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Optimization problem

  1. Jul 28, 2004 #1
    How to find a positive number such that the sum of the number and its reciprocal is as small as possible ?
     
  2. jcsd
  3. Jul 29, 2004 #2
    Let y = x + 1/x. Find the local minimum of y in the first quadrant and you're done. I suggest you also graph y to get a visual.
     
  4. Jul 29, 2004 #3
    thank you.

    I was thinking about that before..
    In the graph it shows that x = +- 1
    but x is +1 when it's local minimum, and x is -1 when it's local maximum.

    So the answer would be x = -1

    Am I right ?
     
  5. Jul 29, 2004 #4

    Galileo

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    You were looking for a positive number, right? :smile:
     
  6. Jul 29, 2004 #5
    Oh calculus is going to make me feels sick :)
     
  7. Jul 29, 2004 #6

    Gokul43201

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    Both -1 and +1 are local minima. Ther maxima are at 0 and [tex] \infty[/tex]
     
    Last edited: Jul 29, 2004
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