Absolute Minimums and Maximums — Question 1

Question 1 – Solution

Strategy. On a closed, bounded rectangle, compare values at interior critical points, one-variable boundary critical points, and corners.

Step 1: Interior $$f_x=2x-2,\qquad f_y=2y-4.$$ The sole interior critical point is $(1,2)$, where $$f(1,2)=1+4-2-8=\boxed{-5}.$$

Step 2: Boundaries On $x=0$ and $x=3$, $$f(0,y)=y^2-4y,\qquad f(3,y)=y^2-4y+3.$$ Both have a boundary critical value at $y=2$, giving $-4$ and $-1$. On $y=0$ and $y=4$, $$f(x,0)=f(x,4)=x^2-2x,$$ whose critical point $x=1$ gives $-1$. The four corner values are $$f(0,0)=0,\quad f(3,0)=3,\quad f(0,4)=0,\quad f(3,4)=3.$$

Step 3: Comparison The smallest candidate value is $-5$ and the largest is $3$. Hence $$\boxed{f_{\min}=-5\text{ at }(1,2)},$$ $$\boxed{f_{\max}=3\text{ at }(3,0)\text{ and }(3,4)}.$$ Completing the square, $f=(x-1)^2+(y-2)^2-5$, independently confirms the comparison.