James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

966 Chapter 14 Partial Derivatives
To find the extreme values guaranteed by Theorem 8, we note that, by Theorem 2, if
f has an extreme value at sx 1 , y 1 d, then sx 1 , y 1 d is either a critical point of f or a boundary
point of D. Thus we have the following extension of the Closed Interval Method.
9 To find the absolute maximum and minimum values of a continuous function
f on a closed, bounded set D:
1. Find the values of f at the critical points of f in D.
2. Find the extreme values of f on the boundary of D.
3. The largest of the values from steps 1 and 2 is the absolute maximum value;
the smallest of these values is the absolute minimum value.
EXAMPLE 7 Find the absolute maximum and minimum values of the function
f sx, yd − x 2 2 2xy 1 2y on the rectangle D − hsx, yd | 0 < x < 3, 0 < y < 2j.
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 in (9), we first find the critical points. These occur when
f x − 2x 2 2y − 0 f y − 22x 1 2 − 0
y
(0, 2)
L£
(2, 2)
(3, 2)
so the only critical point is s1, 1d, and the value of f there is f s1, 1d − 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 shown in Figure 12. On L 1 we have y − 0 and
L¢ L
(0, 0)
FIGURE 12 12
L¡
7et140712
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MasterID: 01630
9
(3, 0)
x
f sx, 0d − x 2 0 < x < 3
This is an increasing function of x, so its minimum value is f s0, 0d − 0 and its maximum
value is f s3, 0d − 9. On L 2 we have x − 3 and
f s3, yd − 9 2 4y 0 < y < 2
This is a decreasing function of y, so its maximum value is f s3, 0d − 9 and its minimum
value is f s3, 2d − 1. On L 3 we have y − 2 and
f sx, 2d − x 2 2 4x 1 4 0 < x < 3
By the methods of Chapter 4, or simply by observing that f sx, 2d − sx 2 2d 2 , we see
that the minimum value of this function is f s2, 2d − 0 and the maximum value is
f s0, 2d − 4. Finally, on L 4 we have x − 0 and
0
D
L¡
3
0 L
FIGURE 13
f(x, f sx, y)=≈-2xy+2y
yd − x 2 2 1 2y
2
f s0, yd − 2y 0 < y < 2
with maximum value f s0, 2d − 4 and minimum value f s0, 0d − 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 s1, 1d − 1 at the critical point and
conclude that the absolute maximum value of f on D is f s3, 0d − 9 and the absolute
minimum value is f s0, 0d − f s2, 2d − 0. Figure 13 shows the graph of f .
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