Lagrange Multipliers: Domain must be open set?

Question

Find minimum area of a square with vertices on the unit sphere $S^1$.

Consider the square vertex in the first quadrant.

Then the area function is $f(x,y) = 4xy$ and the constraint is ${g(x,y) = x^2-y^2-1=0}$

Further: $grad f(x,y) = (4y,4x), grad g(x,y) = (2x,2y)$

When finding the maximum and considering only positive values for x,y (which means Lagrange Multiplier $\lambda > 0$), one obtains $x=y$.

But finding the minimum and considering now $x,y\geq 0$ (so the minimum is at $(1,0)$ or $(0,1)$) does not work (no such $\lambda$ exists), because for $(x,y) = (1,0)$ $grad f = (0,4)$, $grad g=(2,0)$ are linearly independent. I cannot see which requirements are violated.

It seems that the domain has to be an open set (? If so, why ?)

Thanks in advance

Answer 1

Yes, it has to be an open set. It's just like finding a local extreme of a differentiable function $f$ of one variable. If the domain is an open interval (or an uniton of open intervals), you just have to check those points at which $f'$ is zero. But if the domain of $f$ is a closed interval $[a,b]$, you have to check also what happens at $a$ and $b$.