Constrained Optimization

Question

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?

Answer 1

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.