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786
CHAPTER 15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)
27.
Let
Q
be the point on an ellipse closest to a given point
P
outside the ellipse. It was known to the Greek
mathematician Apollonius (third century BCE)that
PQ
is perpendicular to the tangent at
Q
(Figure 13). Explain in words
why this conclusion is a consequence of the method of Lagrange multipliers.
Hint:
The circles centered at
P
are level
curves of the function to be minimized.
P
Q
FIGURE 13
SOLUTION
Let
P
=
(
x
0
,
y
0
)
. The distance
d
between the point
P
and a point
Q
=
(
x
,
y
)
on the ellipse is minimum
where the square
d
2
is minimum (since the square function
u
2
is increasing for
u
≥
0). Therefore, we want to minimize
the function
f
(
x
,
y
,
z
)
=
(
x
−
x
0
)
2
+
(
y
−
y
0
)
2
+
(
z
−
z
0
)
2
subject to the constraint
g
(
x
,
y
)
=
x
2
a
2
+
y
2
b
2
=
1
The method of Lagrange indicates that the solution
Q
is the point on the ellipse where
∇
f
=
λ
∇
g
, that is, the point
on the ellipse where the gradients
∇
f
and
∇
g
are parallel. Since the gradient is orthogonal to the level curves of the
function,
∇
g
is orthogonal to the ellipse
g
(
x
,
y
)
=
1, and
∇
f
is orthogonal to the level curve of
f
passing through
Q
.
But this level curve is a circle through
Q
centered at
P
, hence the parallel vectors
∇
g
and
∇
f
are orthogonal to the
ellipse and to the circle centered at
P
respectively. We conclude that the point
Q
is the point at which the tangent to the
ellipse is also the tangent to the circle through
Q
centered at
P
. That is, the tangent to the ellipse at
Q
is perpendicular
to the radius
of the circle.
28.
Antonio has $5.00 to spend on a lunch consisting of hamburgers ($1.50 each) and French fries ($1.00 per order).
Antonio’s satisfaction from eating
x
1
hamburgers and
x
2
orders of French fries is measured by a function
U
(
x
1
,
x
2
)
=
√
x
1
x
2
. How much of each type of food should he purchase to maximize his satisfaction (assume that fractional amounts
of each food can be purchased)?
Antonio has $5.00 to spend on the lunch, hence the total cost 1
.
5
x
1
+
x
2
must satisfy
1
.
5
x
1
+
x
2
=
5
We thus want to maximize the function
U
(
x
1
,
x
2
)
=
√
x
1
x
2
subject to the constraint
g
(
x
,
y
)
=
1
.
5
x
1
+
x
2
=
5 with
x
1
>
0,
x
2
>
0.
Step 1.
Write out the Lagrange Equations. The gradient vectors are
∇
U
=
1
2
D q
x
2
x
1
,
q
x
1
x
2
E
and
∇
g
= h
1
.
5
,
1
i
, hence the
Lagrange Condition
∇
U
=
∇
g
gives the following equations:
1
2
±
x
2
x
1
=
1
.
5
1
2
±
x
1
x
2
=
⇒
x
2
x
1
=
9
2
x
1
x
2
=
4
2
Step 2.
Solve for
x
1
and
x
2
using the constraint. The two equations in step 1 give
2
=
x
2
9
x
1
=
x
1
4
x
2
Therefore,
4
x
2
2
=
9
x
2
1
x
2
2
=
9
4
x
2
1
⇒
x
2
=
3
2
x
1
We now substitute
x
2
=
3
2
x
1
in the constraint 1
.
5
x
1
+
x
2
=
5 and solve for
x
1
.Weget
1
.
5
x
1
+
3
2
x
1
=
5
3
x
1
=
5
⇒
x
1
=
5
3
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View Full DocumentSECTION
15.8
Lagrange Multipliers: Optimizing with a Constraint
(ET Section 14.8)
787
We fnd
x
2
by the relation
x
2
=
3
2
x
1
:
x
2
=
3
2
·
5
3
=
5
2
We obtain the critical point:
µ
5
3
,
5
2
¶
Step 3.
Conclusions. We conclude that Antonio should have
5
3
hamburgers and
5
2
orders oF Fries, to maximize his sat
isFaction. Notice that
U
(
x
1
,
x
2
)
=
√
x
1
x
2
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 Winter '08
 GANGliu
 Math

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