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Lagrange again :facepalm:

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Seems straightforward enough, Lagrangian optimization

    2. Relevant equations
    Find the max of x^-1 + y^-1 subject to the constraint m=x+y

    3. The attempt at a solution

    At first I thought no problems, x*=y*=m/2, however:

    Using the Lagrangian formula yields derivatives as follows:
    wrt x: -x^-2 - lambda
    wrt y: -y^-2 - lambda
    lambda: m-x-y

    Putting the coefficients into a bordered Hessian seems to give a positive def. matrix implying a minimum? Is this a trick question or is it possible to maximize?
  2. jcsd
  3. Sep 8, 2010 #2


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    Yes, x=y=(1/2) is a local minimum. You can also just use your constraint to eliminate y in 1/x+1/y and graph it as a function of x. As x ranges over all possible values you can see there is no global max or global min. There are vertical asymptotes.
  4. Sep 8, 2010 #3
    Thanks for the reply, you are making sense to me. I'm still confused as to why I was asked to maximize it, when there are no maximum points?
  5. Sep 8, 2010 #4


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    Typo? Always a possibility.
  6. Sep 8, 2010 #5
    I'm thinking you are probably right.

    On an exam no less :S.

    Guess I should have known better. Thanks again.
  7. Sep 9, 2010 #6
    Although, I've been reading that extreme values can occur at endpoints, stationary points, and singular points. So obviously this function has some singular points, but I do not know enough to go any further.

    Could there be a max at the asymptotes?
  8. Sep 9, 2010 #7


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    As I said, graph 1/x+1/(m-x) to see what the function looks like. Yes, it's singular at x=0 and x=m. No maxes there.
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