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

962 Chapter 14 Partial Derivatives
z
Next we calculate the second partial derivatives and Dsx, yd:
x
FIGURE 44
z=x$+y$-4xy+1
− x 4 1 y 4 2 1 1
y
f xx − 12x 2 f x y − 24 f yy − 12y 2
Dsx, yd − f xx f yy 2 s f x y d 2 − 144x 2 y 2 2 16
Since Ds0, 0d − 216 , 0, it follows from case (c) of the Second Derivatives Test that
the origin is a saddle point; that is, f has no local maximum or minimum at s0, 0d.
Since Ds1, 1d − 128 . 0 and f xx s1, 1d − 12 . 0, we see from case (a) of the test that
f s1, 1d − 21 is a local minimum. Similarly, we have Ds21, 21d − 128 . 0 and
f xx s21, 21d − 12 . 0, so f s21, 21d − 21 is also a local minimum.
The graph of f is shown in Figure 4.
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MasterID: 01622
A contour map of the function f in
Example 3 is shown in Figure 5. The
level curves near s1, 1d and s21, 21d
are oval in shape and indicate that as
we move away from s1, 1d or s21, 21d
in any direction the values of f are
increasing. The level curves near s0, 0d,
on the other hand, resemble hyper bolas.
They reveal that as we move away from
the origin (where the value of f is 1), the
values of f decrease in some directions
but increase in other directions. Thus
the contour map suggests the presence
of the minima and saddle point that we
found in Example 3.
y
_0.5
0
0.5
1 0.9 1.1
1.5
2
3
x
FIGURE 5
TEC In Module 14.7 you can use
contour maps to estimate the locations
of critical points.
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EXAMPLE 4 Find and classify the critical points of the function
MasterID: 01623 f sx, yd − 10x 2 y 2 5x 2 2 4y 2 2 x 4 2 2y 4
Also find the highest point on the graph of f.
SOLUtion The first-order partial derivatives are
f x − 20xy 2 10x 2 4x 3 f y − 10x 2 2 8y 2 8y 3
So to find the critical points we need to solve the equations
4 2xs10y 2 5 2 2x 2 d − 0
5 5x 2 2 4y 2 4y 3 − 0
From Equation 4 we see that either
x − 0 or 10y 2 5 2 2x 2 − 0
In the first case (x − 0), Equation 5 becomes 24ys1 1 y 2 d − 0, so y − 0 and we
have the critical point s0, 0d.
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