Answer
Consider the continuous function $f\left( {x,y} \right) = x + 2y$.
Since there is no solution for ${f_x} = 0$ and ${f_y} = 0$, we conclude that there is no critical point for $f$. Hence, there is no global maximum on $D$.
This does not contradict Theorem 3 since $D$ is not bounded.
Work Step by Step
We have the domain $D = \left\{ {\left( {x,y} \right):x + y \ge 0,x + y \le 1} \right\}$.
The boundary points of $D$ are on the lines $x+y=0$ and $x+y=1$. Since $D$ contains all its boundary points, it is closed. However, $D$ is not bounded because we can always choose a point $\left( {x,y} \right)$ such that $D$ contains point arbitrarily far from the origin. Therefore, Theorem 3 is not applicable to $D$.
Let us choose a continuous function $f\left( {x,y} \right) = x + 2y$.
The partial derivatives are
${f_x} = 1$, ${\ \ \ }$ ${f_y} = 2$
Since there is no solution for ${f_x} = 0$ and ${f_y} = 0$, we conclude that there is no critical point for $f$. Hence, there is no global maximum on $D$. This does not contradict Theorem 3 since $D$ is not bounded.
