780
C H A P T E R
15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)
20.
Show that the point
(
x
0
,
y
0
)
closest to the origin on the line
ax
+
by
=
c
has coordinates
x
0
=
ac
a
2
+
b
2
,
y
0
=
bc
a
2
+
b
2
SOLUTION
We need to minimize the distance
d
(
x
,
y
)
=
x
2
+
y
2
subject to the constraint
g
(
x
,
y
)
=
ax
+
by
=
c
.
Notice that the distance
d
(
x
,
y
)
is at a minimum at the same points where the square of the distance
d
2
(
x
,
y
)
is at a
minimum (since the function
u
2
is increasing for
u
≥
0). Therefore, we may find the minimum of
f
(
x
,
y
)
=
x
2
+
y
2
subject to the constraint
ax
+
by
=
c
.
Step 1.
Write out the Lagrange Equations. The gradient vectors are
∇
f
=
2
x
,
2
y
and
∇
g
=
a
,
b
, hence the Lagrange
Condition
∇
f
=
λ
∇
g
is
2
x
,
2
y
=
λ
a
,
b
or
2
x
=
λ
a
2
y
=
λ
b
Step 2.
Solve for
λ
in terms of
x
and
y
. The Lagrange equations give
λ
=
2
x
a
and
λ
=
2
y
b
Step 3.
Solve for
x
and
y
using the constraint. We equate the two expressions for
λ
and solve for
y
in terms of
x
:
2
x
a
=
2
y
b
⇒
y
=
b
a
x
We now substitute
y
=
bx
a
in the equation of the constraint
ax
+
by
=
c
and solve for
x
:
ax
+
b
·
b
a
x
=
c
a
+
b
2
a
x
=
c
a
2
+
b
2
a
x
=
c
⇒
x
=
ac
a
2
+
b
2
We find
y
using the relation
y
=
bx
a
:
y
=
b
a
·
ac
a
2
+
b
2
=
bc
a
2
+
b
2
The critical point is thus
x
0
=
ac
a
2
+
b
2
,
y
0
=
bc
a
2
+
b
2
(1)
Step 4.
Conclusions. It is clear geometrically that the problem has a minimum value and it does not have a maximum
value. Therefore the minimum occurs at the critical point. We conclude that the point closest to the origin on the line
ax
+
by
=
c
is given by (1). To show that the vector
x
0
,
y
0
is perpendicular to the line, we write the line in vector form
as
x
−
x
0
,
y
−
y
0
·
a
,
b
=
0. Thus,
a
,
b
is perpendicular to the line. Since
x
0
,
y
0
=
c
a
2
+
b
2
a
,
b
, then
x
0
,
y
0
is
parallel to
a
,
b
, and thus also perpendicular to the line.
21.
Find the maximum value of
f
(
x
,
y
)
=
x
a
y
b
for
x
,
y
≥
0 on the unit circle, where
a
,
b
>
0 are constants.
SOLUTION
We must find the maximum value of
f
(
x
,
y
)
=
x
a
y
b
(
a
,
b
>
0) subject to the constraint
g
(
x
,
y
)
=
x
2
+
y
2
=
1.
Step 1.
Write out the Lagrange Equations. We have
∇
f
=
ax
a
−
1
y
b
,
bx
a
y
b
−
1
and
∇
g
=
2
x
,
2
y
. Therefore the
Lagrange Condition
∇
f
=
λ
∇
g
is
ax
a
−
1
y
b
,
bx
a
y
b
−
1
=
λ
2
x
,
2
y
or
ax
a
−
1
y
b
=
2
λ
x
bx
a
y
b
−
1
=
2
λ
y
(1)
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S E C T I O N
15.8
Lagrange Multipliers: Optimizing with a Constraint
(ET Section 14.8)
781
Step 2.
Solve for
λ
in terms of
x
and
y
. If
x
=
0 or
y
=
0,
f
has the minimum value 0. We thus may assume that
x
>
0
and
y
>
0. The equations (1) imply that
λ
=
ax
a
−
2
y
b
2
,
λ
=
bx
a
y
b
−
2
2
Step 3.
Solve for
x
and
y
using the constraint. Equating the two expressions for
λ
and solving for
y
in terms of
x
gives
ax
a
−
2
y
b
2
=
bx
a
y
b
−
2
2
ax
a
−
2
y
b
=
bx
a
y
b
−
2
ay
2
=
bx
2
y
2
=
b
a
x
2
⇒
y
=
b
a
x
We now substitute
y
=
b
a
x
in the constraint
x
2
+
y
2
=
1 and solve for
x
>
0. We obtain
x
2
+
b
a
x
2
=
1
(
a
+
b
)
x
2
=
a
x
2
=
a
a
+
b
⇒
x
=
a
a
+
b
We find
y
using the relation
y
=
b
a
x
:
y
=
b
a
a
a
+
b
=
ab
a
(
a
+
b
)
=
b
a
+
b
We obtain the critical point:
a
a
+
b
,
b
a
+
b
Extreme points can also occur where
∇
g
=
0
, that is, 2
x
,
2
y
=
0
,
0 or
(
x
,
y
)
=
(
0
,
0
)
. However, this point is not on
the constraint.
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 Winter '08
 GANGliu
 Critical Point, Optimization, lagrange equations

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