- #1

gevni

- 25

- 0

\[ 2 \times A \times (\frac{T}{x^3})=0 \]

Here A,T,x are greater then 0. I don't know how to proceed further, how to solve it for x? Can you please guide me?

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- MHB
- Thread starter gevni
- Start date

- #1

gevni

- 25

- 0

\[ 2 \times A \times (\frac{T}{x^3})=0 \]

Here A,T,x are greater then 0. I don't know how to proceed further, how to solve it for x? Can you please guide me?

- #2

MarkFL

Gold Member

MHB

- 13,302

- 11

Can you post the original problem in its entirety?

- #3

gevni

- 25

- 0

The original formula is

\[ cost= A\times x+ B \times(n-x)+ C\times \frac{T}{x} \]

As I am only interested in minimum value of this formula. So I took 2nd derivative of it. That results in ;

\[ 2 \times C \times (\frac{T}{x^3})=0 \]

Here, A,B,C,T,n and x are greater then zero and A,B,C,T,n are known variables. I don't know how to solve it for x after 2nd derivative ?

- #4

skeeter

- 1,104

- 1

$y' = 0$ at $x = \sqrt{\dfrac{CT}{A-B}}$, assuming $A > B$

$y'' = \dfrac{2CT}{x^3} > 0$, so the minimum occurs at the x-value stated above

- #5

gevni

- 25

- 0

$y' = 0$ at $x = \sqrt{\dfrac{CT}{A-B}}$, assuming $A > B$

$y'' = \dfrac{2CT}{x^3} > 0$, so the minimum occurs at the x-value stated above

Thank you soooo much for helping. I was confused about whether I have to use

$y' = 0$ at $x = \sqrt{\dfrac{CT}{A-B}}$

or

$y' = 0$ at $x = - \sqrt{\dfrac{CT}{A-B}}$

And your explanation made it easy to understand.

- #6

skeeter

- 1,104

- 1

Thank you soooo much for helping. I was confused about whether I have to use

$y' = 0$ at $x = \sqrt{\dfrac{CT}{A-B}}$

or

$y' = 0$ at $x = - \sqrt{\dfrac{CT}{A-B}}$

And your explanation made it easy to understand.

you stated ...

Here, A,B,C,T,nand x are greater then zero

positive square root, correct?

- #7

HOI

- 923

- 2

This is the same as \[\frac{2AT}{x^3}= 0 \].

\[ 2 \times A \times (\frac{T}{x^3})=0 \]

Here A,T,x are greater then 0. I don't know how to proceed further, how to solve it for x? Can you please guide me?

But a fraction is 0 only when the

- #8

HOI

- 923

- 2

A minimum for f(x) can occur where the first derivative is 0 and the second derivative is positive, not where the second derivative is 0!

The original formula is

\[ cost= A\times x+ B \times(n-x)+ C\times \frac{T}{x} \]

As I am only interested in minimum value of this formula. So I took 2nd derivative of it. That results in ;

\[ 2 \times C \times (\frac{T}{x^3})=0 \]

Here, A,B,C,T,n and x are greater then zero and A,B,C,T,n are known variables. I don't know how to solve it for x after 2nd derivative ?

- #9

gevni

- 25

- 0

yesyou stated ...

positive square root, correct?

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