- #1

- 52

- 0

**[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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter physicsed
- Start date

- #1

- 52

- 0

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

- 277

- 2

- #3

- 52

- 0

1 + ln x =0

-1 = ln x

thats about all i can do

-1 = ln x

thats about all i can do

- #4

- 412

- 4

did you say there that

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

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

- #5

- 1,753

- 1

physicsed ... you took the antiderivative ...

- #6

- 277

- 2

you have to take the derivative, not the integral

- #7

- 52

- 0

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

- #8

- 277

- 2

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

- #9

- 52

- 0

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

- 41,847

- 966

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

- 52

- 0

whaaat

Last edited:

- #12

- 52

- 0

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

- 52

- 0

how are they equall?

- #15

- 555

- 0

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

- #16

- 52

- 0

so, i was right?

- #17

- 555

- 0

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

- 52

- 0

[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

- 52

- 0

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

- 0

[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

- 52

- 0

am soo stupid....thanks

- #23

- 555

- 0

No problem, it happens ;)

Share: