Least Possible Sum of Recipricols of Two Positive Ints = 9

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Discussion Overview

The discussion revolves around finding the least possible sum of the reciprocals of two positive integers that add up to 9. Participants explore various methods and reasoning related to this mathematical problem.

Discussion Character

  • Mathematical reasoning, Exploratory, Debate/contested

Main Points Raised

  • One participant suggests listing all possibilities to identify a general pattern.
  • Another participant proposes that the least sum of reciprocals occurs when the two integers are as equal as possible, citing an example of 1/4 + 1/5.
  • A later post reiterates the original question and introduces a related problem about maximizing the area of a rectangle given its perimeter.
  • One participant mentions that a square with each side equal to perimeter/4 maximizes the area.
  • Another participant discusses differentiating the function 1/x + 1/(9-x) to find critical points, concluding that the maximal and minimal values occur at x=0 and x=4.5, suggesting that 4 and 5 are the closest integers.

Areas of Agreement / Disagreement

Participants present various methods and insights, but there is no clear consensus on the best approach or the final answer regarding the least sum of reciprocals.

Contextual Notes

Some assumptions about the nature of the integers and the method of finding the least sum of reciprocals are not explicitly stated, and there are unresolved mathematical steps in the differentiation approach.

Itachi
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The sum of two positive integers is 9. What is the least possible sum of their recipricols?
 
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you can do it by listing all the possibilities, that'll show you what the general pattern is.
 
1/4 + 1/5. The general formala simply says make the numbers as equal as possible.
 
Oh man! ... mathman..I italic[jus] worked that out!...was about to post it!..oh well...:D
 
Itachi said:
The sum of two positive integers is 9. What is the least possible sum of their recipricols?

OR "given the perimeter of a rectangle, how do you maximize its area?"
 
A square with each side of length perimeter/4.
 
I guess you could write 1/x + 1/(9-x) =s, differentiate and set equal to zero getting:

x^2 = (9-x)^2. The maximal value is x=0, and the minimal value is x=4.5, or they are the same. So 4 and 5 are the closest.
 

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