Absolute Minimums and Maximums — Question 9

Question 9 – Solution

Strategy. Determine the finite interval where the line lies above the parabola, then reduce each boundary to one variable.

Step 1: Region The boundaries meet when $$x^2=2-x,$$ so $(x+2)(x-1)=0$. The intersection points are $(-2,4)$ and $(1,1)$, and $$-2\le x\le 1,\qquad x^2\le y\le 2-x.$$

Step 2: Candidate search Since $f_y=1$, there are no interior critical points. On the lower boundary $y=x^2$, $$f=2x^2,$$ whose minimum is $0$ at $x=0$ and whose largest endpoint value is $8$ at $x=-2$. On the upper boundary $y=2-x$, $$f=x^2-x+2=\left(x-\frac 12\right)^2+\frac 74.$$ Its boundary minimum is $7/4$ at $x=1/2$; its endpoint values are $8$ and $2$.

Step 3: Comparison Therefore $$\boxed{f_{\min}=0\text{ at }(0,0)}, \qquad \boxed{f_{\max}=8\text{ at }(-2,4)}.$$ Because $f=x^2+y$ increases with $y$, it is also geometrically consistent that vertical comparisons reduce to the lower curve for the minimum and upper curve for the maximum.