I was thinking whether there is some sufficient conditions to have always binding constraint for the following types of optimization problem:
$ \max_{x, y \in [0,1]x[0,1]} f(x,y; \alpha) \qquad \text{ s. t. } \qquad g(x,y; \alpha)>=0 $
where $f$ and $g$ are continuous and differentiable in whole domain and $\alpha$ is some model parameters.
Do you guys have any idea?
I first rephrase the problem $$ \max_{x, y} f(x, y; \alpha) \quad \text{s.t.} \quad \begin{align} c_1(x, y) \geq 0 \\ c_2(x, y) \geq 0 \\ c_3(x, y) \geq 0 \\ c_4(x, y)\geq 0 \\ g(x, y) \geq 0 \end{align} $$ where $c_0(x, y) = x$, $c_1(x, y) = 1-x$, $c_2(x, y) = y$ and $c_3(x, y) = 1-y$. If $f$ can be expressed in the form $$ f(x, y) = a_x x + a_y y + b, $$ i.e. $f$ is linear and $g$ can be expressed in the form $$g(x, y) = k_1 x + k_2 y + d $$ i.e. $g$ is linear then what we have is a Linear Programming (LP) problem. If there exists $(x_0, y_0)$ allowed by all the constraint, i.e. our permissible set is nonempty we know that the maximum of the function exists and must lie on either of the vertices where at least two of the constraints are active. In our case we have the following candidates: $(0, 0), (0, 1), (1,0), (1, 1)$ or at the intersection where $g(x, y) = 0 = c_i(x, y)$ for some $i = 1, ..., 4$. Thus we can conclude that if the problem is an LP and is not solved by $(0, 0), (0, 1), (1,0), (1, 1)$ then $g(x, y)$ must be active.
Of course the assumption of linearity both on the function and the constraint is quite strong, but I fail to find any relaxed constraint that ensures that the optimal $(x^*, y^*)$ is found when $g(x, y)$ is active.