Optimization problem

1. Jul 28, 2004

rumaithya

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

2. Jul 29, 2004

e(ho0n3

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. Jul 29, 2004

rumaithya

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. Jul 29, 2004

Galileo

You were looking for a positive number, right?

5. Jul 29, 2004

rumaithya

Oh calculus is going to make me feels sick :)

6. Jul 29, 2004

Gokul43201

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

Last edited: Jul 29, 2004
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