How to solve for x using 2nd derivative?

[gevni]

Hi, I am trying to find the minimum root (x) of one formula. For that, I took 2nd derivative and got this equation.

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

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?

 

[MarkFL]

Can you post the original problem in its entirety?

 

[gevni]

Ok! so I have a formula for cost calculation. I have to find the value of x where that formula will gives minimum value as cost should not be equal to zero, it has some minimum value.
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 ?

 

[skeeter]

$y = Ax + B(n-x) + \dfrac{CT}{x}$ will have a minimum at the x-value where $y' = 0$ and $y'' > 0$

$y' = 0$ 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

 

[gevni]

$y = Ax + B(n-x) + \dfrac{CT}{x}$ will have a minimum at the x-value where $y' = 0$ and $y'' > 0$

$y' = 0$ 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.

 

[skeeter]

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,n and x are greater then zero

positive square root, correct?

 

[HOI]

Hi, I am trying to find the minimum root (x) of one formula. For that, I took 2nd derivative and got this equation.

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

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?

This is the same as \[\frac{2AT}{x^3}= 0 \].
But a fraction is 0 only when the numerator is 0. There is no value of x that makes it 0!

 

[HOI]

Ok! so I have a formula for cost calculation. I have to find the value of x where that formula will gives minimum value as cost should not be equal to zero, it has some minimum value.
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 ?

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!

 

[gevni]

you stated ...



positive square root, correct?

yes