## Calculus (3rd Edition)

$$global~max:2,\ \ \ \ global~min: 0$$
Given $$f(x, y)=x+y, \quad 0 \leq x \leq 1, \quad 0 \leq y \leq 1$$ The maximum of $x$ is $1$ and of $y$ is $1$. The maximum of $x+y$ is $1+1$ and the minimum is $0+0$. Hence, the global maximum of $f$ on the given set is $$f (1, 1) = 1 + 1 = 2$$ and the global minimum is $$f (0, 0) = 0 + 0 = 0$$