Optimize Positive Number Sum of Number & Reciprocal

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Discussion Overview

The discussion revolves around finding a positive number such that the sum of the number and its reciprocal is minimized. Participants explore mathematical approaches, including calculus and graphical analysis, to identify the local minimum of the function defined by this sum.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant suggests defining the function as y = x + 1/x and finding its local minimum in the first quadrant.
  • Another participant notes that the graph indicates local minima at x = ±1, but emphasizes that only x = +1 is relevant for the positive solution.
  • A later reply questions the validity of x = -1 as a solution, pointing out the requirement for a positive number.
  • One participant expresses frustration with calculus, indicating a personal challenge with the topic.
  • Another participant mentions that both -1 and +1 are local minima, while maxima occur at 0 and ∞.

Areas of Agreement / Disagreement

Participants generally agree on the approach of using calculus and graphical methods, but there is disagreement regarding the interpretation of local minima and the relevance of negative values in the context of finding a positive number.

Contextual Notes

The discussion does not resolve the mathematical steps necessary to confirm the local minima or maxima, nor does it clarify the implications of the findings for the specific requirement of a positive number.

rumaithya
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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 ?
 
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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.
 
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 ?
 
rumaithya said:
So the answer would be x = -1
Am I right ?

You were looking for a positive number, right? :smile:
 
Oh calculus is going to make me feels sick :)
 
Both -1 and +1 are local minima. Ther maxima are at 0 and [tex]\infty[/tex]
 
Last edited:

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