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One last optimization problems

  1. Apr 21, 2008 #1
    [SOLVED] one last optimization problems

    1. The problem statement, all variables and given/known data

    find two positive numbers such that the sum of the number and its reciprocal is as small as possible.

    2. Relevant equations

    x+(1/x) = s

    f' =1 + ln x

    3. The attempt at a solution

    lost
     
  2. jcsd
  3. Apr 21, 2008 #2
    you've got it right so far. What happens to the derivative when something is at a minimum or a maximum?
     
  4. Apr 21, 2008 #3
    1 + ln x =0
    -1 = ln x

    thats about all i can do
     
  5. Apr 21, 2008 #4
    did you say there that
    d/dx (1/x) = ln (x)?
     
  6. Apr 21, 2008 #5
    physicsed ... you took the antiderivative ...
     
  7. Apr 21, 2008 #6
    you have to take the derivative, not the integral
     
  8. Apr 21, 2008 #7
    do u want me to use the quo rule for 1/x which is -1/x^2???
     
  9. Apr 21, 2008 #8
    that is the correct answer, but you don't have to use the quotient rule to find it...
    1/x = x^(-1) ... d/dx (1/x) = (-1)x^(-2) = -(1/x^2)
     
  10. Apr 22, 2008 #9
    thats what i got to so far
    (x^2 + 1)/ x

    [tex] 0=({{2x^2}-{x^2}+{1}})/{x^2} [/tex]

    which is undifined...
    NO answer
     
    Last edited: Apr 22, 2008
  11. Apr 22, 2008 #10

    HallsofIvy

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    You are missing ( ) in that. The derivative of (x2+ 1)/x, using the quotient rule, is
    [tex]\frac{(x^2+ 1)'(x)- (x^2+ 1)(x)'}{x^2}= \frac{(2x)- (x^2+1)}{x^2}= \frac{x^2- 1}{x^2}= 0[/itex]
    when x= 1 or -1.

    Although it would be simpler, as others have pointed out, to write the function as x+ x-1 so its derivative is 1- x-1 which is 0 when x= 1 or -1.
     
  12. Apr 22, 2008 #11
    whaaat
     
    Last edited: Apr 22, 2008
  13. Apr 22, 2008 #12


    [tex] \frac{(2x)- (x^2+1)}{x^2}= \frac{x^2- 1}{x^2} [/tex]

    how are those equal?
     
    Last edited: Apr 22, 2008
  14. Apr 22, 2008 #13
    A slight mistake there (probably a typo), it should be:
    [tex]\frac{(2x^2)- (x^2+1)}{x^2}= \frac{x^2- 1}{x^2}[/tex]

    [tex](\frac{d}{dx}(x^2+1)) (x) = (2x)(x) = 2x^2[/tex]
     
    Last edited: Apr 22, 2008
  15. Apr 22, 2008 #14
    how are they equall?
     
  16. Apr 22, 2008 #15
    As I said, HallsOfIvy made a little mistake (typo probably) in the calculation of the derivative.
    He forgot to multiply the derivative of (x^2 + 1) with x.
     
  17. Apr 22, 2008 #16
    so, i was right?
     
  18. Apr 22, 2008 #17
    No.

    You're answer was:
    [tex]\frac{2x^2 - x^2 + 1}{x^2} = 0[/tex] which indeed does not have a solution.

    The correct answer however should be:
    [tex]\frac{2x^2 - (x^2 + 1)}{x^2} = 0[/tex] (there is one very important difference!)
     
  19. Apr 22, 2008 #18
    [tex]
    \frac{2x^2- (x^2+ 1)}{x^2}= \frac{x^2-1}{x^2}
    [/tex]

    how are they equall?
     
  20. Apr 22, 2008 #19
    (Is the forum messing up or are we just posting too fast?? :P)

    [tex]\frac{2x^2 - (x^2 + 1)}{x^2} = \frac{2x^2 - x^2 - 1}{x^2} = \frac{x^2 - 1}{x^2}[/tex]

    [tex]\frac{x^2 - 1}{x^2} = 0 \ \rightarrow \ x^2 - 1 = 0 \ \rightarrow \ x = \pm \sqrt(1) = \pm 1[/tex]
     
    Last edited: Apr 22, 2008
  21. Apr 22, 2008 #20
    [tex]
    2x^2 - x^2 - 1 = x^2 - 1
    [/tex]

    it might sound childish, but how is that equal?

    you can't factor anything or simplify it
     
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