Problem Set 4: Equality-Constrained Maximization
1. Consider the maximization problem
x
1
+
x
2
1.
The Lagrangian is
L
=
x
1
+
x
2
−
λ
(
x
2
1
+
x
2
2
−
1)
with FONCs
1
−
λ
2
x
∗
1
= 0
1
−
λ
2
x
∗
2
= 0
−
(
x
∗
1
2
+
x
∗
2
2
−
1) = 0
The first two equations imply that
x
∗
1
=
x
∗
2
. The third equation implies that
x
∗
1
2
+
x
∗
2
2
= 1, or
x
∗
k
=
±
radicalbigg
1
2
So there are two potential candidates,
parenleftBigg
radicalbigg
1
2
,
radicalbigg
1
2
parenrightBigg
,
parenleftBigg
−
radicalbigg
1
2
,
−
radicalbigg
1
2
parenrightBigg
,
The Bordered Hessian is
0
−
2
x
1
−
2
x
2
−
2
x
1
−
2
λ
0
−
2
x
2
0
−
2
λ
Since
λ
appears in it, we’ll need to compute it as well, which is different from many examples
we saw in class. Also different is that the
x
i
’s appear along the top and left-hand borders. The
determinant of the third principal minor of the Bordered Hessian is
2
x
1
(4
x
1
λ
) + 2
x
2
(4
λx
2
) = 8
λ
(
x
2
1
+
x
2
2
)
So that the sign only depends on the multiplier,
λ
. Since
λ
=
1
2
x
∗
1
=
1
2
x
∗
2
it will be positive if both terms are positive, and negative if both terms are negative.
Consequently, the critical point
parenleftBigg
radicalbigg
1
2
,
radicalbigg
1
2
parenrightBigg
1

is a maximum while
parenleftBigg
−
radicalbigg
1
2
,
−
radicalbigg
1
2
parenrightBigg
is actually a minimum.
Since the two critical points where
x
∗
and
y
∗
have the same sign are local maxima and yield
the same value of the objective function, they are both global maxima.
2. Consider the maximization problem
radicalbigg
r
a
radicalbigg
r
b
radicalbigg
r
b
parenrightbigg
,
parenleftbiggradicalbigg
r
a
,
−
radicalbigg
r
b
parenrightbigg
,
parenleftbigg
−
radicalbigg
r
a
,
radicalbigg
r
b
parenrightbigg
,
parenleftbigg
−
radicalbigg
r
a
,
−
radicalbigg
r
b
parenrightbigg
,

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- Fall '12
- Johnson
- p1