# Maximum, function with three variables

• Apr 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 \$\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
• Apr 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 \$\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?

Quote:

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
• Apr 11th 2011, 02:39 AM
Schdero

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?
• Apr 11th 2011, 04:28 AM
CaptainBlack
Quote:

Originally Posted by Schdero