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How to solve for x using 2nd derivative?

gevni

New member
May 3, 2020
25
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

Administrator
Staff member
Feb 24, 2012
13,774
Can you post the original problem in its entirety?
 

gevni

New member
May 3, 2020
25
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

Well-known member
MHB Math Helper
Mar 1, 2012
655
$y = Ax + B(n-x) + \dfrac{CT}{x}$ 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

New member
May 3, 2020
25
$y = Ax + B(n-x) + \dfrac{CT}{x}$ 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

Well-known member
MHB Math Helper
Mar 1, 2012
655
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?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
426
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!
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
426
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

New member
May 3, 2020
25