How to solve a Karush-Kuhn-Tucker example 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.

I will confess that I don't know this "KKT" method! I have to solve it with basic "Calculus methods".
We want to maximize f(x,y)= xy subject to the condition that $x+ y^2\le 2$.  The graph of $x= 2- y^2$ is a parabola with vertex (2, 0) that crosses the y-axis at $(0, \sqrt{2})$ and $(0, -\sqrt{2})$.  $\nabla f= y\vec{i}+ x\vec{j}= 0$ at (0, 0).  The maximum of f occurs either at (0, 0) (and f(0,0)= 0) or on that parabola.  On the parabola $x= 2- y^2$, $f(x,y)= f(2- y^2, y)= (2- y^2)y= 2y- y^3$.  Setting $F(y)= 2y- y^3$, $F'= 2- 3y^2= 0$.  $y^2= 2/3$, y= 2/3 or y= -2/3.  If y= 2/3, x= 2- 4/9= 14/9, f(x,y)= f(14/9,2/3)= 28/27.   If y= -2/3, x= 2- 14/9= 4/9, f(x, y)= f(4/9, -2/3)= -8/27.
Of those three values, -8/27, 0, and 28/27, the largest is 28/27 which occurs at (14/8, 2/3
).
