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Homework Help: LaGrange multipliers with natural base

  1. Mar 8, 2010 #1
    1. The problem statement, all variables and given/known data
    f(x,y,z)=exy and x5+y5=64

    Find Max and Min


    2. Relevant equations
    ∇F = <yexy, xexy>
    λ∇G = <5x4λ, 5y4λ>



    3. The attempt at a solution

    yexy = 5x4λ
    xexy = 5y4λ
    x5+y5=64

    No idea where to go from here...
     
  2. jcsd
  3. Mar 8, 2010 #2

    Dick

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    I would start by dividing the first equation by the second.
     
  4. Mar 8, 2010 #3
    Alright, dividing the first by the second, I get y/x = y4/x^4, or y=x

    plugging that in, we get 2x5=64, or x=2

    since x=2, y must equal 2

    so one solution is e^4, and that was correct as the maximum, but I am having trouble finding the minimum.

    I really appreciate the help.
     
  5. Mar 8, 2010 #4

    Dick

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    I get y/x=(x/y)^4. Which is a little different. That should warn you to be concerned about the cases where x=0 or y=0. They don't work here. Still it's still something to think about in general. But have you sketched a graph of x^5+y^5=64? It's unbounded. You can get a greatest lower bound for exp(xy), but does it have a minimum?
     
  6. Mar 8, 2010 #5
    Ok, so since the graph is unbounded, then there is no minimum?
     
  7. Mar 8, 2010 #6

    Dick

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    No, the graph can be unbounded in general and you can have a minimum. But what happens in this case? Tell me why there isn't a minimum.
     
  8. Mar 8, 2010 #7
    well, the function can approach e-infinity, or 0, correct?
     
  9. Mar 8, 2010 #8

    Dick

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    Exactly. But only as x->infinity. There is no point where it actually reaches 0. So I would say 'no minimum'. I was sort of torn about this for a while.
     
  10. Mar 8, 2010 #9
    Awesome, that makes sense. I really appreciate it. Thank you
     
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