Fritz John condition

Question

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

Answer 1

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 $x=2$

Answer 2

We have $x^2-4x+4 \leq 0 \iff (x-2)^2 \le 0 \iff x=2.$ Hence you have to minimize the function $h(y)=2 +y^2$.

But $ \min \{h(y): y \in \mathbb R\}=2.$