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