UNCWilmington
ECN 321
Department of Economics and Finance
Dr. Chris Dumas
Consumer Decisions in
Factor Markets
Example Problems
Consumer Labor/Leisure Model  Example Problem
This is a simple model of a consumer's decision regarding how much to work or "labor."
This model is the
foundation for many models of labor supply, unemployment, and household income determination.
U is utility
L is labor hours per day
Z is leisure hours per day
w is the hourly wage rate
(Note:
w is a given constant)
I is income per day
Consumer gets utility (U) from leisure (Z) and income (I).
Consumer wants more of both Z and I, but there
is a tradeoff: the more he labors (the larger is L) the larger is I, but the smaller is Z.
Suppose consumer
survey data and statistical regression analyses indicate that a typical consumer's tastes regarding various
combinations of I and Z are described by the following utility function:
U = 20·I + 8Z  (1/10)I
2
 (1/4)Z
2
In addition, if you stop to ponder a moment, you would eventually realize that the following constraints are
key in this situation:
Time Constraint:
L + Z = 24
The consumer has 24 hours each day to allocate b/n L and Z.
Income Identity:
I = w·L
Note: An
identity
is an equation that is true by definition.
For example, the definition of income is
"(wage rate)(labor hours)."
Identity equations are often used to help simplify problems.
Usually,
identity equations are used to eliminate some of the variables in a problem through substitution, as
we will see below.
The consumer's optimization problem can be written now as:
max
U = 20·I + 8Z  (1/10)I
2
 (1/4)Z
2
I,Z
subject to:
L + Z = 24
We can remove L from the problem (and thereby simplify the problem) by using the Income Identity.
Rearranging the Income Identity, we find that L = I/w.
We now substitute I/w for L in the Time Constraint,
and the problem becomes:
max
U = 20·I + 8Z  (1/10)I
2
 (1/4)Z
2
I,Z
subject to:
I/w + Z = 24
(Note: L is now removed from the problem, and w is a given constant.)
This problem has the following key features:
* Nonlinear
programming problem
* Two
choice variables
* One
constraint
Because this is a nonlinear programming problem with a constraint, we will use Lagrange's method as the
solution method.
Converting the problem into an equivalent Lagrangian problem, we rewrite the objective
function as a Lagrangian expression, denoted "L."
In doing so, we incorporate the constraint into the
1
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UNCWilmington
ECN 321
Department of Economics and Finance
Dr. Chris Dumas
Lagrangian expression by introducing the new choice variable
λ
into the problem as a Lagrangian
multiplier:
max
L = 20·I + 8Z  (1/10)I
2
 (1/4)Z
2
+
λ
(24  I/w  Z)
I,Z,
λ
subject to:
nothing
F.O.C.'s:
(1)
0
)
w
/
1
(
I
)
10
/
1
(
2
20
I
L
=
λ


=
∂
∂
(2)
0
Z
)
4
/
1
(
2
8
Z
L
=
λ


=
∂
∂
(3)
0
Z
w
/
I
24
L
=


=
λ
∂
∂
The FOC equations provide us with three equations in three unknowns, I, Z and
λ
.
Solve these three
equations for the solution values of I, Z and
λ
.
There are several ways to work the algebra to solve these
three equations.
One way is illustrated below:
Combine FOC's (1) and (2) to eliminate
λ
:
λ
λ
=


)
w
/
1
(
Z
)
4
/
1
(
2
8
I
)
10
/
1
(
2
20
)
w
/
1
(
Z
)
4
/
1
(
2
8
I
)
10
/
1
(
2
20
=


]
Z
)
4
/
1
(
2
8
)[
w
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 Fall '08
 Dumas
 Economics, Microeconomics, Optimization, Debt, Household income in the United States, Department of Economics and Finance, Dr. Chris Dumas

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