LABOR NOTES, PART THREE: CONSTRAINED OPTIMIZATION
Everything in economics is a constrained optimization problem. Remember the
definition you first heard in intro micro: “economics is the study of the behavior of
individuals who satisfy their wants, given limited resources.” You’d like to watch
television, you’d like to make straight As, you’d like to go out with your buddies,
you’d like to sleep—but face it, there are only twenty-four hours in the day. This
means that you have to make a decision about how to spend those limited hours to
produce the most happiness.
In this class, we’re going to talk about the following problems:
• Labor/leisure trade-off: You like free time, you like to buy stuff, but
you have to work (and give up free time) to buy stuff.
• Time allocation: You want to use your time to have fun, produce things
around the house (like meals, clean rooms), and to work—but you only
have a certain amount of time in each day.
• Fertility: You want to have children, which means taking time off from
work; you want to provide well for your children (clothing, education, cars),
which means working—there’s a quality/quantity tradeoff.
• Human capital: Getting more education means you earn more, but
getting more education reduces the number of years that you can work.
• Intra-household allocation: You want to spend money on a new car,
but your spouse wants to spend it on a television.
All of these can be written as constrained optimization problems.
There are two components to any constrained optimization problem: the objective
and the constraint. First, you have to figure out what you are maximizing: it’s a
function of some variables. Second, you have to figure out what are the limitations
on these variables.
Labor notes, page 16
In the classic problem, the consumer enjoys two goods, x1 and x2 . He is trying to
maximize his happiness—“utility”—from these. However, he has only w dollars to
spend, so his total purchases ( x1 times its price plus x2 times its price) cannot exceed
his wealth. We would write this problem as follows:
max x1 , x2 U(x1 , x2 ) subject to p1 ! x1 + p2 ! x2 = w
(The subscripted x1 and x2 are the things he gets to choose. Often we will omit this,
if it is clear.) There are three ways to solve this problem. All produce the same
result, so which technique you use is largely a matter of preference.
3.1
Substitution
This is the simplest method, although it can lead to complicated algebra. This is the
idea behind it: if you have a function of one variable, a maximum occurs where the
derivative (that is, its derivative) is equal to zero. We can reduce the objective to a
function of only one variable, by substituting the budget constraint into it.
Step 1: Solve the constraint for one variable in terms of the other.
Step 2: Substitute this into the objective function.
Step 3: Take a derivative with respect to the one remaining choice variable,
and set it equal to zero.
Step 4: Solve for this choice variable.
Step 5: Using the constraint, solve for the other choice variable.
Here’s how it works. Suppose that the utility function is U(x1 , x2 ) = ln x1 + ln x2 .
First, note that if the individual spends all his money—which he will, since he derives
no pleasure from unspent money—then it will be the case that x2 = (w ! x1 p1 ) p2 .
Substituting this into the utility function, we have:
max x1 ln x1 + ln ( (w ! p1 x1 ) p2 )
Labor notes, page 17
That’s the objective, already taking the constraint into account. To maximize a
function of one variable, we take its derivative with respect to x1 and set that equal
to zero:
1
p1
!
=0
*
x1 w ! p1 x1*
Note that it’s only the optimal choice of x1 that solves that condition, which I denote
by x1* . Now we solve for x1* . Move the second term to the right-hand side of the
equation, and multiply both sides by x1* (w ! p1 x1* ) :
x1* p1 = w ! x1* p1
Collecting terms and dividing through, we have:
x1* = 12 (w p1 )
This is half of the answer. To find x2* , we go back to the solved budget constraint:
x2* = (w ! x1* p1 ) p2 = (w ! 12 (w p1 ) p1 ) p2 = 12 (w p2 )
These are the individual’s optimal choices of x1* and x2* .
3.2
Marginal rate of substitution
Recall from graphical analysis that an optimum occurs where the marginal rate of
substitution between two goods equals their relative prices (or where the slope of the
indifference curve equals the slope of the budget constraint). We exploit this
intuition in the second technique for solving constrained optimization problems.
Step 1: Write out the optimality condition in terms of marginal rates of
substitutions and relative prices (or slopes, if you prefer).
Step 2: Replace marginal rate of substitution with ratio of marginal utilities
(or marginal products).
Step 3: Take derivatives of the objective function to find the marginal
utilities, and substitute this into the optimality condition.
Labor notes, page 18
Step 4: Using this equation and the budget constraint, solve for the choice
variables.
We start off by writing down that the MRS between the goods equals their price
ratio, and then writing that the MRS depends on relative marginal utilities:
!U(x1* , x2* ) / !x1 p1
MRS =
=
!U(x1* , x2* ) / !x2 p2
The trickiest part of this is remembering which marginal utility goes in the
numerator and which in the denominator. A graph might help you keep it straight.
(Usually, if “1” is in the numerator on the left, it is in the numerator on the right; if
“2” is in the denominator on the left, it is in the denominator on the right.) Then we
use the utility function to calculate marginal utilities:
!U(x1* , x2* ) !x1 = 1 x1*
!U(x1* , x2* ) !x2 = 1 x2*
Substituting those into the optimality condition, we have:
!U(x1 , x2 ) / !x1 1 / x1 p1 !U(x1* , x2* ) / !x1 1 / x1* p1
=
=
=
=
!U(x1 , x2 ) / !x2 1 / x2 p2 !U(x1* , x2* ) / !x2 1 / x2* p2
Then we rearrange to get that:
x1* p1 = x2* p2
Along with the original budget constraint, w = x1 p1 + x2 p2 , we can solve this to get
the same answer as before: x1* = 12 (w p1 ) and x2* = 12 (w p2 ) .
3.3
Lagrangians
The third technique involves the use of Lagrangians, which you may have
encountered if you’ve taken a multivariate calculus class. These are somewhat
technical and unintuitive, but easy once you get the hang of them. Don’t worry if
you don’t get them—either of the other techniques will work fine.
Step 1: Rearrange the constraint so that it equals zero.
Labor notes, page 19
Step 2: Define a new function, L, to equal the objective function plus a
variable ! times the rearranged budget constraint.
Step 3: Take the derivatives of L with respect to each of the choice variables
(separately) as well as with respect to ! , and set all equal to zero.
Step 4: Solve these equations for the choice variables.
From the previous problem, we can subtract one side off of both sides of the budget
constraint to get:
w ! x1 p1 ! x2 p2 = 0
Then we define the Lagrangian function to be:
L(x1 , x2 , ! ) = ln x1 + ln x2 + ! (w " x1 p1 " x2 p2 )
We take derivatives with respect to x1 , x2 , and ! , and we set them equal to zero:
!L(x1* , x2* , " ) !x1 = 1 x1* # " p1 = 0
!L(x1* , x2* , " ) !x2 = 1 x2* # " p2 = 0
!L(x1* , x2* , " ) !" = w # p1 x1* # p2 x2* = 0
All we have to do is solve these for x1* , and x2* . Usually the easiest step is to eliminate
the “multiplier” ! first. Rearranging the first two conditions gives us:
! = 1 (x1* p1 )
! = 1 (x2* p2 )
Combining these, we have:
x1* p1 = x2* p2
Finally, we use the third condition, w ! x1* p1 ! x2* p2 = 0 to solve. We should get the
same answer as with the previous techniques.
Labor notes, page 20
3.4
Corner solutions
Often there’s an understanding that the choice variables must be non-negative. For
example, you can’t buy “negative 12 widgets,” and you can’t spend “negative 3 hours”
doing any activity. However, these solution techniques can find optima outside the
domain of permissible values. Consider the problem:
max x1 , x2 {ln x1 + x2 } subject to p1 ! x1 + p2 ! x2 = w
If we solve this using any of the techniques above (you can do this for practice,
mimicking the steps I just went through), you’ll end up with the solution:
x1* = p2 p1
x2* = ( w p2 ! 1)
What happens if w < p2 ? This predicts that you purchase a negative amount of x2 ,
which isn’t possible. It also predicts that you spend more than your wealth on the
first good. If you graph this, you’ll find that the tangency occurs outside the
northeast quadrant. With two goods, it will always be true that the optimal choice is
at the corner nearest to where the tangency occurs. Simply set the “negative” value
to zero, and say that the individual spends all his money on the other good. The
correct way to write this solution is:
x1* = p2 p1 , x2* = ( w p2 ! 1)
if w " p2 ;
*
*
x1 = w p1 , x2 = 0
if w ! p2 .
Sometimes we’ll find corner solutions very interesting, and sometimes we’ll ignore
them. When people choose not to work at all, they are at a corner solution. Labor
force participation is an interesting topic—what determines who stays at home,
whether college students get a job? On the other hand, for prime-aged male workers,
it’s essentially a non-issue. Virtually everyone works, so we’ll study what determines
how much they work.
One final note: to quote Tom Lehrer, “in the new math, the important thing is to
understand what you’re doing, rather than to get the right answer.” On exams, I’m
much more interested to see that you know how to solve the problem. Fumbling on
algebra (as you’ll see me do in class, I’m sure) is a fairly minor sin. Show your work
clearly, and I’ll understand when x ! x = 2x .
Labor notes, page 21