Absolute Minimums and Maximums — Question 5

Question 5 – Solution

Strategy. Check the annulus interior, then write each boundary circle as $x=r\cos\theta$, $y=r\sin\theta$ with fixed $r$.

Step 1: Interior $$f_x=2x-3, \qquad f_y=2y.$$ The critical point is $(3/2,0)$, whose radius $3/2$ lies strictly between $1$ and $2$. Its value is $$f(3/2,0)=\frac 94-\frac 92=\boxed{-\frac 94}.$$

Step 2: Circular boundaries On a circle of radius $r$, $$f=r^2-3r\cos\theta.$$ For $r=1$, its range is $[-2,4]$. For $r=2$, its range is $[-2,10]$. The value $10$ occurs on the outer circle when $\cos\theta=-1$, at $(-2,0)$.

Step 3: Comparison Comparing $-9/4$, the boundary minimum $-2$, and the boundary maximum $10$ gives $$\boxed{f_{\min}=-\frac 94\text{ at }(3/2,0)}, \qquad \boxed{f_{\max}=10\text{ at }(-2,0)}.$$ Both boundary components were checked, so the hole in the region causes no missing candidates.