Absolute Minimums and Maximums — Question 2

Question 2 – Solution

Strategy. Treat the interior and the circular boundary separately; on the boundary the radial term is constant.

Step 1: Interior $$f_x=2x-2,\qquad f_y=2y.$$ The only solution is $(1,0)$, which lies on the boundary rather than in the interior. Thus there are no interior critical points.

Step 2: Boundary Set $x=\cos\theta$, $y=\sin\theta$. Then $$f(\cos\theta,\sin\theta)=1-2\cos\theta.$$ The smallest value occurs when $\cos\theta=1$ and the largest when $\cos\theta=-1$. Therefore $$\boxed{f_{\min}=-1\text{ at }(1,0)}, \qquad \boxed{f_{\max}=3\text{ at }(-1,0)}.$$

Step 3: Geometric verification $$f(x,y)=(x-1)^2+y^2-1.$$ Thus $f+1$ is squared distance from $(x,y)$ to $(1,0)$. Within the disk, that distance is smallest at $(1,0)$ and largest at the diametrically opposite point $(-1,0)$. This confirms both locations and values.