Maximum, function with three variables

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April 9th 2011, 11:28 PM
Schdero

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

Thank you

Schdero
April 11th 2011, 01:28 AM
CaptainBlack

Quote:

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 $(B-2m)/A$ (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?

Quote:

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

Thank you

Schdero

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

CB
April 11th 2011, 02:39 AM
Schdero

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

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?
April 11th 2011, 04:28 AM
CaptainBlack

Quote:

Originally Posted by Schdero

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

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