SECTION
15.8
Lagrange Multipliers: Optimizing with a Constraint
(ET Section 14.8)
765
2
6
−
2
26
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.
Exercises
1.
Use Lagrange multipliers to Fnd the extreme values of the function
f
(
x
,
y
)
=
2
x
+
4
y
subject to the constraint
g
(
x
,
y
)
=
x
2
+
y
2
−
5
=
0.
(a)
Show that the Lagrange equation
∇
f
=
λ
∇
g
gives
x
=
1and
y
=
2.
(b)
Show that these equations imply
6=
0and
y
=
2
x
.
(c)
Use the constraint equation to determine the possible critical points
(
x
,
y
)
.
(d)
Evaluate
f
(
x
,
y
)
at the critical points and determine the minimum and maximum values.
SOLUTION
(a)
The Lagrange equations are determined by the equality
∇
f
=
∇
g
.WeFndthem:
∇
f
=

f
x
,
f
y
®
= h
2
,
4
i
,
∇
g
=

g
x
,
g
y
®
= h
2
x
,
2
y
i
Hence,
h
2
,
4
i =
h
2
x
,
2
y
i
or
(
2
x
)
=
2
(
2
y
)
=
4
⇒
x
=
1
y
=
2
(b)
The Lagrange equations in part (a) imply that
0. The Frst equation implies that
x
=
1
and the second equation
gives
y
=
2
. Therefore
y
=
2
x
.
(c)
We substitute
y
=
2
x
in the constraint equation
x
2
+
y
2
−
5
=
0 and solve for
x
and
y
.Thisgives
x
2
+
(
2
x
)
2
−
5
=
0
5
x
2
=
5
x
2
=
1
⇒
x
1
1
,
x
2
=
1
Since
y
=
2
x
,wehave
y
1
=
2
x
1
2,
y
2
=
2
x
2
=
2. The critical points are thus
(
−
1
,
−
2
)
and
(
1
,
2
).
Extreme values can also occur at the points where
∇
g
= h
2
x
,
2
y
i = h
0
,
0
i
.However
,
(
0
,
0
)
is not on the constraint.
(d)
We evaluate
f
(
x
,
y
)
=
2
x
+
4
y
at the critical points, obtaining
f
(
−
1
,
−
2
)
=
2
·
(
−
1
)
+
4
·
(
−
2
)
10
f
(
1
,
2
)
=
2
·
1
+
4
·
2
=
10
Since
f
is continuous and the graph of
g
=
0 is closed and bounded, global minimum and maximum points exist. So
according to Theorem 1, we conclude that the maximum of
f
(
x
,
y
)
on the constraint is 10 and the minimum is
−
10.
2.
±ind the extreme values of
f
(
x
,
y
)
=
x
2
+
2
y
2
subject to the constraint
g
(
x
,
y
)
=
4
x
−
6
y
=
25.
(a)
Show that the Lagrange equations yield 2
x
=
4
,4
y
6
.
(b)
Show that if
x
=
0or
y
=
0, then
=
0 and the Lagrange equations give
x
=
y
=
0. Since
(
0
,
0
)
does not satisfy
the constraint, you may assume that
x
and
y
are nonzero.