Absolute maximum and minimum values of a function over a region
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.
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$For a continuous function over a closed region, the extreme values will occur either at a critical point or at a point on the boundary. Parameterize the boundary with $x=\cos t, y=\sin t$ and compute the critical points as a function of $t$. Compare the function value with the $-1$ at the critical point you already found.
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