# How to solve for x using 2nd derivative?

• MHB
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?

Gold Member
MHB
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 $x = - \sqrt{\dfrac{CT}{A-B}}$

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 $x = - \sqrt{\dfrac{CT}{A-B}}$

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