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UNC-Wilmington
Department of Economics and Finance
ECN 321
Dr. Chris Dumas
Consumer Choice
with Non-linear Objective Functions and No Constraints
Example Problems
In this handout, we examine consumer utility maximization problems that can be classified as
non-linear programming problems without constraints. Recall that if one or more of the
equations in an optimization problem is non-linear, then we cannot use linear programming to
solve the problem; instead, we must turn to non-linear programming. In the examples below, the
objective function is a non-linear equation; this makes the problem a non-linear programming
problem.
When there are no constraints in a non-linear optimization problem, we are able to use "the
classical calculus method” to solve the problem (as described below in this handout), and we do
not need to use Lagrange's Method (the subject of the next handout) to solve the problem. On the
other hand, when there are constraints in the problem, we cannot use the classical calculus
method to solve the problem, and we must turn to Lagrange's Method.
Example (1)
* Nonlinear programming problem
* One choice variable
* No constraints (note: when we have no constraints, we DON'T need to use Lagrange's method)
(note on the note above: Lagrange’s method is discussed in the next handout)
max U = 40 + 10X1 - X12
X1
subject to: no constraints
To maximize U, first find the derivative of the objective function with respect to the choice variable:
dU
dU
 10  2X 1 (Note:
is the marginal utility of X1, typically denoted MUX1.)
dX 1
dX1
dU
Note that as X1 becomes larger, marginal utility,
, becomes smaller, reflecting diminishing marginal utility.
dX1
The FOC condition says that to find the value of X1 that maximizes U, set the first derivative equal to zero:
F.O.C.:
dU
 10  2X 1  0
dX 1
Solve FOC for X1:
X1* = 10/2 = 5
Solution: X1* = 5
The "star" on X1 means "the particular value of X1 that maximizes U."
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UNC-Wilmington
Department of Economics and Finance
ECN 321
Dr. Chris Dumas
Example (2)
* Nonlinear programming problem
* Two choice variables
* No interaction among the choice variables
* No constraints (note: when we have no constraints, we DON'T need to use Lagrange's method)
Note: Interaction among the choice variables is NOT present in the utility function below (that is, the utility provided
by consuming X1 does NOT depend on the amount of X2 consumed, and vice versa; that is, the effects of X1 and X2
on utility are INDEPENDENT of one another). We know this because the equation for
X2, and the equation for
U
does NOT contain
X 1
U
does NOT contain X1. In the next example, Example (3), we will consider a
X 2
situation in which there IS interaction among the choice variables, which makes the problem more difficult.
max U = 2X1 - X12 + 4X2 - X22
X 1, X 2
subject to: no constraints
F.O.C.'s:
U
U
 2  2X 1  0 (Note:
is the marginal utility of X1, typically denoted MUX1.)
X 1
X 1
U
U
 4  2X 2  0 (Note:
(2)
is the marginal utility of X2, typically denoted MUX2.)
X 2
X 2
(1)
Solve FOC (1) for X1:
Solve FOC (2) for X2:
X1* = 2/2 = 1
X2* = 4/2 = 2
Solution: (X1* = 1, X2* = 2)
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UNC-Wilmington
Department of Economics and Finance
ECN 321
Dr. Chris Dumas
Example (3)
* Nonlinear programming problem
* Two choice variables
* Interaction among the choice variables
* No constraints (note: when we have no constraints, we DON'T need to use Lagrange's method)
Note: Interaction among the choice variables is present in the utility function below (that is, the utility provided by
consuming X1 depends on the amount of X2 consumed, and vice versa; that is, the effects of X1 and X2 on utility are
not independent of one another). We know this because the equation for
for
U
DOES contain X2, and the equation
X 1
U
DOES contain X1. When the choice variables interact, the marginal utility of each variable will depend
X 2
on the levels of both variables.
max U = 2X1 + 3X2 - X12 - X22 - X1X2
X 1, X 2
subject to: no constraints
F.O.C.'s:
U
U
 2  2X 1  X 2  0 (Note:
is the marginal utility of X1, typically denoted MUX1.)
X 1
X 1
U
U
 3  2X 2  X 1  0 (Note:
(2)
is the marginal utility of X2, typically denoted MUX2.)
X 2
X 2
(1)
The FOC equations provide us with two equations in two unknowns. Solve these two equations for the solution
values of X1 and X2. There are several ways to work the algebra to solve these two equations. One way is illustrated
below:
Solve FOC (1) for X2:
2 - 2X1 - X2 = 0
2 - 2X1 = X2
X2 = 2 - 2X1
Substitute the equation for X2 derived above into FOC (2) and solve for X1:
3 - 2X2 - X1 = 0
3 - 2(2 - 2X1) - X1 = 0
3 - 4 + 4X1 - X1 = 0
- 1 + 3X1 = 0
3X1 = 1
X1* = 1/3 (Note: we typically use a star "*" to denote the solution value of a variable.)
Substitute X1* back into FOC (1) and solve for X2:
2 - 2X1 - X2 = 0
2 - 2(1/3) - X2 = 0
X2* = 4/3 = 1.33
(Note: we typically use a star "*" to denote the solution value of a variable.)
Hence, the solution is: X1* = 1/3 ,
X2* = 4/3 = 1.33 .
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