# Optimization problem

How to find a positive number such that the sum of the number and its reciprocal is as small as possible ?

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 ?

Galileo
Homework Helper
rumaithya said:
So the answer would be x = -1
Am I right ?
You were looking for a positive number, right?

Oh calculus is going to make me feels sick :)

Gokul43201
Staff Emeritus
Both -1 and +1 are local minima. Ther maxima are at 0 and $$\infty$$