# Thread: Maximum, function with three variables

1. ## Maximum, function with three variables

Hi everybody!
I have a function, which encompassed formerly two variables and three constants. I found the maximum of this function. For some analysis I'm doing, I have to make one of the constants a variable though.
If i do the partial derivative regarding the new variable, i get $\displaystyle (B-2m)/A$ (A,B = constants; m = variable; A,B,m >0; m<B). I wonder now, where the maximum of this new function with three variables is.
What i did is i set $\displaystyle (B-2m)/A = 0$, what's true for $\displaystyle m=B/2$. This itself is a linear function with a negative slope, but the function is always positive before $\displaystyle m=B/2$. It starts from B and does down linearly untill $\displaystyle m=B/2$. So theoretically the overall function increases untill $\displaystyle m=B/2$, after this point, the partial derivative gets negative ($\displaystyle ((B-2m)/A)< 0$ for $\displaystyle m>B/2$) and henceforth starts from then on decreasing the overall function.

Is it correct that i thus say, the maximum of the function with three varibales is the same as the one with two, except for $\displaystyle m=B/2$?

Thank you

Schdero

2. Originally Posted by Schdero
Hi everybody!
I have a function, which encompassed formerly two variables and three constants. I found the maximum of this function. For some analysis I'm doing, I have to make one of the constants a variable though.
If i do the partial derivative regarding the new variable, i get $\displaystyle (B-2m)/A$ (A,B = constants; m = variable; A,B,m >0; m<B). I wonder now, where the maximum of this new function with three variables is.
Partial derivative of what?

What i did is i set $\displaystyle (B-2m)/A = 0$, what's true for $\displaystyle m=B/2$. This itself is a linear function with a negative slope, but the function is always positive before $\displaystyle m=B/2$. It starts from B and does down linearly untill $\displaystyle m=B/2$. So theoretically the overall function increases untill $\displaystyle m=B/2$, after this point, the partial derivative gets negative ($\displaystyle ((B-2m)/A)< 0$ for $\displaystyle m>B/2$) and henceforth starts from then on decreasing the overall function.

Is it correct that i thus say, the maximum of the function with three varibales is the same as the one with two, except for $\displaystyle m=B/2$?

Thank you

Schdero
You need to post sufficient information so that we can see what exactly you are talking about.

CB

3. Overall Function=$\displaystyle M((b-M)/a)>((b-d)(ad+bc))/(a+c)^2$

Maximum of the function f(a,c) is at a and c = infinity, as then fa=0 fc=0, (faa)(fcc)-fac^2 > 0.

Is there any further information you need to judge the corectness of my implication, that the maximum of f(a,c,m) lies at a and c = infinity, m=b/2?

4. Originally Posted by Schdero
Overall Function=$\displaystyle M((b-M)/a)>((b-d)(ad+bc))/(a+c)^2$

Maximum of the function f(a,c) is at a and c = infinity, as then fa=0 fc=0, (faa)(fcc)-fac^2 > 0.

Is there any further information you need to judge the corectness of my implication, that the maximum of f(a,c,m) lies at a and c = infinity, m=b/2?
What is the function you are trying to maximise, and what are the vaiables and constraints?

CB