MHB Minimize a function: Find value of x that result in lowest value of formula

[gevni]

Hi, I have this formula, What I want is to find the value of "x" (without trying all possibilities) so that the result of the formula will be the lowest possible value under the constraint when x !=0, and x<n. Here, values of A,B,C, Q, R,n are already known and fixed.

$$f(x)=A\left(((\frac{Q}{n-x}-\frac{R}{x})+(n-x))n\right)+Bnx+C((\frac{R}{x})n)$$

How can I find the value of x? I know that x can be between 1 to n-1. But how do I continue from there? I was thinking there must be a way to calculate it instead of trying a lot of possibilities. Like some derived equation for optimal value of x that made the whole formula sum the lowest possible.

 

[gevni]

Hi, I have this formula, What I want is to find the value of "x" (without trying all possibilities) so that the result of the formula will be the lowest possible value under the constraint when x !=0, and x<n. Here, values of A,B,C, Q, R,n are already known and fixed.

$$f(x)=A\left(((\frac{Q}{n-x}-\frac{R}{x})+(n-x))n\right)+Bnx+C((\frac{R}{x})n)$$

How can I find the value of x? I know that x can be between 1 to n-1. But how do I continue from there? I was thinking there must be a way to calculate it instead of trying a lot of possibilities. Like some derived equation for optimal value of x that made the whole formula sum the lowest possible.

Update: I remove the constant n and then take the 1st and 2nd derivative. Now my formula is like that;

$$f(x)=A\left((\frac{Q}{n-x}-\frac{R}{x})+(n-x)\right)+Bx+C(\frac{R}{x})$$

I took first and 2nd derivative as
$$ f′(x)= A\left(\dfrac{R}{x^2}+\dfrac{Q}{\left(n-x\right)^2}-1\right)-\dfrac{CR}{x^2}+B $$

$$ f′′(x)= A(\dfrac{2Q}{(n−x)^3}−\dfrac{2R}{x^3})+\dfrac{2CR}{x^3} $$

What would be the next step to find value of x that will minimize this equation?

 

[MarkFL]

Okay, what you want to do now is equate your first derivative to zero to find your critical values, that is those values of \(x\) that may optimize the function. What do you find?

 

[gevni]

Okay, what you want to do now is equate your first derivative to zero to find your critical values, that is those values of \(x\) that may optimize the function. What do you find?

Thanks you for the reply. For finding the critical points for x do I need to put the values of known constant in the formula?

 

[MarkFL]

Thanks you for the reply. For finding the critical points for x do I need to put the values of known constant in the formula?

No, you can get your critical values in terms of the parameters. There's no need to plug those in.

 

[MarkFL]

Let's follow up...we want:

$$f'(x)=A\left(\frac{R}{x^2}+\frac{Q}{(n-x)^2}-1\right)-\frac{CR}{x^2}+B=0$$

$$f'(x)=\frac{AR}{x^2}+\frac{AQ}{(n-x)^2}-A-\frac{CR}{x^2}+B=0$$

$$f'(x)=\frac{AR(n-x)^2+AQx^2-Ax^2(n-x)^2-CR(n-x)^2+Bx^2(n-x)^2}{x^2(n-x)^2}=0$$

As you have stated that \(x\ne0\) and \(x<n\), this implies:

$$AR(n-x)^2+AQx^2-Ax^2(n-x)^2-CR(n-x)^2+Bx^2(n-x)^2=0$$

As this is a 4th degree polynomial, I would at this point consider substituting in for the parameters (as you asked about doing) and then using software or a numeric technique to approximate the roots.

 

[gevni]

Let's follow up...we want:

$$f'(x)=A\left(\frac{R}{x^2}+\frac{Q}{(n-x)^2}-1\right)-\frac{CR}{x^2}+B=0$$

$$f'(x)=\frac{AR}{x^2}+\frac{AQ}{(n-x)^2}-A-\frac{CR}{x^2}+B=0$$

$$f'(x)=\frac{AR(n-x)^2+AQx^2-Ax^2(n-x)^2-CR(n-x)^2+Bx^2(n-x)^2}{x^2(n-x)^2}=0$$

As you have stated that \(x\ne0\) and \(x<n\), this implies:

$$AR(n-x)^2+AQx^2-Ax^2(n-x)^2-CR(n-x)^2+Bx^2(n-x)^2=0$$

As this is a 4th degree polynomial, I would at this point consider substituting in for the parameters (as you asked about doing) and then using software or a numeric technique to approximate the roots.

Thanks Mark! As I am interested in extrema points so I took 2nd derivative not the first 1 and simplify it into cubic equation of degree 3. And then using Cardano’s formula I found the roots and took real root as my minimum value of x.