Optimization problem

  • Thread starter rumaithya
  • Start date
  • #1
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How to find a positive number such that the sum of the number and its reciprocal is as small as possible ?
 

Answers and Replies

  • #2
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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.
 
  • #3
20
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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 ?
 
  • #4
Galileo
Science Advisor
Homework Helper
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rumaithya said:
So the answer would be x = -1
Am I right ?
You were looking for a positive number, right? :smile:
 
  • #5
20
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Oh calculus is going to make me feels sick :)
 
  • #6
Gokul43201
Staff Emeritus
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Both -1 and +1 are local minima. Ther maxima are at 0 and [tex] \infty[/tex]
 
Last edited:

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