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

Section 14.7 Maximum and Minimum Values 963
In the second case s10y 2 5 2 2x 2 − 0d, we get
6 x 2 − 5y 2 2.5
and, putting this in Equation 5, we have 25y 2 12.5 2 4y 2 4y 3 − 0. So we have to
solve the cubic equation
7 4y 3 2 21y 1 12.5 − 0
Using a graphing calculator or computer to graph the function
tsyd − 4y 3 2 21y 1 12.5
_3 2.7
as in Figure 6, we see that Equation 7 has three real roots. By zooming in, we can find
the roots to four decimal places:
y < 22.5452 y < 0.6468 y < 1.8984
FIGURE 66
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(Alternatively, we could have used Newton’s method or solved numerically using a
calculator or computer to locate these roots.) From Equation 6, the corresponding
x-values are given by
x − 6s5y 2 2.5
If y < 22.5452, then x has no corresponding real values. If y < 0.6468, then
x < 60.8567. If y < 1.8984, then x < 62.6442. So we have a total of five critical
points, which are analyzed in the following chart. All quantities are rounded to two
decimal places.
Critical point Value of f f xx D Conclusion
s0, 0d 0.00 210.00 80.00 local maximum
s62.64, 1.90d 8.50 255.93 2488.72 local maximum
s60.86, 0.65d 21.48 25.87 2187.64 saddle point
Figures 7 and 8 give two views of the graph of f and we see that the surface opens
downward. [This can also be seen from the expression for f sx, yd: the dominant terms
are 2x 4 2 2y 4 when | x | and | y | are large.] Comparing the values of f at its local maximum
points, we see that the absolute maximum value of f is f s62.64, 1.90d < 8.50. In
other words, the highest points on the graph of f are s62.64, 1.90, 8.50d.
z
z
TEC Visual 14.7 shows several
families of surfaces. The surface in
Figures 7 and 8 is a member of one
of these families.
x
y
x
y
FIGURE 77 FIGURE 88 ■
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