I have serious problems with understanding these problems of the absolute maximums and minima of the functions given in the indicated domain $D$:
$$f(x,y) = 1+ x+2y, D = \{(x,y);x \geq 0,y \geq 0,x + y \leq 1\}$$
$$ f(x,y) = x^3 + y^3 −3xy, D = \{(x,y);0≤ x ≤2,−1≤ y ≤2 \}$$
In the first I do not know how to apply the inequality to the function to be optimized. In the second, even worse, I do not understand how to apply what I have been taught considering only the equalities in all domain.
Regarding the problem
$$ \min(\max) f(x,y) = x^3+y^3-3 x y \ \ \text{s. t.}\ \ D = \{(x,y);0≤ x ≤2,−1≤ y ≤2 \} $$
introducing the slack variables $s_k$ we have the lagrangian
$$ L(x,y,\lambda,s) = f+\lambda_1(x-s_1^2)+\lambda_2(2-x-s_2^2)+\lambda_3(y+1-s_3^2)+\lambda_4(2-y-s_4^2) $$
The stationary points are determined by solving
$$ \nabla L = 0 = \left\{ \begin{array}{rcl} 3 x^2+\lambda_1&=&\lambda_2+3 y \\ 3 y^2+\lambda_3&=&\lambda_4+3 x \\ \lambda_1 s_1&=&0 \\ \lambda_2 s_2&=&0 \\ \lambda_3 s_3&=&0 \\ \lambda_4 s_4&=&0 \\ s_1^2&=&x \\ s_2^2+x&=&2 \\ s_3^2&=&y+1 \\ s_4^2+y&=&2 \\ \end{array} \right. $$
giving
$$ \left[ \begin{array}{ccccccccccc} f & x & y & s_1^2 & s_2^2 & s_3^2 & s_4^2 & \lambda_1 & \lambda_2 & \lambda_3 & \lambda_4\\ -1 & 0 & -1 & 0 & 2 & 0 & 3 & -3 & 0 & -3 & 0 \\ 0 & 0 & 0 & 0 & 2 & 1 & 2 & 0 & 0 & 0 & 0 \\ 8 & 0 & 2 & 0 & 2 & 3 & 0 & 6 & 0 & 0 & 12 \\ -1 & 1 & 1 & 1 & 1 & 2 & 1 & 0 & 0 & 0 & 0 \\ 13 & 2 & -1 & 2 & 0 & 0 & 3 & 0 & 15 & 3 & 0 \\ 4 & 2 & 2 & 2 & 0 & 3 & 0 & 0 & 6 & 0 & 6 \\ 8-4 \sqrt{2} & 2 & \sqrt{2} & 2 & 0 & 1+\sqrt{2} & 2-\sqrt{2} & 0 & 12-3 \sqrt{2} & 0 & 0 \\ 8-4 \sqrt{2} & \sqrt{2} & 2 & \sqrt{2} & 2-\sqrt{2} & 3 & 0 & 0 & 0 & 0 & 12-3 \sqrt{2} \\ \end{array} \right] $$
Here when $s_k = 0$ indicates that the corresponding restriction is active. Follows a plot showing the feasible region $\mathcal{D}$ with level curves as well as the stationary points (red dots). Black vectors indicates the objective function gradient and red vectors indicates the active restrictions gradient at each stationary point. With those elements you can qualify each point using the KKT criterion.