skeeter said:$y = Ax + B(n-x) + \dfrac{CT}{x}$ 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 said: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.
Here, A,B,C,T,n and x are greater then zero
This is the same as \[\frac{2AT}{x^3}= 0 \].gevni said: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?
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 said: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 ?
yesskeeter said:you stated ...
positive square root, correct?
The 2nd derivative is used to find the concavity of a function, which helps determine the direction of the function's curvature. This information is useful in solving for x because it can help identify the maximum or minimum points of the function.
To find the 2nd derivative, you first need to take the derivative of the function to get the 1st derivative. Then, take the derivative of the 1st derivative to get the 2nd derivative. Alternatively, you can use the power rule or product rule to find the 2nd derivative.
Yes, the 2nd derivative can be used to solve for x in any type of function, as long as the function is continuous and differentiable. This means that the function must have a defined value at every point and its derivative must exist at every point.
The steps for solving for x using the 2nd derivative are as follows:
The 2nd derivative test is a method for determining the nature of a critical point (maximum or minimum) using the 2nd derivative of a function. It states that if the 2nd derivative is positive at a critical point, the point is a minimum, and if the 2nd derivative is negative, the point is a maximum. This information can be used to solve for x by finding the critical points and evaluating the function at those points to determine the maximum or minimum values.