# How to solve for x using 2nd derivative?

#### gevni

##### New member
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

Staff member
Can you post the original problem in its entirety?

• topsquark and anemone

#### gevni

##### New member
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
$y = Ax + B(n-x) + \dfrac{CT}{x}$ 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 and topsquark

#### gevni

##### New member
$y = Ax + B(n-x) + \dfrac{CT}{x}$ 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

##### Well-known member
MHB Math Helper
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?

• gevni

#### Country Boy

##### Well-known member
MHB Math Helper
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
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 and topsquark