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LaGrange multipliers with natural base

  • Thread starter whiteway
  • Start date
  • #1
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Homework Statement


f(x,y,z)=exy and x5+y5=64

Find Max and Min


Homework Equations


∇F = <yexy, xexy>
λ∇G = <5x4λ, 5y4λ>



The Attempt at a Solution



yexy = 5x4λ
xexy = 5y4λ
x5+y5=64

No idea where to go from here...
 

Answers and Replies

  • #2
Dick
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I would start by dividing the first equation by the second.
 
  • #3
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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.
 
  • #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?
 
  • #5
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Ok, so since the graph is unbounded, then there is no minimum?
 
  • #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.
 
  • #7
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well, the function can approach e-infinity, or 0, correct?
 
  • #8
Dick
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well, the function can approach e-infinity, or 0, correct?
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.
 
  • #9
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Awesome, that makes sense. I really appreciate it. Thank you
 

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