Fritz John condition Consider the following problem :

$$\min x+y^2$$
$$\text{s.t.}~~x^2-4x+4 \leq 0 $$

The question is : Find all the Fritz-Jons points for this problem.$$$$
Solution : let $f(x,y)=x+y^2$ and $g(x,y)=x^2-4x+4$ we can observe that g is differentiable and continuous at $(\bar x, \bar y)$. And let $(\bar x, \bar y)$ be a feasible solution
\begin{align}
\nabla f(\bar x,\bar y)&amp;=(1,2 \bar y)^T \cr
 \nabla g(\bar x,\bar y)&amp;=(2 \bar x-4,0)^T
\end{align}
Therefore, If $\bar x$ solves the above problem locally, then there exists $\lambda_1,\lambda_2 \in \mathcal{R}$ such that :
$$ \lambda_1(1,2 \bar y)^T+\lambda_2(2 \bar x-4,0)^T=0\Leftrightarrow
\begin{cases}\lambda_1+\lambda_2(2 \bar x-4)=0 \cr
\lambda_12\bar y =0 \end{cases}$$

I'm stuck here
Let $$f(x,y)=x+y^2$$ and $$\frac{\partial f(x,y)}{\partial x}=1$$ so we have to search the extrema on the curve $$x^2-4x+4=(x-2)^2=0$$ and this is $x=2$
Let $$f(x,y)=x+y^2$$ and $$\frac{\partial f(x,y)}{\partial x}=1$$ so we have to search the extrema on the curve $$x^2-4x+4=(x-2)^2=0$$ and this is $x=2$