Find the absolute extrema of f on the closed bounded set S in the plane of f(x,y)=2x^2-y^2?

Question

S in the disk x^2+y^2 is less than or equal to 1.

Answer 1

f(x, y) = 2x^2 - y^2 - it's obvious that the only internal critical point is (0, 0) and this is a saddle point. So the only extrema will be on the boundary x^2 + y^2 = 1. In this case we can substitute to get g(x) = 2x^2 - (1-x^2) = 3x^2 - 1, with -1 ≤ x ≤ 1.
Now g'(x) = 6x = 0 at x = 0, and g"(x) = 6 implies we have absolute minima at (0, 1) and (0, -1). Also g(1) = g(-1) = 2, so we have absolute maxima at (1, 0) and (-1, 0).
So the absolute minimum of f is -1 at (0, ± 1) and the absolute maximum is 2 at (±1, 0).