Dick

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    That's a whole different problem. Nothing to do with lagrange multipliers. Start with the with (3,3,3) point. That gives you 9. The only way you can do better is to make say x=2. Then 1/y+1/z must be 1/2. So y and z can't be too large. Just keep eliminating things that don't work. It's really just a guessing game.
     
  6. Apr 5, 2013 #5
    There has to be a better way to solve this problem that I'm not seeing, other than guessing. When I started this problem I used the lagrange multipliers method to solve for the critical points. This led to the solution set {(1,1,-1), (1,-1,1), (-1,1,1), (3,3,3)}. Now I know that just because there is a critical point that does not mean it is a max or min. Also, the lagrange multipliers method will not always reveal the max or min as illustrated in this problem. But I can't leave it at the only way to find the solution is to guess, because how would I know that I'm not missing another solution.
     
  7. Apr 5, 2013 #6

    Dick

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    It's not quite just guessing. You know (3,3,3) gives you 9. That might be a max. If x, y and z are all greater then or equal to 3, then you can't satisfy the constraint (except in the (3,3,3) case. So one of them must be less than 3. It doesn't matter which one is less than 3 and the solutions are symmetric, so pick that one to be x. x=1 doesn't work. So x must be 2. Now you've got the smaller problem of maximizing y+z subject to the constraint 1/y+1/z=1/2. Start playing the same game.

    If you string all this together you've got a proof that you have the correct solution and you haven't missed anything if the reasoning is solid.

    That the lagrange method gave you integers for the critical points is an accident. In a more general problem the minima might not have been integers, then you'd be stuck with finding the minimum integer solution using the same sort of elimination game.
     
  8. Apr 5, 2013 #7

    Bacle2

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    You can always reduce this to a 2-variable problem , by solving for either of x,y or z in your constraint equation and then substituting into the objective function.
     
  9. Apr 6, 2013 #8

    epenguin

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    The question is completely symmetrical, switch x, y, and z around any way and nothing changes.
    Suggesting that x=y=z has to be a solution. You can at least convince yourself of this geometrically, the try work it up to a nice argument. Look at the case, as was suggested, of just two variables. Generally be geometrical.

    I think it is not giving too much away to ask couldn't one or two of them be infinite consistent with the constraint? If they count that as a maximum.
     
  10. Apr 6, 2013 #9

    Bacle2

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    But then,don't you lose the geometry if some values are infinite? Or, will youuse ,say, some
    projective geometry? If you see x+y+z as a plane, how do you see it when two of them take infinite values?
     
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