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This is mostly intuition, but think of what segments the set consists of and note them down. Then, for each of the sets of the segments, determine the critical points -(e.g. for set $A = \{ (x, y) \in \R^2 \divides x = 0, 0 \leq y \leq 3 \}$, we compute the critical points of $f(0, y)$) +(e.g. for set $A = \{ (x, y) \in \R^2 \divides x = 0, 0 \leq y \leq 3 \}$, we compute the critical points of $f(0, y)$). + +This can be done as follows if only one variable remains: $\frac{\dx}{\dx y} f(0, y)$ using Analysis I conditions ($\frac{\dx}{\dx x}$ for $x$ variable of course), +i.e. if derivative cannot be $0$, there is no critical point there, else find solution for $x$ or $y$. For cases where $x$ and $y$ are both not $0$, we have to parametrize the set (e.g. for set $C = \{ (x, y) \in \R^2 \divides 3x + y = 3, 0 \leq x \leq 1 \}$, we have $\gamma(t) = (t, 3 - 3t)$ and compute the critical points of $f(\gamma(t))$) + +Finally, evaluate if the points are minima or maxima. It is often easiest to compute $f(x, y)$ at these points to see, +where the lowest value is the global minimum and the highest value the global maximum (obviously). +Always consider the corners as possible maxima or minima (if some corners are critical points, all are highly likely to be). + +The tangent plane at a critical point of a function $f : \R^n \rightarrow \R$, is of the form $\{ (x, y, z) \dividees z = \text{const} \}$, with $z = f(x_0)$.