Given the function $f(x,y) = \begin {cases} \frac{6}{7}(x^2 + \frac{xy}{2})&\text{ , } (x,y) \in (0,1)\times(0,2)\\ 0&\text{ , } (x,y) \notin (0,1)\times(0,2) \end{cases} $
I have tried finding the point where the tangent plane is parallel to the axis plane. Solving the $\begin {cases} \frac{\partial f}{\partial x}f(x,y) = 0 \\ \frac{\partial f}{\partial y}f(x,y) = 0\end {cases}$ equation system gave me the result is the $(0, 0)$ point. I got the same result checking where the $\nabla f = \vec{0}$. However using the "D-test", which I found on the internet, showed that it is actually a minimum point of the function. The problem is that the maximum point is not in a point where the tangent plane is parallel to the axis plane, rather at the edge of this $(0,1)\times(0,2)$ area. I know by checking the plot of the function that this maximum point is in $(1, 2)$, however I would like to know how could I possibly calculate it by myself.