Given the function: $f(x, y) = x^2 + y^2 – 6x – 4y +11$
Determine the global extrema for f in the set $M = \{(x, y) \in \mathbb{R} | x^2 + y^2 -4x \le 5\}$
So I know that continuous functions on a compact set have both a maximum and a minimum there. But I'm a bit stuck here. Am I supposed to solve this by using the hessian matrix? And how would I go about showing that the local extrema are also global extrema, is it because of the compactness?
$\nabla{f(x,y)} = (\partial_x f,\partial_y f) = (2x – 6, 2y -4)$
$\nabla{f(x,y) = (0,0)}$
$2x -6 = 0 \\2y – 4 = 0$
Stationary point a = [3, 2]
Best Answer
To find the global extrema of continuous $f$ on a compact subset $M$ of $\mathbb{R}^2$, you do two things:
1. Interior. Find all critical points of $f$ in the interior of $M$.
2. Boundary. Find potential global extrema of $f$ on the boundary of $M$. If this boundary is given by a constraint (e.g., $x^2+y^2-4x=5$, as yours is), then you may use Lagrange multipliers.
Evaluate $f$ at all the points found in the two steps above. The largest (respectively, smallest) of these values is the global maximum (resp., global minimum) value of $f$ on $M$.
You have completed part 1.
(Note: Since you're asking about global extrema on a compact subset $M$ of $\mathbb{R}^2$, there is absolutely no reason to do anything for local extrema. If you wanted to know what happens at the critical point $(3,2)$, then fine---test it using the second-derivative test. But you are asking about global extrema; the second-derivative test is useless for this task.)