Problem Set 4: Equality-Constrained Maximization
1. Consider the maximization problem
max
x
1
,x
2
x
1
+
x
2
subject to
x
2
1
+
x
2
2
= 1
Sketch the constraint set and contour lines of the objective function. Find all critical points of the
Lagrangian. Verify whether each critical point is a local maximum or a local minimum. Find the
global maximum.
You should sketch the constraint set and contour lines/indifference curves. The constraint
is a circle of radius one, and the objective is a linear hyperplane with slope
−
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
max
x,y
xy
subject to
a
2
x
2
+
b
2
y
2
=
r
Sketch the constraint set and contour lines of the objective function. Find all critical points of the
Lagrangian. Verify whether each critical point is a local maximum or a local minimum. Find the
global maximum. How does the value of the objective vary with
r
?
a
? How does
x
∗
respond to a
change in
r
; show this using the closed-form solutions and the IFT.
Sketch. The constraint set is an ellipse and the objective function is “Cobb-Douglas”.
The Lagrangian is
L
(
x, λ
) =
xy
−
λ
(
a
2
x
2
+
b
2
y
2
−
r
)
with FONCs
y
−
λax
= 0
x
−
λby
= 0
−
(
a
2
x
2
+
b
2
y
2
−
r
) = 0
The first two equations imply that
y/x
=
ax/
(
by
), or
by
2
=
ax
2
. Using the third equation, we
get
ax
2
=
r
or
x
∗
=
±
radicalbigg
r
a
implying that
y
∗
=
±
radicalbigg
r
b
so we have four critical points,
parenleftbiggradicalbigg
r
a
,
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