764
C H A P T E R
15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)
We thus showed that if the Fermat point exists, then
A
<
120
◦
. Similarly, one shows also that
B
and
C
must be
smaller than 120
◦
. We conclude that if one of the angles in
ABC
is equal or greater than 120
◦
, then the Fermat point
does not exist. In that case, the minimum value of
f
(
x
,
y
)
occurs at a point where
f
x
or
f
y
do not exist, that is, at one of
the points
A
,
B
, or
C
.
15.8 Lagrange Multipliers: Optimizing with a Constraint
(ET Section 14.8)
Preliminary Questions
1.
Suppose that the maximum of
f
(
x
,
y
)
subject to the constraint
g
(
x
,
y
)
=
0 occurs at a point
P
=
(
a
,
b
)
such that
∇
f
P
=
0. Which of the following are true?
(a)
∇
f
P
is tangent to
g
(
x
,
y
)
=
0 at
P
.
(b)
∇
f
P
is orthogonal to
g
(
x
,
y
)
=
0 at
P
.
SOLUTION
(a)
Since the maximum of
f
subject to the constraint occurs at
P
, it follows by Theorem 1 that
∇
f
P
and
∇
g
P
are
parallel vectors. The gradient
∇
g
P
is orthogonal to
g
(
x
,
y
)
=
0 at
P
, hence
∇
f
P
is also orthogonal to this curve at
P
.
We conclude that statement (b) is false (yet the statement can be true if
∇
f
P
=
(
0
,
0
)
).
(b)
This statement is true by the reasoning given in the previous part.
2.
Figure 8 shows a constraint
g
(
x
,
y
)
=
0 and the level curves of a function
f
. In each case, determine whether
f
has
a local minimum, local maximum, or neither at the labeled point.
4
3
2
1
1
2
3
4
A
B
g
(
x
,
y
)
=
0
g
(
x
,
y
)
=
0
∇
f
∇
f
FIGURE 8
SOLUTION
The level curve
f
(
x
,
y
)
=
2 is tangent to the constraint curve at the point
A
. A close level curve that
intersects the constraint curve is
f
(
x
,
y
)
=
1, hence we may assume that
f
has a local maximum 2 under the constraint
at
A
. The level curve
f
(
x
,
y
)
=
3 is tangent to the constraint curve. However, in approaching
B
under the constraint,
from one side
f
is increasing and from the other side
f
is decreasing. Therefore,
f
(
B
)
is neither local minimum nor
local maximum of
f
under the constraint.
3.
On the contour map in Figure 9:
(a)
Identify the points where
∇
f
=
λ
∇
g
for some scalar
λ
.
(b)
Identify the minimum and maximum values of
f
(
x
,
y
)
subject to
g
(
x
,
y
)
=
0.
2
6
−
2
2
6
Graph of
g
(
x
,
y
)
=
0
Contour plot of
f
(
x
,
y
)
(contour interval 2)
−
2
−
4
−
6
y
x
FIGURE 9
Contour map of
f
(
x
,
y
)
; contour interval 2.
SOLUTION
(a)
The gradient
∇
g
is orthogonal to the constraint curve
g
(
x
,
y
)
=
0, and
∇
f
is orthogonal to the level curves of
f
.
These two vectors are parallel at the points where the level curve of
f
is tangent to the constraint curve. These are the
points
A
,
B
,
C
,
D
,
E
in the figure:
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S E C T I O N
15.8
Lagrange Multipliers: Optimizing with a Constraint
(ET Section 14.8)
765
2
6
−
2
2
6
g
(
x
,
y
)
=
0
∇
f
A
,
∇
g
A
A
E
C
D
B
−
2
−
6
−
6
(b)
The minimum and maximum occur where the level curve of
f
is tangent to the constraint curve. The level curves
tangent to the constraint curve are
f
(
A
)
= −
4
,
f
(
C
)
=
2
,
f
(
B
)
=
6
,
f
(
D
)
= −
4
,
f
(
E
)
=
4
Therefore the global minimum of
f
under the constraint is
−
4 and the global maximum is 6.
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 Winter '08
 GANGliu
 Critical Point, Optimization, lagrange equations, Extreme value, Constraint

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