Why Lagrange multipliers don't help to find the minimum of $f(x,y)=x^2+y^2$ with the constraint $y=1$? Please help me understand why the following doesn't work.
Say I want to find the minimum of the function $f(x,y)=x^2+y^2$ with the constraint $y=1$. So I declare the helper function $g(x,y)=(y-1)^2=0$. And by using the Lagrange multipliers method, what I get is $F=x^2+y^2+\lambda(y-1)^2, F_x=2x, F_y=2y+2\lambda(y-1)$ and
\begin{cases}
2x=0\\
2y+2\lambda(y-1)=0\\
(y-1)^2=0.
\end{cases}
The above has no solutions, although obviously $(0, 1)$ is a minimum value. 
 A: One of the requirements of the method is that $\nabla g \not= 0$ on the surface $g(x,y) = 0$.  But in your case $\nabla g(x,y) = (0,2(y-1))$ which is zero at every point on the surface $(y-1)^2 = 0$.
Moral of the story:  use $g(x,y) = y-1$ instead.
A: A slightly different perspective (but same idea), you are penalizing a constraint $(y-1)^2=0$, which is nonconvex, so strong duality will not apply (and the Lagrange solution will not give the optimum). In contrast, $y = 1$ is a convex constraint, so everything is good.
A: We want to find a critical point for $x^2+y^2$ given that $y=1$.

Proper Approach
This translates to the problem of finding when
$$
2x\delta x+2y\delta y=0\tag{1}
$$
is true for any $\delta(x,y)$ that fixes $y-1$, that is,
$$
0\,\delta x+1\,\delta y=0\tag{2}
$$
Thus, for any $\delta(x,y)$ that is perpendicular to $(0,1)$, $\delta(x,y)$ must be perpendicular to $(x,y)$. This means that $(x,y)\parallel(0,1)$, and therefore, $x=0$.

Problematic Approach
If we replace $(2)$ by the constraint that $\delta(x,y)$ stabilizes $(y-1)^2$, we get
$$
0\,\delta x+2(y-1)\,\delta y=0\tag{3}
$$
Unfortunately, since we have the constraint $y=1$, $(3)$ gives us that $0=0$. That is, $(3)$ gives us no information.
