λw
+
ν
Note that neither
c
= 0 or
r
= 0 allows a solution.
Thus
μ
=
ν
= 0 by complementary slackness.
Moreover,
r
=
T
implies
c
= 0, so it is also impossible.
Thus
ρ
= 0 by complementary slackness.
However,
λ
= 1
/pc >
0, so
pc
+
rw
=
wT
is the only binding constraint.
Constraint qualification is clearly satisfied with this one constraint. We solve the firstorder equa
tions, obtaining
c
=
2
wT
3
p
and
r
=
T
3
.
Finally, we need only look at the determinant of the bordered Hessian because there is one constraint
and two unknowns. The bordered Hessian is
H
=
0
p
w
p

c

2
0
w
0

2
r

2
.
Its determinant is 2(
p/r
)
2
+4(
cw
)
2
>
0. Since
n
= 2, det
H
(

1)
2
>
0, which implies we have a maximum.
5. Let
A
be an
n
×
n
positive definite matrix. Define
N
(
x
) =
√
x
T
Ax
. Is
N
a norm? That is, does it obey the
three conditions a norm must obey?
Answer:
It is a norm.
1) It is absolutely homogeneous of degree 1 by construction.
2) It is clearly
nonnegative, and if
N
(
x
) = 0, the quadratic form
x
T
Ax
= 0. Since this form is positive definite,
x
= 0.
Finally, the triangle inequality must be shown.
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 Spring '08
 STAFF
 Economics, Critical Point, Derivative, Optimization, Continuous function, λ

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