I'm trying to solve this problem by KKT
$$\begin{align*} \text{min} & \quad x y \\ \text{s.t} & \quad x^2 + y^2 \le 1 \end{align*}$$
One of the regularity conditions is linear independence constraint constraint qualification - LICQ, which is
The gradients of the active inequality constraints and the gradients of the equality constraints are linearly independent at $x^{*}$.
Could you please verify if I correctly apply the KKT's theorem? Thank you so much for your help!
My attempt:
Let $f(x,y) = x y$, $g(x,y) = x^2+y^2-1$, and $\mathcal K= \{(x,y) \in \mathbb R^2 \mid g(x,y) \le 0\}$. Because $\mathcal K$ is compact and $f$ is continuous, the problem has a solution.
Moreover, $\nabla f (x,y) = \begin{pmatrix} y \\ x \\ \end{pmatrix}$ and $\nabla g (x,y) = \begin{pmatrix} 2x \\ 2y \\ \end{pmatrix}$. Consider the system $$\begin{cases} g(x,y) &= 0\\ \mu \nabla g (x,y) &= 0\\ \end{cases} \iff \begin{cases} x^2+y^2 &= 1\\ \mu \begin{pmatrix} 2x \\ 2y \\ \end{pmatrix} &= 0\\ \end{cases} \implies \mu =0$$
Hence LICQ is satisfied. Consider the system
$$\begin{cases} \mu \ge 0 \\ \mu g(x,y)=0 \\ \nabla f (x,y)+\mu \nabla g (x,y) =0 \end{cases} \iff \begin{cases} \mu \ge 0 \\ \mu (x^2+y^2-1)=0 \\ \begin{pmatrix} y \\ x \\ \end{pmatrix} + \mu \begin{pmatrix} 2x \\ 2y \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \end{pmatrix} \end{cases} \iff \begin{cases} \mu \ge 0 \\ \mu (x^2+y^2-1)=0 \\ 2x \mu = -y \\ 2y \mu = -x \end{cases}$$
Solving this system, we get $$(\mu,x,y) \in \left \{(0,0,0), \left (1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ), \left (1,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}} \right ), \left (-1,\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}} \right ), \left (1,\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ) \right \}$$
Consequently, $f (0,0) = 0$, $f \left (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ) = \frac{1}{2}$, $f \left (\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}} \right ) = \frac{-1}{2}$, $f \left (\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}} \right ) = \frac{1}{2}$, and $f \left (\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ) = \frac{-1}{2}$.
Hence the solution to the problem is $$\left\{\left (\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}} \right ),\left (\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ) \right\}$$
Hint: You can also consider $$f(x,\pm\sqrt{1-x^2})=x\times(\pm\sqrt{1-x^2})$$