Let $ n,m \in \mathbb{N} \text{ and } m \geq n \gt 1$
Let $ f(n,m) = \frac{m-n+1}{2(n-1)}$ and therefor:
$\frac{\partial f}{\partial m}(m,n) = \frac{1}{2(n-1)}$
$\frac{\partial f}{\partial n}(m,n) = -\frac{m}{2(n-1)²}$
Since the two partial derivatives do not have a solution such that they both equal to zero, how can I get the local maximum? Is there a problem in my reasoning?
Thanks in advance
If your domain is discrete, then you have to define the notion of "local." One way to think of it is the $8$ lattice points/neighbors that surround the given point $(m,n)$ and then you can say $(m,n)$ is a local maximum if it's greater than or equal to those $8$ values $f$ is evaluated at. But that would not work because incrementing $m$ by $1$ will always increase your function.
More generally, there is no such notion even if you extend the immediate neighborhood to a greater a raduis because $f$ is strictly increasing in $m:$ $$f(m+1,n) - f(m,n) = \dfrac{1}{2(n-1)} > 0.$$
As a last resort, it might be possible to define some weird, discrete topology for which a neighborhood of $m$ does not include $m+1,$ but I am not aware of such things.