# Calculus - How to minimize x^2 + (1/x^2)

1. May 16, 2014

### yoyo16

1. The problem statement, all variables and given/known data

Find a positive number such that the sum of the square of the number and its reciprocal is a minimum.

2. Relevant equations

3. The attempt at a solution
y=X^2+(1/X^2)
Derivative=2x-2x^(-3)
0=2x-2x^3
2/x^3=2x
2=2x^4
1=x^4
x=1

The answer I got was 1 but I'm not sure if this is right. Can someone please tell me if I did it right or wrong. If wrong, can you point out my mistake. Thanks :)

2. May 16, 2014

### vela

Staff Emeritus
[strike]Looks good except that you missed a possible solution. What other value of x satisfies x^4=1?[/strike]

Oops, never mind. I didn't notice you were looking for a positive number.

You just now need to verify that the critical point corresponds to a minimum.

3. May 16, 2014

### yoyo16

Is it possible for the minimum to be a smaller value or would the smallest value be 1?

4. May 16, 2014

### LCKurtz

If the graph is concave up for $x>0$ your relative minimum will be an absolute minimum on that interval. Have you checked that?

5. May 16, 2014

### Ray Vickson

Your function is strictly convex in the convex region {x > 0}, so any stationary point is automatically a global minimum within the region.

6. May 16, 2014

### yoyo16

Oh ok so since it is absolute it would be 1 right?