# Optimation with e

1. Nov 27, 2005

### seiferseph

I've been having some trouble with this problem, heres what i did

i solve for the derivative f'(x) = 1/x^2 - 2lnx/x^2
i solve for the critical points to be only one, lnx = 1/2 so x = e^1/2 (or sqrt(e) )

now, since the range is between e and e^2, and since the e and e^2 are the only critical points on the interval, i thought one had to be the min and one had to be the max. i find the minimum to be f(e) = 1/e and i thought the max was f(e^2), but it turns out the minimum is f(e^3/2) = 2/e^3/2

where did i make a mistake? e^3/2 must be a critical point somehow right?

Last edited: Nov 27, 2005
2. Nov 27, 2005

### Muzza

You made a mistake when you calculated f'.

3. Nov 27, 2005

### HallsofIvy

Staff Emeritus
1. Yes, you did miscalculate f' but it wouldn't change your answer.

2. What makes you think 2/e3/2 is the minimum? It's larger than 1/e.

4. Nov 27, 2005

### seiferseph

i tried again and calculated

f'(x) = 3 - 2lnx / x^2

now this yields critical point x = e^3/2 and it works, thanks

5. Nov 27, 2005

### seiferseph

1/e is the min and 2/e3/2 is the maximum, i had them mixed up