This is a problem example taken from professor Robert Israel:
$$\max f(x,y)=xy \quad \text{subject to }\quad x+y^2\leq2, \quad x,y\geq0 \quad \quad (1)$$
The solution begins by writing the KKT conditions for this problem, and then one reach the conclusion that the global optimum is $(x^*,y^*)=(4/3,\sqrt{2/3})$.
However the linear independence constraint qualification (LICQ) fails everywhere, so in principle the KKT approach cannot be used directly. I have seen multiple examples solved like this, and I don't understand why this is legitimate.
It seems to me that the correct approach would be to reason as follows:
Suppose $(x^*,y^*)$ is a constrained local maximum for the problem. Then we see that neither of the nonnegativity constraints can bind at $(x^*,y^*)$, because otherwise every neighborhood of $(x^*,y^*)$ would contain a feasible point $(x,y)$ with $f(x,y)>0=f(x^*,y^*)$. Therefore $(x^*,y^*)$ is a constrained local maximum for the problem
$$\max f(x,y)=xy \quad \text{subject to }\quad x+y^2\leq2 \quad \quad (2)$$
which satisfies the LICQ condition everywhere. Using the KKT conditions gives $(x^*,y^*)=(4/3,\sqrt{2/3})$.
For sufficiency, we note that $f$ is continuous and that the feasible set in $(1)$ is compact, so $f$ attains a global maximum at some feasible $(x^*,y^*)$. Combined with the previous necessity argument, we conclude that $(x^*,y^*)=(4/3,\sqrt{2/3})$ is the unique local and global maximum for the problem.
Am I missing something? Thanks a lot for your help.
CQ is to be tested for active constraints (that become equalities) at a point only. Here you have three constraints:
Case 1: only one constraint of those three is active: gradients are non-zero, i.e. LICQ ok.
Case 2: $x=y=0$, $x+y^2<2$: gradients $(1,0)$ and $(0,1)$ are linearly independent, i.e. LICQ ok.
Case 3: $x=x+y^2-2=0$, $y>0$, and Case 4: $y=x+y^2-2=0$, $x>0$ - left as an exercise.
Case 5: all three are active: not possible.