Now if
μ
0
>
1,
μ
1
=
μ
0

1
>
0, so
x
= 0 by complementary slackness.
It follows that
py
= 10
(
y
= 10
/p
) by the budget constraint, and then 10 =
py
= 1
/
4
pμ
2
0
, so
μ
0
= 1
/
√
40
p
. This can only be a
solution if 1
/
√
40
p <
1, which is equivalent to 1
>
40
p
.
But if
μ
0
= 1,
μ
1
= 1

μ
0
= 0. The complementary slackness condition for
x
is satisfied, so
x
≥
0.
Then
y
= 1
/
4
p
2
, and
x
= 10

1
/
4
p
≥
0. This requires 40
p
≥
1.
Summing up, if 0
< p <
1
40
, (
x, y
) = (0
,
10) is the only maximum. If
p
=
1
40
, (
x, y
) = (0
,
10) is still
the only maximum. Finally, if
p >
1
40
, the maximum is (10

1
4
p
,
1
4
p
2
).
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MATHEMATICAL ECONOMICS FINAL, DECEMBER 11, 2000
Page 2
4. Find and classify all critical points of the function
f
(
x, y
) = (
x
2
+
y
2
)
e

x
2
.
Answer:
The firstorder conditions are [2
x

2
x
(
x
2
+
y
2
)]
e

x
2
and 2
ye

x
2
= 0. It follows that
y
= 0
and so 2
x

2
x
3
= 0. The critical points are (1
,
0), (0
,
0), and (

1
,
0).
The Hessian is:
[(2

6
x
2

2
y
2
)

2
x
(2
x

2
x
3

2
xy
2
)]
e

x
2

4
xye

x
2

4
xye

x
2
2
e

x
2
This yields:
H
(0
,
0) =
2
0
0
2
H
(1
,
0) =
H
(

1
,
0) =

4
e

1
0
0
2
e

1
.
It follows that
H
(0
,
0) is positive definite because
H
1
= 2
>
0 and
H
2
= 4
>
0. Thus (0
,
0) is a local
minimum (in fact, it is global). Moreover,
H
(1
,
0) =
H
(

1
,
0) is indefinite because
H
2
=

8
e

1
<
0.
Thus (1
,
0) and (

1
,
0) are saddlepoints. The function has no maximum.
5. Consider the matrix
A
=
3
2
1
3
1
0
3
3
1
a
) What is the rank of
A
?
Answer:
The rank of
A
is 3.
One way to see this is to compute the determinant.
It is
(3)(1)(1) + (3)(3)(1) + (3)(2)(0)

(3)(1)(1)

(3)(2)(1)

(3)(3)(0) = 3 + 9

3

6 = 3. Since it
is nonzero, the matrix is invertible. As it has 3 rows, it has rank 3.
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