# One last optimization problems

1. Apr 21, 2008

### physicsed

[SOLVED] one last optimization problems

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

find two positive numbers such that the sum of the number and its reciprocal is as small as possible.

2. Relevant equations

x+(1/x) = s

f' =1 + ln x

3. The attempt at a solution

lost

2. Apr 21, 2008

### lzkelley

you've got it right so far. What happens to the derivative when something is at a minimum or a maximum?

3. Apr 21, 2008

### physicsed

1 + ln x =0
-1 = ln x

thats about all i can do

4. Apr 21, 2008

### rootX

did you say there that
d/dx (1/x) = ln (x)?

5. Apr 21, 2008

### rocomath

physicsed ... you took the antiderivative ...

6. Apr 21, 2008

### lzkelley

you have to take the derivative, not the integral

7. Apr 21, 2008

### physicsed

do u want me to use the quo rule for 1/x which is -1/x^2???

8. Apr 21, 2008

### lzkelley

that is the correct answer, but you don't have to use the quotient rule to find it...
1/x = x^(-1) ... d/dx (1/x) = (-1)x^(-2) = -(1/x^2)

9. Apr 22, 2008

### physicsed

thats what i got to so far
(x^2 + 1)/ x

$$0=({{2x^2}-{x^2}+{1}})/{x^2}$$

which is undifined...

Last edited: Apr 22, 2008
10. Apr 22, 2008

### HallsofIvy

Staff Emeritus
You are missing ( ) in that. The derivative of (x2+ 1)/x, using the quotient rule, is
$$\frac{(x^2+ 1)'(x)- (x^2+ 1)(x)'}{x^2}= \frac{(2x)- (x^2+1)}{x^2}= \frac{x^2- 1}{x^2}= 0[/itex] when x= 1 or -1. Although it would be simpler, as others have pointed out, to write the function as x+ x-1 so its derivative is 1- x-1 which is 0 when x= 1 or -1. 11. Apr 22, 2008 ### physicsed whaaat Last edited: Apr 22, 2008 12. Apr 22, 2008 ### physicsed [tex] \frac{(2x)- (x^2+1)}{x^2}= \frac{x^2- 1}{x^2}$$

how are those equal?

Last edited: Apr 22, 2008
13. Apr 22, 2008

### Nick89

A slight mistake there (probably a typo), it should be:
$$\frac{(2x^2)- (x^2+1)}{x^2}= \frac{x^2- 1}{x^2}$$

$$(\frac{d}{dx}(x^2+1)) (x) = (2x)(x) = 2x^2$$

Last edited: Apr 22, 2008
14. Apr 22, 2008

### physicsed

how are they equall?

15. Apr 22, 2008

### Nick89

As I said, HallsOfIvy made a little mistake (typo probably) in the calculation of the derivative.
He forgot to multiply the derivative of (x^2 + 1) with x.

16. Apr 22, 2008

### physicsed

so, i was right?

17. Apr 22, 2008

### Nick89

No.

$$\frac{2x^2 - x^2 + 1}{x^2} = 0$$ which indeed does not have a solution.

The correct answer however should be:
$$\frac{2x^2 - (x^2 + 1)}{x^2} = 0$$ (there is one very important difference!)

18. Apr 22, 2008

### physicsed

$$\frac{2x^2- (x^2+ 1)}{x^2}= \frac{x^2-1}{x^2}$$

how are they equall?

19. Apr 22, 2008

### Nick89

(Is the forum messing up or are we just posting too fast?? :P)

$$\frac{2x^2 - (x^2 + 1)}{x^2} = \frac{2x^2 - x^2 - 1}{x^2} = \frac{x^2 - 1}{x^2}$$

$$\frac{x^2 - 1}{x^2} = 0 \ \rightarrow \ x^2 - 1 = 0 \ \rightarrow \ x = \pm \sqrt(1) = \pm 1$$

Last edited: Apr 22, 2008
20. Apr 22, 2008

### physicsed

$$2x^2 - x^2 - 1 = x^2 - 1$$

it might sound childish, but how is that equal?

you can't factor anything or simplify it