I am taking an intro real analysis course and we just covered constrained optimization.I remember the Lagrange multiplier method from multi variable calc, but I want to understand the reasoning behind the theorem a bit more. As an example, in Multi-variable, a problem such as optimize $f(x,y)=x-y$ subject to $x^2-y^2=2$ was considered a degenerate case. This is because if you actually use Lagrange multipliers, you get $1=2\lambda x$, $-1=-2\lambda y$ which gives $x=y$. Obviously, with the constraint, this is not useful. I was trying to understand why the theory broke down here, and it occurred to me that the hyperbola $x^2-y^2=2$ is not compact! I believe the theory breaks down when the constraint is not compact (Heine Borell applies here and this curve is unbounded). So, I believe the maximum is $(x,y)=(\sqrt{2}, 0)$ but I see no obvious way to find the min (assuming their is one!)
Can someone give me some intuition as to why this theorem is breaking down?
There is no "breakdown", nor a "degenerate case" here. Lagrange's method has truly reported that the given objective function $f(x,y):=x-y$ has no conditionally stationary point on the hyperbola $\gamma: \ x^2-y^2=2$. (The geometric reason behind this is that $\gamma$ has no points where the tangent is parallel to the level lines of $f$.)
Of course the question remains whether $f$ assumes global extrema on $\gamma$. Now it is easily seen that in the points $\bigl(x,-\sqrt{x^2-2}\bigr)\in\gamma$ the function $f$ assumes arbitrarily large values when $x\to\infty$. It follows that a global maximum does not exist, and by symmetry a global minimum does not exist either. As $\gamma$ is not compact we should not be too surprised.