744
CHAPTER 15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)
23.
Use a computer algebra system to fnd numerical approximations to the critical points oF
f
(
x
,
y
)
=
(
1
−
x
+
x
2
)
e
y
2
+
(
1
−
y
+
y
2
)
e
x
2
Use ±igure 18 to determine whether they correspond to local minima or maxima.
z
x
y
FIGURE 18
Plot oF the Function
f
(
x
,
y
)
=
(
1
−
x
+
x
2
)
e
y
2
+
(
1
−
y
+
y
2
)
e
x
2
.
SOLUTION
The critical points are the solutions oF
f
x
(
x
,
y
)
=
0and
f
y
(
x
,
y
)
=
0. We compute the partial derivatives:
f
x
(
x
,
y
)
=
(
−
1
+
2
x
)
e
y
2
+
³
1
−
y
+
y
2
´
e
x
2
·
2
x
f
y
(
x
,
y
)
=
³
1
−
x
+
x
2
´
e
y
2
·
2
y
+
(
−
1
+
2
y
)
e
x
2
Hence, the critical points are the solutions oF the Following equations:
(
2
x
−
1
)
e
y
2
+
2
x
³
1
−
y
+
y
2
´
e
x
2
=
0
(
2
y
−
1
)
e
x
2
+
2
y
³
1
−
x
+
x
2
´
e
y
2
=
0
Using a CAS we obtain the Following solution:
x
=
y
=
0
.
27788, which From the fgure is a local minimum.
24.
Use the contour map in ±igure 19 to determine whether the critical points
A
,
B
,
C
,
D
are local minima, maxima, or
saddle points.
1
1
0
0
2
3
−
1
−
1
−
2
−
3
3
2
1
0
−
3
−
2
−
1
3
1
02
−
3
−
1
−
2
A
C
D
B
FIGURE 19
The nearby level curves at
A
and
C
are closed curves encircling
A
and
C
.Aswemovetowa
rds
A
the
Function increases in all directions, while moving towards
C
the Function decreases in all directions. We conclude that
the Function has a local maximum at
A
and a local minimum at
C
. The level curves through
B
and
D
consist oF two
curves intersecting at these points respectively. These curves divide the neighborhoods near
B
and
D
into Four regions.
In some oF the regions the Function is increasing and in others it is decreasing as we move towards
B
or
D
. This implies
that
B
and
D
are saddle points.
25.
Which oF the Following domains are closed and which are bounded?
(a)
{
(
x
,
y
)
∈
R
2
:
x
2
+
y
2
≤
1
}
(b)
{
(
x
,
y
)
∈
R
2
:
x
2
+
y
2
<
1
}
(c)
{
(
x
,
y
)
∈
R
2
:
x
≥
0
}
(d)
{
(
x
,
y
)
∈
R
2
:
x
>
0
,
y
>
0
}
(e)
{
(
x
,
y
)
∈
R
2
:
1
≤
x
≤
4
,
5
≤
y
≤
10
}
(f)
{
(
x
,
y
)
∈
R
2
:
x
>
0
,
x
2
+
y
2
≤
10
}
(a)
{
(
x
,
y
)
∈
R
2
:
x
2
+
y
2
≤
1
}
: This domain is bounded since it is contained, For instance, in the disk
x
2
+
y
2
<
2.
The domain is also closed since it contains all oF its boundary points, which are the points on the unit circle
x
2
+
y
2
=
1.