S E C T I O N
15.7
Optimization in Several Variables
(ET Section 14.7)
729
Point
S
in Figure (B) is neither a local extremum nor a saddle point of
f
.
1
3
0
−
1
−
3
S
Figure (B)
3.
Let
f
(
x
,
y
)
be a continuous function on a domain
D
in
R
2
. Determine which of the following statements are true:
(a)
If
D
is closed and bounded, then
f
takes on a maximum value on
D
.
(b)
If
D
is neither closed nor bounded, then
f
does not take on a maximum value of
D
.
(c)
f
(
x
,
y
)
need not have a maximum value on
D
= {
(
x
,
y
)
:
0
≤
x
,
y
≤
1
}
.
(d)
A continuous function takes on neither a minimum nor a maximum value on the open quadrant
{
(
x
,
y
)
:
x
>
0
,
y
>
0
}
.
SOLUTION
(a)
This statement is true. It follows by the Theorem on Existence of Global Extrema.
(b)
The statement is false. Consider the constant function
f
(
x
,
y
)
=
2 in the following domain:
x
1
y
D
= {
(
x
,
y
)
:
0
<
x
≤
1
,
0
≤
y
<
∞}
Obviously
f
is continuous and
D
is neither closed nor bounded. However,
f
takes on a maximum value (which is 2) on
D
.
(c)
The domain
D
= {
(
x
,
y
)
:
0
≤
x
,
y
≤
1
}
is the following rectangle:
x
1
y
1
D
= {
(
x
,
y
)
:
0
≤
x
,
y
≤
1
}
D
is closed and bounded, hence
f
takes on a maximum value on
D
. Thus the statement is false.
(d)
The statement is false. The constant function
f
(
x
,
y
)
=
c
takes on minimum and maximum values on the open
quadrant.
Exercises
1.
Let
P
=
(
a
,
b
)
be a critical point of
f
(
x
,
y
)
=
x
2
+
y
4
−
4
xy
.
(a)
First use
f
x
(,
x
,
y
)
=
0 to show that
a
=
2
b
. Then use
f
y
(
x
,
y
)
=
0 to show that
P
=
(
0
,
0
)
,
(
2
√
2
,
√
2
)
, or
(
−
2
√
2
,
−
√
2
)
.
(b)
Referring to Figure 15, determine the local minima and saddle points of
f
(
x
,
y
)
and find the absolute minimum
value of
f
(
x
,
y
)
.
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730
C H A P T E R
15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)
x
y
z
FIGURE 15
SOLUTION
(a)
We find the partial derivatives:
f
x
(
x
,
y
)
=
∂
∂
x
x
2
+
y
4
−
4
xy
=
2
x
−
4
y
f
y
(
x
,
y
)
=
∂
∂
y
x
2
+
y
4
−
4
xy
=
4
y
3
−
4
x
Since
P
=
(
a
,
b
)
is a critical point,
f
x
(
a
,
b
)
=
0. That is,
2
a
−
4
b
=
0
⇒
a
=
2
b
Also
f
y
(
a
,
b
)
=
0, hence,
4
b
3
−
4
a
=
0
⇒
a
=
b
3
We obtain the following equations for the critical points
(
a
,
b
)
:
a
=
2
b
a
=
b
3
Equating the two equations, we get
2
b
=
b
3
b
3
−
2
b
=
b
(
b
2
−
2
)
=
0
⇒
⎧
⎨
⎩
b
1
=
0
b
2
=
√
2
b
3
= −
√
2
Since
a
=
2
b
, we have
a
1
=
0,
a
2
=
2
√
2,
a
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 Fall '09
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 Critical Point, Multivariable Calculus, Optimization, S E V E R A L VA R

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