OPMT 5701
Applications of Lagrangian
Utility Maximization with a simple rationing constraint
Consider a familiar problem of utility maximization with a budget constraint:
Maximize
U
=
U
(
x, y
)
subject to
B
=
P
x
x
+
P
y
y
and
x
>
x
But where a ration on
x
has been imposed equal to
x.
We now have two constraints. The Lagrange method
easily allows us to set up this problem by adding the second constraint in the same manner as the
fi
rst. The
Lagrange becomes
Max
x,y
U
(
x, y
) +
λ
1
(
B
−
P
x
x
−
P
y
y
) +
λ
2
(
x
−
x
)
However, in the case of more than one constraint, it is possible that one of the constraints is nonbinding.
In the example we are using here, we know that the budget constraint will be binding but it is not clear if
the ration constraint will be binding. It depends on the size of
x.
The two possibilities are illustrated in
fi
gure one. In the top graph, we see the standard utility maximiza
tion result with the solution at point E. In this case the ration constraint,
x
, is larger than the optimum
value
x
∗
.
In this case the second constraint could have been ignored.
In the bottom graph the ration constraint is binding. Without the constraint, the solution to the max
imization problem would again be at point E. However, the solution for
x
violates the second constraint.
Therefore the solution is determined by the intersection of the two constraints at point
E’
Procedure:
This type of problem requires us to vary the
fi
rst order conditions slightly. Cases where constraints may or
not be binding are often referred to as
KuhnTucker
conditions.
The KuhnTucker conditions are
L
x
=
U
x
−
P
x
λ
1
−
λ
2
= 0
x
≥
0
L
y
=
U
y
−
P
y
λ
1
= 0
y
≥
0
and
L
λ
1
=
B
−
P
x
x
−
P
y
y
≥
0
λ
1
≥
0
L
λ
2
=
x
−
x
≥
0
λ
2
≥
0
Now let us interpret the KuhnTucker conditions for this particular problem. Looking at the Lagrange
U
(
x, y
) +
λ
1
(
B
−
P
x
x
−
P
y
y
) +
λ
2
(
x
−
x
)
We require that
λ
1
(
B
−
P
x
x
−
P
y
y
) = 0
therefore either
λ
1
= 0
or
B
−
P
x
x
−
P
y
y
= 0
If we interpret
λ
1
as the marginal utility of the budget (Income), then if the budget constraint is not met
the marginal utility of additional
B
is zero (
λ
1
= 0
).
(2) Similarly for the ration constraint, either
x
−
x
= 0
or
λ
2
= 0
λ
2
can be interpreted as the marginal utility of relaxing the ration constraint.
1
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y
x
U
0
U
1
E
B

P
y
B

P
x
= x*
y*
Case 2: Second constraint
is Binding
x
y
U
0
U
1
Maximize
U = U(x,y)
subject to
B = P
x
x + P
y
y
and x < X
E
B

P
y
B 

P
x
x*
y*
CASE 1: Second
constraint Not Binding
X
X
E’
Figure 1:
Solving by Trial and Error
Solving these types of problems is a bit like detective work. Since there are more than one possible outcomes,
we need to try them all. But before you start, it is important to think about the problem and try to make
an educated guess as to which constraint is more likely to be nonbinding. In this example we can be sure
that the budget constraint will always be binding, therefore we only need to worry about the e
ff
ects of the
ration constraint.
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 Fall '11
 WendyWu
 Economics, Utility, Constraint, lagrange multipliers, Ly, Zy, KuhnTucker conditions

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