# One last optimization problems

[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.

x+(1/x) = s

f' =1 + ln x

## The Attempt at a Solution

lost

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

1 + ln x =0
-1 = ln x

thats about all i can do

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

physicsed ... you took the antiderivative ...

you have to take the derivative, not the integral

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

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)

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

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

which is undifined...

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

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

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

how are those equal?

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

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

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how are they equall?

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.

so, i was right?

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!)

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

how are they equall?

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

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

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

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

it might sound childish, but how is that equal?

you can't factor anything or simplify it

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.

am soo stupid....thanks

No problem, it happens ;)