- #1

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**[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

- Thread starter physicsed
- Start date

- #1

- 52

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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

lost

- #2

- 275

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- #3

- 52

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1 + ln x =0

-1 = ln x

thats about all i can do

-1 = ln x

thats about all i can do

- #4

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did you say there that

d/dx (1/x) = ln (x)?

d/dx (1/x) = ln (x)?

- #5

- 1,752

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physicsed ... you took the antiderivative ...

- #6

- 275

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you have to take the derivative, not the integral

- #7

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

- #8

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1/x = x^(-1) ... d/dx (1/x) = (-1)x^(-2) = -(1/x^2)

- #9

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thats what i got to so far

(x^2 + 1)/ x

[tex] 0=({{2x^2}-{x^2}+{1}})/{x^2} [/tex]

which is undifined...

NO answer

(x^2 + 1)/ x

[tex] 0=({{2x^2}-{x^2}+{1}})/{x^2} [/tex]

which is undifined...

NO answer

Last edited:

- #10

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

You are missing ( ) in that. The derivative of (xthats what i got to so far

(x^2 + 1)/ x

[tex] 0=({{2x^2}-{x^2}+{1}})/{x^2} [/tex]

which is undifined...

NO answer

[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

- #11

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whaaat

Last edited:

- #12

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You are missing ( ) in that. The derivative of (x^{2}+ 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} [/tex]

how are those equal?

Last edited:

- #13

- 555

- 0

A slight mistake there (probably a typo), it should be:

[tex]\frac{(2x^2)- (x^2+1)}{x^2}= \frac{x^2- 1}{x^2}[/tex]

[tex](\frac{d}{dx}(x^2+1)) (x) = (2x)(x) = 2x^2[/tex]

[tex]\frac{(2x^2)- (x^2+1)}{x^2}= \frac{x^2- 1}{x^2}[/tex]

[tex](\frac{d}{dx}(x^2+1)) (x) = (2x)(x) = 2x^2[/tex]

Last edited:

- #14

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how are they equall?

- #15

- 555

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He forgot to multiply the derivative of (x^2 + 1) with x.

- #16

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so, i was right?

- #17

- 555

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You're answer was:

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

The correct answer however should be:

[tex]\frac{2x^2 - (x^2 + 1)}{x^2} = 0[/tex] (there is one very important difference!)

- #18

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[tex]

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

[/tex]

how are they equall?

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

[/tex]

how are they equall?

- #19

- 555

- 0

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

[tex]\frac{2x^2 - (x^2 + 1)}{x^2} = \frac{2x^2 - x^2 - 1}{x^2} = \frac{x^2 - 1}{x^2}[/tex]

[tex]\frac{x^2 - 1}{x^2} = 0 \ \rightarrow \ x^2 - 1 = 0 \ \rightarrow \ x = \pm \sqrt(1) = \pm 1[/tex]

[tex]\frac{2x^2 - (x^2 + 1)}{x^2} = \frac{2x^2 - x^2 - 1}{x^2} = \frac{x^2 - 1}{x^2}[/tex]

[tex]\frac{x^2 - 1}{x^2} = 0 \ \rightarrow \ x^2 - 1 = 0 \ \rightarrow \ x = \pm \sqrt(1) = \pm 1[/tex]

Last edited:

- #20

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

[/tex]

it might sound childish, but how is that equal?

you can't factor anything or simplify it

- #21

- 555

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

- #22

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am soo stupid....thanks

- #23

- 555

- 0

No problem, it happens ;)

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