Consider the following function: $$f(x, y) = \sqrt{1-x^2-y^2} e^{-3x^2-3xy + 2y^2}$$
I have to show that there do exist global max and min, and I have to calculate those points.
attempts
First of all, the domain is $D: x^2+y^2 \leq 1$, which is bounded and closed, so it's compact. Then we can invoke Weierstrass theorem that guarantees the existence of max and min.
Now I search the points in two ways: internal and boundary points. Internal points, all ok. $\nabla f(x, y) = (0, 0)$. A not easy couple of equations arise, but in the end $(x, y) = (0, 0)$ is a solution.
The problem arises at the boundary, because if I restrict $f$ at the boundary, which is $\partial D: x^2+y^2 = 1$, I get
$$f(x, y)\bigg|_{\partial D} = 0$$
And I cannot study that function on those points.
Yet I know that there are points on the boundary where $f$ reaches the minimum (the solution speaks about the points $a = (-1/\sqrt{10}, 3/\sqrt{10})$ and $b = (3/\sqrt{10}, 1/\sqrt{10})$
I don't understand: how we find those points on $\partial D$ if $f$ over here is zero?
You're right - $f = 0$ everywhere on the boundary circle, so there's nothing special about $\frac{1}{\sqrt{10}}(-1, 3)$ or $\frac{1}{\sqrt{10}}(3,1)$ compared to other points on the circle.
But $0$ is the global minimum, which is attained everywhere on the unit circle and nowhere else. Can you see why?