This preview shows pages 1–3. Sign up to view the full content.
econ/ams 11b
winter 2010
Midterm 2 Review — Solutions
1.
Find the critical point(s) of the functions below subject to the given constraint.
a.
h
(
u,v
) = 5
u
+ 8
v
;
2
u
2
+ 3
v
2
= 30.
The Lagrangian in this case is
F
(
u,v,λ
) = 5
u
+ 8
v

λ
(2
u
2
+ 3
v
2

30)
,
and the ﬁrst order equations are
F
u
= 0 =
⇒
5

4
λu
= 0
F
v
= 0 =
⇒
8

6
λv
= 0
F
λ
= 0 =
⇒
2
u
2
+ 3
v
2
= 30
Solving the ﬁrst two equations for
λ
(and cross multiplying) gives
λ
=
5
4
u
=
4
3
v
=
⇒
15
v
= 16
u
=
⇒
v
=
16
u
15
.
Finally, substituting the last expression for
v
into the third equation (the constraint)
gives
2
u
2
+ 3
±
16
u
15
²
2
= 30 =
⇒
450
u
2
+ 768
u
2
= 6750 =
⇒
u
=
±
r
1125
203
.
Thus, there are two critical points:
(
u
*
,v
*
,λ
*
) =
±
r
1125
203
,
r
1280
203
,
r
203
720
!
≈ ±
(2
.
345
,
2
.
511
,
0
.
531)
.
b.
k
(
x,y,z
) =
xy
2
z
3
;
5
x
+ 8
y
+ 12
z
= 300.
Comment:
To simplify the analysis, we will assume the additional condition
xyz
6
= 0.
I.e., none of the variables
x
,
y
or
z
can be zero.
The Lagrangian in this case is
F
(
x,y,z,,λ
) =
xy
2
z
3

λ
(5
x
+ 8
y
+ 12
z

300)
and the ﬁrst order equations are
F
x
= 0 =
⇒
y
2
z
3

5
λ
= 0
F
y
= 0 =
⇒
2
xyz
3

8
λ
= 0
F
z
= 0 =
⇒
3
xy
2
z
2

12
λ
= 0
F
λ
= 0 =
⇒
5
x
+ 8
y
+ 12
z
= 300
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document Solving the ﬁrst three equations for
λ
gives
λ
=
y
2
z
3
5
=
xyz
3
4
=
xy
2
z
2
4
.
Comparing the ﬁrst two expressions above, multiplying by 20 and dividing by
yz
3
, we
ﬁnd that
4
y
= 5
x
=
⇒
y
=
5
x
4
.
Comparing the ﬁrst and third expressions above, multiplying by 20 and dividing by
y
2
z
2
, we ﬁnd that
4
z
= 5
x
=
⇒
z
=
5
x
4
.
Substituting these expressions for
y
and
z
into the constraint gives
5
x
+ 8
5
x
4
+ 12
5
x
4
= 300 =
⇒
30
x
= 300 =
⇒
x
= 10
.
Thus, (excluding the points where
x,y
or
z
are 0), there is one critical point:
(
x
*
,y
*
,z
*
,λ
*
) = (10
,
12
.
5
,
12
.
5
,
61035
.
15625)
.
Comment:
(continued) There are
inﬁnitely many
other critical points for this La
grangian, where one or two of
x
,
y
or
z
are equal to 0. However, at all of these critical
points, the objective function
k
(
x,y,z
) =
xy
2
z
3
is identically 0 (why?), so these critical
points are not very interesting.
2.
A household’s utility function is given by
U
(
q
A
,q
B
,q
C
) = 5 ln
q
A
+12ln
q
B
+18ln
q
C
, where
q
A
,
q
B
and
q
C
are the quantities of goods of type
A
,
B
and
C
that the household consumes
per year. The average prices per unit of these goods are
p
A
= 10,
p
B
= 15 and
p
C
= 25,
respectively, and the household’s disposable income is
Y
d
= $60
,
000.
a.
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 03/15/2010 for the course ECON 11 taught by Professor Yk during the Spring '10 term at University of California, Santa Cruz.
 Spring '10
 yk

Click to edit the document details