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

Section 14.7 Maximum and Minimum Values 961
FIGURE 33
z=¥-≈
− y 2 2 x 2
7et140703
05/05/10
MasterID: 01621
Photo by Stan Wagon, Macalester College
x
z
y
Example 2 illustrates the fact that a function need not have a maximum or minimum
value at a critical point. Figure 3 shows how this is possible. The graph of f is the hyperbolic
paraboloid z − y 2 2 x 2 , which has a horizontal tangent plane (z − 0) at the origin.
You can see that f s0, 0d − 0 is a maximum in the direction of the x-axis but a minimum
in the direction of the y-axis. Near the origin the graph has the shape of a saddle and so
s0, 0d is called a saddle point of f.
A mountain pass also has the shape of a saddle. As the photograph of the geological
formation illustrates, for people hiking in one direction the saddle point is the lowest
point on their route, while for those traveling in a different direction the saddle point is
the highest point.
We need to be able to determine whether or not a function has an extreme value at a
crit ical point. The following test, which is proved at the end of this section, is analogous
to the Second Derivative Test for functions of one variable.
3 Second Derivatives Test Suppose the second partial derivatives of f
are con tinuous on a disk with center sa, bd, and suppose that f x sa, bd − 0 and
f y sa, bd − 0 [that is, sa, bd is a critical point of f ]. Let
D − Dsa, bd − f xx sa, bd f yy sa, bd 2 f f x y sa, bdg 2
(a) If D . 0 and f xx sa, bd . 0, then f sa, bd is a local minimum.
(b) If D . 0 and f xx sa, bd , 0, then f sa, bd is a local maximum.
(c) If D , 0, then f sa, bd is not a local maximum or minimum.
Note 1 In case (c) the point sa, bd is called a saddle point of f and the graph of f
crosses its tangent plane at sa, bd.
Note 2 If D − 0, the test gives no information: f could have a local maximum or
local minimum at sa, bd, or sa, bd could be a saddle point of f.
Note 3 To remember the formula for D, it’s helpful to write it as a determinant:
D − Z
f xx
f yx
f x y
f yy
Z − f xx f yy 2 s f x y d 2
EXAMPLE 3 Find the local maximum and minimum values and saddle points of
f sx, yd − x 4 1 y 4 2 4xy 1 1.
SOLUtion We first locate the critical points:
f x − 4x 3 2 4y
f y − 4y 3 2 4x
Setting these partial derivatives equal to 0, we obtain the equations
x 3 2 y − 0 and y 3 2 x − 0
To solve these equations we substitute y − x 3 from the first equation into the second
one. This gives
0 − x 9 2 x − xsx 8 2 1d − xsx 4 2 1dsx 4 1 1d − xsx 2 2 1dsx 2 1 1dsx 4 1 1d
so there are three real roots: x − 0, 1, 21. The three critical points are s0, 0d, s1, 1d,
and s21, 21d.
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