Solving an obvious problem using Lagrange multipliers/KKT I have the following optimization problem:

$\max_{x,y} xy $ subject to $ x, y \geq 0$ and $y + x^2 \leq 6$.

I know how to solve this problem. By looking at the objective function it is clear that the maximum must lie on $y + x^2 = 6$ and then it can be found as

$\max_x x(6-x^2)$ for $x \in [0, \sqrt{6}]$.

I would like to practice Lagrange multipliers by doing it that way.
So I set up the Lagrangian

$L(x,y, \lambda, \mu, \nu) = xy - \lambda (y + x^2 - 6) + \mu x + \nu y$

and look at the KKT conditions:

$ \frac{\partial L}{\partial x} = y - 2\lambda x + \mu = 0$

$ \frac{\partial L}{\partial y} = x - \lambda + \nu = 0$

$x,y,\lambda, \mu, \nu \geq 0 $

$y + x^2 \leq 6$

$\mu x = \nu y = \lambda (y + x^2 -6) = 0$

I'm able to solve for $x$ and $y$ in terms of the Lagrange multipliers:

$x = \lambda - \nu$ and $y = 2 \lambda^2 - 2 \lambda \nu - \mu$

But after that I'm stuck. I'm not sure how to proceed, and the apparent simplicity of the objective function seems to have vanished.

Should I consider the 8 possible cases from complementary slackness one by one?

(Context: this is an example of the Nash bargaining problem)
You can obtain the solutions by reasoning.

Suppose $\lambda=0$, then $y=-\mu$ and $x=-\nu$. Then $y=\mu=0$ and $x=\nu=0$ due to nonnegativity. Thus we obtain one point that satisfies the KKT conditions.

Now suppose $\lambda&gt;0$, then $y+x^2=6$. Suppose $x=0$, then $y=6$, $\nu=0$, $\lambda=0$, contradiction. So, $x&gt;0$. Suppose $y=0$, then $x=\sqrt{6}$, $\mu=0$, $\lambda=0$, contradiction. So, $y&gt;0$. So, $\mu=\nu=0$. So, $x=\lambda$ and $y=2\lambda x = 2 x^2$. So, $3x^2 = 6$, $x=\sqrt{2}$, $y=4$.

The reason we obtain multiple KKT points is that the problem is not a convex optimization problem. The KKT conditions are therefore necessary but not sufficient. Comparing the objective value for both solutions, the second solution is optimal. If you perform a logarithmic transformation on the objective, the problem becomes convex and the KKT conditions are also sufficient.
We have: $f(x,y) = xy \le x(6-x^2)= 6x-x^3 = g(x)\implies g'(x) = 6-3x^2 = 3(2-x^2)=0 \iff x = \sqrt{2}\implies g''(\sqrt{2})= - 6\sqrt{2} &lt; 0 \implies g_{\text{max}} = g(\sqrt{2})= 4\sqrt{2}= f_{\text{max}}$. Note that $f_{\text{max}}$ occurs at $ x = \sqrt{2}, y = 4$