One last optimization problems

[physicsed]

[SOLVED] one last optimization problems

Homework Statement

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

Homework Equations

x+(1/x) = s

f' =1 + ln x

The Attempt at a Solution

lost

 

[lzkelley]

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

 

[physicsed]

1 + ln x =0
-1 = ln x

thats about all i can do

 

[rootX]

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

 

[rocomath]

physicsed ... you took the antiderivative ...

 

[lzkelley]

you have to take the derivative, not the integral

 

[physicsed]

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

 

[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)

 

[physicsed]

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

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

which is undifined...
NO answer

 

[HallsofIvy]

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

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

which is undifined...
NO answer

You are missing ( ) in that. The derivative of (x²+ 1)/x, using the quotient rule, is
[tex]\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]<br /> when x= 1 or -1.<br /> <br /> 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.[/tex]

 

[physicsed]

whaaat

 

[physicsed]

You are missing ( ) in that. The derivative of (x²+ 1)/x, using the quotient rule, is
[tex]\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]<br /> when x= 1 or -1.<br /> <br /> 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.[/tex]

[tex] <br /> <br /> <br /> $$\frac{(2x)- (x^2+1)}{x^2}= \frac{x^2- 1}{x^2}$$<br /> <br /> how are those equal?[/tex]

 

[Nick89]

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

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

 

[physicsed]

how are they equall?

 

[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.

 

[physicsed]

so, i was right?

 

[Nick89]

No.

You're answer was:
$$\frac{2x^2 - x^2 + 1}{x^2} = 0$$ 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!)

 

[physicsed]

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

how are they equall?

 

[Nick89]

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

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

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

 

[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

 

[Nick89]

Surely you can substract $$1x^2$$ from $$2x^2$$?

$$2x^2 - x^2 = (2-1)x^2 = 1x^2 = x^2$$. I can't write it out any more than that.

 

[physicsed]

am soo stupid...thanks

 

[Nick89]

No problem, it happens ;)