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Utility Maximization Subject to a Budget Constraint
Christopher Makler
Stanford University Department of Economics
Econ 50: Lecture 6
Choice space:
all possible options
Feasible set:
all options available to you
Optimal choice:
Your best choice(s) of the ones available to you
Constrained Optimization
Preferences describe how the agent ranks all options in the choice space.
For example, we'll assume that you could rank all possible colleges
(and other options for what to do after high school) based upon your preferences.
Preference Ranking
Found a startup
Harvard
Stanford
Play Xbox in parents' basement
Cal
Choice space
Feasible set
Optimal
choice!
Found a startup
Stanford
Cal
Harvard
Play XBox in parents' basement
Optimal choice is the highest-ranking option in the feasible set.
The story so far, in two graphs
If we superimpose the budget line on the utility "hill" the nature of the problem becomes clear:
Question: mathematically, how does the utility change as you spend more money on good 1?
Think about what happens
when you spend another dollar on apples
Budget line
Suppose apples (good 1) cost $4/lb,
and bananas (good 2) cost $2/lb.
Approximately what is the change in your utility? Should you do this?
(and one less on bananas)
give up \({1 \over 2}\)lb of bananas
gain \({1 \over 4}\) lb apples
\Delta u \approx
{1 \over 4} \times MU_1
{1 \over 2} \times MU_2
-
Think about what happens
when you spend another dollar on apples
Budget line
Suppose apples (good 1) cost \(p_1\) per pound,
and bananas (good 2) cost \(p_2\) per pound.
Approximately what is the change in your utility? Should you do this?
(and one less on bananas)
give up \({1 \over p_2}\) bananas
gain \({1 \over p_1}\) apples
\Delta u \approx
{MU_1 \over p_1}
{MU_2 \over p_2}
-
"marginal utility per dollar" or "bang for your buck"
{MU \over p}
p_1 = 4
p_2 = 2
u(x_1,x_2) = 3 \ln x_1 + \ln x_2
m = 48
You like apples more, but they're also more expensive.
You think about spending half your money on each good:
that is, buy 6 lbs of apples (good 1) and 12 lbs of bananas (good 2). Is this optimal?
MU_1(x_1,x_2) = {3 \over x_1}
MU_1(6,12) = \frac{1}{2}\ \frac{\text{utils}}{\text{apple}}
p_1 = 4\ \frac{\text{dollars}}{\text{apple}}
= {1 \over 8} \frac{\text{utils}}{\text{dollar}}
{MU_1 \over p_1} =
MU_2(x_1,x_2) = {1 \over x_2}
MU_2(6,12) = \frac{1}{12}\ \frac{\text{utils}}{\text{banana}}
p_2 = 2\ \frac{\text{dollars}}{\text{banana}}
= {1 \over 24} \frac{\text{utils}}{\text{dollar}}
{MU_2 \over p_2} =
What does it mean if you get more "bang for your buck" from good 1 than good 2?
\frac{MU_1}{MU_2} > \frac{p_1}{p_2}
The consumer receives more utility per additional unit of good 1 than the price reflects, relative to good 2.
\frac{MU_1}{p_1} > \frac{MU_2}{p_2}
The consumer receives more
"bang for the buck"
(utils per dollar)
from good 1 than good 2.
Regardless of how you look at it, the consumer would be
better off moving to the right along the budget line --
i.e., consuming more of good 1 and less of good 2.
MRS > \frac{p_1}{p_2}
The consumer is more willing to give up good 2
to get good 1
than the market requires.
p_1 = 4
p_2 = 2
u(x_1,x_2) = 3 \ln x_1 + \ln x_2
m = 48
You like apples more, but they're also more expensive.
You think about spending half your money on each good:
that is, buy 6 lbs of apples (good 1) and 12 lbs of bananas (good 2). Is this optimal?
MU_1(x_1,x_2) = {3 \over x_1}\ \frac{\text{utils}}{\text{apple}}
p_1 = 4\ \frac{\text{dollars}}{\text{apple}}
= 2\ \frac{\text{bananas}}{\text{apple}}
MU_2(x_1,x_2) = {1 \over x_2}\ \frac{\text{utils}}{\text{banana}}
p_2 = 2\ \frac{\text{dollars}}{\text{banana}}
{p_1 \over p_2} =
MRS(x_1,x_2)=
= {3x_2 \over x_1}\ \frac{\text{bananas}}{\text{apple}}
= 6\ \frac{\text{bananas}}{\text{apple}}
= {3 \times 12 \over 6} \frac{\text{bananas}}{\text{apple}}
at (6,12)...
IF...
THEN...
The optimal consumption bundle will be characterized by two equations:
MRS = \frac{p_1}{p_2}
p_1x_1 + p_2x_2 = m
More generally: the optimal bundle may be found using the Lagrange method
(certain conditions hold,
that we're going to
talk about next week)
p_1 = 4
p_2 = 2
u(x_1,x_2) = 3 \ln x_1 + \ln x_2
m = 48
{p_1 \over p_2} = 2
MRS(x_1,x_2) = {3x_2 \over x_1}
Tangency condition: set MRS = price ratio
Constraint:
{3x_2 \over x_1} = 2
4x_1 + 2x_2 = 48
x_2 = {2 \over 3}x_1
Two equations, two unknowns;
solve like you always have!
pollev.com/chrismakler
Suppose a consumer's preferences can be represented by the utility function
u(x_1,x_2) = x_1^{1 \over 2} + 2x_2^{1 \over 2}
p_1 = 1
p_2 = 4
m=32
and their budget constraint is determine by
What is the tangency condition
for this problem?
What is the optimal bundle?
u(x_1,x_2)
\max
The Lagrange Method
x_1,x_2
\text{s.t.}
p_1x_1 + p_2x_2
\le m
Cost of Bundle X
Income
Utility
\max
The Lagrange Method
x_1,x_2
\text{s.t.}
m - p_1x_1 - p_2x_2
Income left over
u(x_1,x_2)
u(x_1,x_2)
Utility
\ge 0
The Lagrange Method
m - p_1x_1 - p_2x_2
Income left over
\mathcal{L}(x_1,x_2,\lambda)=
\lambda
u(x_1,x_2)
u(x_1,x_2)
+
(
)
Utility
(utils)
(dollars)
utils/dollar
\frac{\partial \mathcal{L}}{\partial x_1} = \frac{\partial u}{\partial x_1} - \lambda p_1
First Order Conditions
\frac{\partial \mathcal{L}}{\partial x_2} = \frac{\partial u}{\partial x_2} - \lambda p_2
\frac{\partial \mathcal{L}}{\partial \lambda} = m - p_1x_1 - p_2x_2
= 0 \Rightarrow \lambda = \frac{MU_1}{p_1}
= 0 \Rightarrow \lambda = \frac{MU_2}{p_2}
"Bang for your buck" condition: marginal utility from last dollar spent on every good must be the same!
\text{Also: }\frac{\partial \mathcal{L}}{\partial m} = \lambda = \text{bang for your buck, in }\frac{\text{utils}}{\$}
\text{Set two expressions }
\text{for }\lambda \text{ equal:}
\frac{MU_1}{p_1} = \frac{MU_2}{p_2}
The Lagrange Method
m - p_1x_1 - p_2x_2
\mathcal{L}(x_1,x_2,\lambda)=
\lambda
u(x_1,x_2)
+
(
)
\frac{\partial \mathcal{L}}{\partial x_1} = \frac{3}{x_1} - 4\lambda
First Order Conditions
\frac{\partial \mathcal{L}}{\partial x_2} = \frac{1}{x_2} - 2\lambda
\frac{\partial \mathcal{L}}{\partial \lambda} = 48-4x_1-2x_2
= 0 \Rightarrow \lambda = \frac{3}{4x_1}
= 0 \Rightarrow \lambda = \frac{1}{2x_2}
\text{Set two expressions }
\text{for }\lambda \text{ equal:}
\frac{3}{4x_1} = \frac{1}{2x_2}
The Lagrange Method
48 - 4x_1 - 2x_2
\mathcal{L}(x_1,x_2,\lambda)=
\lambda
3 \ln x_1 + \ln x_2
+
(
)
x_2 = \frac{2}{3}x_1
= 0 \Rightarrow 4x_1 + 2x_2 = 48
Why use Lagrange?
1. It works for many goods.
2. The Lagrange multiplier has an
economic meaning (bang for your buck).
pollev.com/chrismakler
Suppose a consumer's preferences can be represented by the utility function
u(x_1,x_2) = x_1^{1 \over 2} + 2x_2^{1 \over 2}
p_1 = 1
p_2 = 4
m=32
and their budget constraint is determine by
What is the Lagrangian for this problem?
Next time: when Lagrange fails.....
Econ 50 | Fall 25 | Lecture 06
By Chris Makler
Econ 50 | Fall 25 | Lecture 06
Constrained optimization when calculus (the Lagrange method) works
