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Optimation with e

  1. Nov 27, 2005 #1
    [​IMG]

    I've been having some trouble with this problem, heres what i did

    i solve for the derivative f'(x) = 1/x^2 - 2lnx/x^2
    i solve for the critical points to be only one, lnx = 1/2 so x = e^1/2 (or sqrt(e) )

    now, since the range is between e and e^2, and since the e and e^2 are the only critical points on the interval, i thought one had to be the min and one had to be the max. i find the minimum to be f(e) = 1/e and i thought the max was f(e^2), but it turns out the minimum is f(e^3/2) = 2/e^3/2

    where did i make a mistake? e^3/2 must be a critical point somehow right?
     
    Last edited: Nov 27, 2005
  2. jcsd
  3. Nov 27, 2005 #2
    You made a mistake when you calculated f'.
     
  4. Nov 27, 2005 #3

    HallsofIvy

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    1. Yes, you did miscalculate f' but it wouldn't change your answer.

    2. What makes you think 2/e3/2 is the minimum? It's larger than 1/e.
     
  5. Nov 27, 2005 #4
    i tried again and calculated

    f'(x) = 3 - 2lnx / x^2

    now this yields critical point x = e^3/2 and it works, thanks
     
  6. Nov 27, 2005 #5
    1/e is the min and 2/e3/2 is the maximum, i had them mixed up
     
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