Constrained Utility Maximization (and Cost Minimization)
A consumer wishes to maximize their utility function U (x1 ; x2 ) while facing the budget
constraint p1 x1 + p2 x2 = I.
This problem could be written as max U (x1 ; x2 ) s.t. p1 x1 + p2 x2 = I. The fundamental
goal of what we are trying to achieve is reaching the highest possible indi¤erence curve
that still touches the budget constraint. We could approach this problem searching for the
tangency point.
Construct the Lagrangian function:
L(x1 ; x2 ; ) = U (x1 ; x2 ) + (I p1 x1 p2 x2 ). The FOCs are:
@L
= U1
p1 = 0;
@x1
@L
= U2
p2 = 0;
@x2
@L
= I p1 x1 p2 x2 = 0:
@
Take the …rst two equations, move the terms with to the right-hand sides and divide
the top equation by the bottom. We get UU12 = pp12 ( cancels out). This is equivalent
to M RS = M RT . At the optimal point, the indi¤erence curve is tangent to the budget
constraint. Since is eliminated, we are left with two equations: UU12 = pp12 and I p1 x1
p2 x2 = 0; and two unknowns: x1 and x2 . We could solve this system of equations for the
optimal values of x1 and x2 . If we keep p1 , p2 , and I as variables, we will derive the formulas
for consumer demands for goods 1 and 2. Proper demands can only contain prices, income
and parameters of the utility function. They can’t depend on the quantities of other goods,
i.e., you can’t have q2 in the …nal formula for demand of q1 . If we work with speci…c values
for p1 , p2 , and I, we will derive the optimal consumption levels of these goods.
Another way of writing the tangency condition M RS = M RT is Up11 = Up22 . In words,
at the optimal point, marginal utility per dollar has to be identical for all goods. This is a
useful, intuitive way of thinking about the optimal consumption. If Up11 were greater than Up22 ,
we could have sold 1=p2 units of good 2 to generate one extra dollar of income. The loss
of utility from this sale is U2 p12 = Up22 . With this dollar, we could buy 1=p1 units of good 1.
This will bring us extra utility U1 p11 = Up11 . Since Up11 > Up22 , we generated more utility than
we lost with this transfer of one dollar from spending it on good 2 to spending it on good
1.
1
Examples of constrained utility maximization for various utility functions
1) Cobb-Douglas: U = q1 q21
The problem is max q1 q21 s.t. p1 q1 + p2 q2 = I.
Just to illustrate di¤erent approaches, I will skip the Lagrangian and start with the
q 1 q1
tangency condition M RS = M RT . For this utility function, M RS = (1 1 )q 2q = 1 qq21 .
1 2
q2
= pp12 , we obtain q2 = (1 p)p2 1 q1 . Substitute this into
q1
the budget constraint p1 q1 + p2 q2 = I to get p1 q1 + (1 )p1 q1 = I. This is equivalent to
= I or p1 q1 1 = I. Then, q1 = p1I . Substitute this into the formula for q2
p 1 q1 1 + 1
above and simplify to obtain q2 = (1 p2 )I . These are the demands for Cobb-Douglas utility
It is equal to M RT = pp12 . From 1
functions.
Note that both demands could be characterized by the following rule. The percentage
of income spent on each good corresponds to the power of that good in the utility function
( for good 1, 1
for good 2). So, if we know that percentage of income I is spent
on good 1, this means that I is the total amount of money spent on good 1. Then, the
demand for this good is q1 = I=p1 . I will call this way of quickly writing down demands
for this utility function the Cobb-Douglas shortcut.
What about the case when the powers in the utility function do not sum up to 1, i.e.,
U = q1 q2 , where + could be anything? Recall that positive monotonic transformations
of a utility function describe the same preferences. So, raise this function to the power
1
. The preferences represented by U = q1 q2 are exactly the same as the preferences
+
=( + )
=( + )
represented by V = q1
q2
. For this function, the powers sum up to 1, so we
I
could use the Cobb-Douglas shortcut to conclude that the demand functions are q1 = +p1
and q2 =
+
p2
I
.
2
2) Constant Elasticity of Substitution (CES): U = (q1 + q2 )1= , where 0 6=
1
1=
The problem is max(q1 + q2 ) s.t. p1 q1 + p2 q2 = I.
Set up the Lagrangian function L = (q1 + q2 )1= + (I p1 q1 p2 q2 ). The derivative
1
with respect to q1 is 1 (q1 + q2 ) 1 q1 1
p1 = 0. The derivative with respect to q2 is
1
(q1 + q2 )
1
1
q2 1
p2 = 0. Move the terms with
q1
q2
equation by the second to obtain
1
to the right-hand side, divide the …rst
1
= pp12 . Then, qq12 =
1
1
p1
p2
and q1 = q2
1
p1
p2
.
1
Substitute this expression into the budget constraint p1 q1 +p2 q2 = I to obtain p1 q2
!
p2 q2 = I. The left-hand side simpli…es to q2
p1
p2
1
Ip2
p1
1
1
1
+p2
1
function.
Ip1
. Similarly, q1 =
p1
1
1
1
1
+ p2
p
= q2 1
1
+p2
1
p2
1
p1
p2
1
. Then, q2 =
1
+p2
1
. These are consumer demands for the CES utility
3
1
+
3) Perfect Substitutes: U = aq1 + bq2 .
The Lagrangian or tangency method would not work here since the slope of indi¤erence
curves, a=b, will never be equal to the slope of the budget constraint p1 =p2 . Despite the
technique not working here, the idea of getting to the highest possible indi¤erence curve
while satisfying the budget constraint still holds. The graph below illustrates the corner
solution, which occurs at the edge of the budget constraint.
q2
q1
Here, we assumed that the indi¤erence curves are steeper than the budget constraint,
i.e., a=b > p1 =p2 . From the graph, we obtained the solution that the consumer will purchase
only good 1 and will spend no income on good 2. Another way of writing the comparison
of the slopes is a=p1 > b=p2 . This indicates that the marginal utility per dollar is higher for
good 1, so this is a more intuitive way of reasoning that the consumer will only purchase
good 1. In summary, if a=p1 > b=p2 (or a=b > p1 =p2 ), the consumer demands are q1 = I=p1
and q2 = 0. If a=p1 < b=p2 (or a=b < p1 =p2 ), the consumer purchases only good 2, so the
demands are q1 = 0 and q2 = I=p2 . If a=p1 = b=p2 (or a=b = p1 =p2 ), the consumer can
purchase any bundle on the budget constraint.
4
4) Perfect Complements: U = min(aq1 ; bq2 )
Here, we also can not use the Lagrangian or tangency method. As we could see from
the graph below, the optimal point will be at the corner of the indi¤erence curve where
MRS is unde…ned.
q2
q1
At the corner of each indi¤erence curve, we don’t have excess of either good, so aq1 = bq2 .
Then, q2 = ab q1 . Substitute this into the budget constraint to obtain p1 q1 + p2 ab q1 = I. The
I
solution is q1 = p1 +p
a . Substitute this expression into the formula for q2 and simplify to
2
b
obtain q2 = p bI+p . These are the demands for perfect complements.
1a
2
5
p
5) Quasilinear preferences: U = q1 + 2 q2
p
The problem is max q1 + 2 q2 s.t. p1 q1 + p2 q2 = I.
p
Set up the Lagrangian function L = q1 + 2 q2 + (I p1 q1 p2 q2 ). The derivative with
p2 = 0. Move the
respect to q1 is 1
p1 = 0. The derivative with respect to q2 is p1q2
terms with to the right-hand side and divide the …rst equality by the second to obtain
p
q2 = p1 =p2 . Then, q2 = p21 =p22 . Substitute this expression into the budget constraint to
I p2 =p
p2
obtain p1 q1 + p2 p21 = I. Then, q1 = p11 2 . This solution works if q1 is non-negative. If it
2
is negative, it means we found a tangency point outside the …rst quadrant (point A in the
graph below).
A
q2
B
q1
In that case, since we can’t have negative quantities, we should think of a goal of
reaching the highest indi¤erence curve subject to the budget constraint. We will have a
corner solution –point B in the graph above, where q1 = 0. Then, all income is spent on
good 2, so q2 = I=p2 .
I p2 =p
In summary, if I p21 =p2 > 0, then q1 = p11 2 and q2 = p21 =p22 . And if I p21 =p2 0,
then q1 = 0 and q2 = I=p2 .
6
Finally, we will consider the problem of constrained expenditure minimization. It answers the question what is the minimum expenditure level E that is necessary to maintain
a particular level of utility U . Formally,
min E = p1 q1 + p2 q2 s.t. U (q1 ; q2 ) = U .
We could try solving this problem by …nding the tangency point (point A in the …gure
below) through the Lagrangian method.
q2
A
q1
Consider Cobb-Douglas preferences: min E = p1 q1 + p2 q2 s.t. q1 q21 = U .
Set up the Lagrangian function L(x1 ; x2 ; ) = p1 q1 +p2 q2 + (U q1 q21 ). The derivative
with respect to q1 is p1
q1 1 q21 = 0. The derivative with respect to q2 is p2
(1
)q1 q2 = 0. Move the terms with to the right-hand side, divide the …rst equation by the
second to obtain pp21 = (1 q2)q1 . Note that this is the same expression as we obtained while
solving the constrained utility maximization problem for Cobb-Douglas above. Indeed, this
is the tangency condition M RS = M RT and it is the same regardless of whether we are
working with utility maximization or expenditure minimization. Use this equation to solve
for q2 in terms of q1 : q2 = (1 p)p2 1 q1 .
At this point, the two constrained optimization problems diverge. For utility maximization, we substitute this value of q2 into the budget constraint p1 q1 + p2 q2 = I and obtain
the solution for q1 and q2 in terms of p1 , p2 , and I. For expenditure minimization, we
substitute this value of q2 into the utility constraint (q1 q21 = U ) and solve for q1 and q2
in terms of p1 , p2 ; and U .
After the substitution, we get q1
(1
)p1 q1
p2
1
= U . Then, q1 = U
p1 1
p2
1
. Sub-
stitute this into the expression for q2 and simplify to obtain q2 = U pp21 1
. While
we found the optimal consumption levels, frequently we would be interested in the expenditure level that corresponds to this consumption. To compute it, substitute the
values for q1 and q2 into the objective function E = p1 q1 + p2 q2 . For this example,
1
1
1
1
E = U p1 p12
+ U p1 p12
= U p1 p21
+ 1
.
1
7
Problem Set 5 provides a variety of practice problems on constrained utility maximization and expenditure minimization.
8