SOLUTION Since \(f\) is a polynomial, it is continuous on
the closed, bounded rectangle \(D\), so
Theorem 8 tells us there is both an absolute maximum and an absolute
minimum. According to step 1, we first find the critical points. These
occur when \[
f_x = 2x - 2y = 0 \quad f_y = -2x + 2 = 0
\] so the only critical point is \((1,
1)\), and the value of \(f\)
there is \(f(1, 1) = 1\). In step 2 we
look at the values of \(f\) on the
boundary of \(D\), which consists of
the four line segments \(L_1, L_2, L_3,
L_4\). On \(L_1\) we have \(y = 0\) and \[
f(x, 0) = x^2 \quad 0 \le x \le 3
\] This is an increasing function of \(x\), so its minimum value is \(f(0, 0) = 0\) and its maximum value is
\(f(3, 0) = 9\). On \(L_2\) we have \(x
= 3\) and \[
f(3, y) = 9 - 4y \quad 0 \le y \le 2
\] This is a decreasing function of \(y\), so its maximum value is \(f(3, 0) = 9\) and its minimum value is
\(f(3, 2) = 1\). On \(L_3\) we have \(y
= 2\) and \[
f(x, 2) = x^2 - 4x + 4 \quad 0 \le x \le 3
\] By observing that \(f(x, 2) = (x -
2)^2\), we see that the minimum value of this function is \(f(2, 2) = 0\) and the maximum value is
\(f(0, 2) = 4\). Finally, on \(L_4\) we have \(x
= 0\) and \[
f(0, y) = 2y \quad 0 \le y \le 2
\] with maximum value \(f(0, 2) =
4\) and minimum value \(f(0, 0) =
0\). Thus, on the boundary, the minimum value of \(f\) is 0 and the maximum is 9. In step 3 we
compare these values with the value \(f(1, 1)
= 1\) at the critical point and conclude that the absolute
maximum value of \(f\) on \(D\) is \(f(3, 0)
= 9\) and the absolute minimum value is \(f(0, 0) = f(2, 2) = 0\). Figure shows the
graph of \(f\).
