The prompt is to find the absolute maximum and minimum values of the function $f(x, y) = x^2 + 2x – y^2$ on the upper half of a unit disc, $D =\{(x, y): x^2 + y^2 \leq 1, y \geq 0 \}$
I began by finding the critical points of the region
$$f_x = 2x + 2$$
$$f_y = -2y$$
$$f_{xx} = 2$$
$$f_{yy} = -2$$
$$f_{xy} = 0$$
Finding critical points from first and second equations,
$$2x + 2 = 0$$
$$-2y = 0$$
$$x = -1, y = 0$$
I'm not sure how to proceed further, any help would be appreciated.
Best Answer
From your calculation, you have shown that there is no critical point in the interior of the half-disk, but the left endpoint of the semicircular boundary is one. The index of that point is
$$ D \ \ = \ \ f_{xx}·f_{yy} \ - \ (f_{xy})^2 \ \ = \ \ 2·(-2) \ - \ 0^2 \ = \ -4 \ \ < \ \ 0 \ \ , $$
which identifies $ \ (-1 \ , \ 0 ) \ $ as a saddle-point. This makes sense since the function is "concave upward" in the $ \ x-$ direction about $ \ x \ = \ -1 \ $ , due to the $ \ x^2 \ + \ 2x \ = \ (x + 1)^2 \ - \ 1 \ $ terms, but is "concave downward" in the $ \ y-$ direction because of the $ \ -y^2 \ $ term. As we will need it for comparisons later, we compute that $ f(-1,0) \ = \ (-1)^2 \ + \ 2(-1) \ - \ 0^2 \ = \ -1 \ \ . $
The lower boundary of the region is the $ \ x-$ axis over the interval $ \ [ \ -1 \ , \ 1 \ ] \ $ . Inserting $ \ y \ = \ 0 \ $ into the function expression gives $ \ f(x,0) \ = \ (x + 1)^2 \ - \ 1 \ \ . $ We have already found $ \ f(-1,0) \ $ to be the minimum on this line segment; the right endpoint has the value $ f(1,0) \ = \ (1+1)^2 \ - \ 1 \ = \ 3 \ \ . $
It remains to examine the upper semicircular boundary. There are a number of ways we might set this up, but to avoid having to work with a square-root or trigonometric identities, we will use the circle equation to write $ \ y^2 \ = \ 1 \ - \ x^2 \ $ and express our function on this section of the boundary as
$$ \ \phi(x) \ = \ (x + 1)^2 \ - \ 1 \ - \ (1 - x^2) \ \ = \ \ (x + 1)^2 \ + \ x^2 \ - \ 2 \ \ . $$
[This gives us the correct values for the function at the endpoints of the region:
$$ \phi(-1) \ \ = \ \ ([-1] + 1)^2 \ + \ (-1)^2 \ - \ 2 \ \ = \ -1 \ \ \ , \ \ \ \phi(1) \ \ = \ \ (1 + 1)^2 \ + \ 1^2 \ - \ 2 \ \ = \ \ 3 \ \ . \ ] $$
We find $ \ \phi'(x) \ = \ 2·(x + 1) \ + \ 2x \ \ = \ \ 0 \ \ \ \Rightarrow \ \ \ 4x \ \ = \ -2 \ \ \ \Rightarrow \ \ \ x \ = \ -\frac{1}{2} \ \ , $ for which $ \phi \left( -\frac{1}{2} \right) \ \ = \ \ \left( \ \left[-\frac{1}{2} \right] + 1 \ \right)^2 \ + \ \left(-\frac{1}{2} \right)^2 \ - \ 2 \ \ = \ \ \frac{1}{4} \ + \ \frac{1}{4} \ - \ 2 \ \ = \ -\frac{3}{2} \ \ . $
Hence, we observe that the absolute maximum of the function on this closed half-disc is $ f(1,0) \ = \ 3 \ $ and the absolute minimum is $ f \left( -\frac{1}{2} \ , \ \frac{\sqrt{3}}{2} \right) \ = \ -\frac{3}{2} \ \ . $ The saddle point at $ \ (-1 \ , \ 0 ) \ $ has no special role in this region.