Maximization problems - Using implicit function to avoid KKT I am wondering if the following procedure to solve a maximization problem in - let's say - two variables with inequality and non-negative constraints works.
More specifically, let's assume something like the following concrete example:
$ \text{max } F(x,y)=x^{\frac{2}{3}} y^{\frac{1}{3}}$
$ \text{sub } G(x,y)=x^2 + y^2 \leq 2 \hspace{0.5cm}\text{and } x,y\geq 0$
I thought that, instead of using Kuhn-Tucker conditions, which would bring me to a real nightmare, it would be better to go for the following steps:


*

*I realize that the objective function is increasing in $\mathbb{R}^3_{+}$.

*For this reason the constraint has to work as an equality constraint and it is binding.

*I take the implicit function of $F(x,y)$  of $y$ in terms of $x$ and I put it equal to the same thing, for the inequality constraint.

*I create a system with the previous equation and the inequality constraint expressed as an equality constraint.

*I focus on the positive value of $x$ and $y$ and that's it.


Does it sound reasonable?
Any feedback is welcome!
 A: This sound reasonable. The precise justification for your step 1 could go along the following lines: Assume that $(x,y)$ is a (feasible) minimizer of $F$. 
Obviously, $F(x,y) > 0$.
Assume $r^2 := x^2 + y^2 < 2$. Then, take $(\sqrt2 \, x/r,\sqrt 2\,y/r)$. Obviously, this point is still feasible and $F(\sqrt 2 \, x/r,\sqrt 2\,y/r) = F(x,y) \, \sqrt 2/r > F(x,y)$. Contradiction.
Hence, $x^2 + y^2 = 2$ holds for the minimizer.
A: Your assumptions of (1) and (2) are correct and then you can use a more formal way. To maximize $F(x,y)$ the first differential must be zero
$$dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy=\frac 23 \frac{y^{1/3}}{x^{1/3}}dx+\frac 13 \frac{x^{2/3}}{y^{2/3}}dy=0$$
Due to constraint $dx$ and $dy$ are related in a way that $x$ and $y$ must always be on the curve $G(x,y)=2$. To find the relation we use the first differential
$$dG=\frac{\partial G}{\partial x}dx+\frac{\partial G}{\partial y}dy=2x\,dx+2y\,dy=0\Rightarrow dy=-\frac xy dx$$
Now we can use this relation to eliminate $dy$ in $dF$
$$dF=\frac 23 \frac{y^{1/3}}{x^{1/3}}dx-\frac 13 \frac{x^{2/3}}{y^{2/3}}\frac xy dx=0$$
$$dF=\bigg(\frac 23 \frac{y^{1/3}}{x^{1/3}}-\frac 13 \frac{x^{2/3}}{y^{2/3}}\frac xy\bigg) dx=0$$
The above equation can be satisfied for any $dx$ (or in any directions) if
$$\frac 23 \frac{y^{1/3}}{x^{1/3}}-\frac 13 \frac{x^{2/3}}{y^{2/3}}\frac xy=0$$
$$2 \frac{y^{1/3}}{x^{1/3}}=\frac{x^{5/3}}{y^{5/3}}$$
$$2 y^2=x^2\Rightarrow y=\frac x{\sqrt 2}$$
Now we satisfied that objective $F(x,y)$ is maximized along the curve $G(x,y)$. We need to satisfy one more condition that $G(x,y)=2$ by using the relation between $y$ and $x$
$$x^2+y^2=x^2+\frac{x^2}2=2\Rightarrow x=\frac 2{\sqrt 3}$$
and
$$y=\frac x{\sqrt 2}=\frac {\sqrt 2}{\sqrt 3}$$
So you neither used Lagrange multipliers nor KKT.
