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One last optimization problems

  • Thread starter physicsed
  • Start date
52
0
[SOLVED] one last optimization problems

1. Homework Statement

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

2. Homework Equations

x+(1/x) = s

f' =1 + ln x

3. The Attempt at a Solution

lost
 

Answers and Replies

275
2
you've got it right so far. What happens to the derivative when something is at a minimum or a maximum?
 
52
0
1 + ln x =0
-1 = ln x

thats about all i can do
 
351
2
did you say there that
d/dx (1/x) = ln (x)?
 
1,750
1
physicsed ... you took the antiderivative ...
 
275
2
you have to take the derivative, not the integral
 
52
0
do u want me to use the quo rule for 1/x which is -1/x^2???
 
275
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)
 
52
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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
 
Last edited:
HallsofIvy
Science Advisor
Homework Helper
41,738
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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
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.
 
52
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whaaat
 
Last edited:
52
0
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} [/tex]

how are those equal?
 
Last edited:
554
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]
 
Last edited:
52
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how are they equall?
 
554
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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.
 
52
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so, i was right?
 
554
0
No.

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!)
 
52
0
[tex]
\frac{2x^2- (x^2+ 1)}{x^2}= \frac{x^2-1}{x^2}
[/tex]

how are they equall?
 
554
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]
 
Last edited:
52
0
[tex]
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
 
554
0
Surely you can substract [tex]1x^2[/tex] from [tex]2x^2[/tex]?

[tex]2x^2 - x^2 = (2-1)x^2 = 1x^2 = x^2[/tex]. I can't write it out any more than that.
 
52
0
am soo stupid....thanks
 
554
0
No problem, it happens ;)
 

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