780
CHAPTER 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
)
=
p
x
2
+
y
2
subject to the constraint
g
(
x
,
y
)
=
+
=
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 Fnd the minimum of
f
(
x
,
y
)
=
x
2
+
y
2
subject to the constraint
+
=
c
.
Step 1.
Write out the Lagrange Equations. The gradient vectors are
∇
f
= h
2
x
,
2
y
i
and
∇
g
= h
a
,
b
i
, hence the Lagrange
Condition
∇
f
=
λ
∇
g
is
h
2
x
,
2
y
i =
h
a
,
b
i
or
2
x
=
a
2
y
=
b
Step 2.
Solve for
in terms of
x
and
y
. The Lagrange equations give
=
2
x
a
and
=
2
y
b
Step 3.
Solve for
x
and
y
using the constraint. We equate the two expressions for
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
+
=
c
and solve for
x
:
+
b
·
b
a
x
=
c
Ã
a
+
b
2
a
!
x
=
c
a
2
+
b
2
a
x
=
c
⇒
x
=
a
2
+
b
2
We Fnd
y
using the relation
y
=
a
:
y
=
b
a
·
a
2
+
b
2
=
bc
a
2
+
b
2
The critical point is thus
x
0
=
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
+
=
c
is given by (1). To show that the vector
h
x
0
,
y
0
i
is perpendicular to the line, we write the line in vector form
as
h
x
−
x
0
,
y
−
y
0
i·h
a
,
b
i=
0. Thus,
h
a
,
b
i
is perpendicular to the line. Since
h
x
0
,
y
0
c
a
2
+
b
2
h
a
,
b
i
,then
h
x
0
,
y
0
i
is
parallel to
h
a
,
b
i
, and thus also perpendicular to the line.
21.
±ind 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.
We must Fnd 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
=
D
a
−
1
y
b
,
a
y
b
−
1
E
and
∇
g
= h
2
x
,
2
y
i
. Therefore the
Lagrange Condition
∇
f
=
∇
g
is
D
a
−
1
y
b
,
a
y
b
−
1
E
=
h
2
x
,
2
y
i
or
a
−
1
y
b
=
2
x
a
y
b
−
1
=
2
y
(1)