# The Edgeworth Box Framework

Christopher Makler

Stanford University Department of Economics

Econ 51: Lecture 5

## Today's Agenda

Part 1: Efficiency

Part 2: Equity

From preferences to allocations

Pareto improvements

Pareto efficiency and the "contract curve"

Some applications

The Utility Possibilities Frontier

Social preferences

Altruism

Fairness

## From Preferences to Allocations

An endowment is a vector saying how much of different goods an agent has.

### Example

Alison has 120 oreos and 20 twizzlers

Bob has 80 oreos and 80 twizzlers

## Canonical Constrained Optimization Problem

f(x_1,x_2)
\text{s.t. }
g(x_1,x_2) = k
k - g(x_1,x_2) = 0
\mathcal{L}(x_1,x_2,\lambda)=
\displaystyle{\max_{x_1,x_2}}
f(x_1,x_2)
k - g(x_1,x_2)
+ \lambda\ (
)

Suppose $$g(x_1,x_2)$$ is monotonic (increasing in both $$x_1$$ and $$x_2$$).

Then $$k - g(x_1,x_2)$$ is negative if you're outside of the constraint,
positive if you're inside the constraint,
and zero if you're along the constraint.

OBJECTIVE

FUNCTION

CONSTRAINT

\mathcal{L}(x_1,x_2,\lambda)=
f(x_1,x_2)
k - g(x_1,x_2)
+ \lambda\ (
)
\displaystyle{\partial \mathcal{L} \over \partial x_1} =

FIRST ORDER CONDITIONS

\displaystyle{\partial \mathcal{L} \over \partial x_2} =
\displaystyle{\partial \mathcal{L} \over \partial \lambda} =
\displaystyle{\partial f \over \partial x_1}
\displaystyle{\partial f \over \partial x_2}
k - g(x_1,x_2)
=0
- \lambda\ \times
\displaystyle{\partial g \over \partial x_1}
\displaystyle{\partial g \over \partial x_2}
- \lambda\ \times
=0
=0

3 equations, 3 unknowns

Solve for $$x_1$$, $$x_2$$, and $$\lambda$$

## It finds the point along the constraint where thelevel set of the objective function passing through that point is tangent to the constraint

\mathcal{L}(x_1,x_2,\lambda)=
f(x_1,x_2)
k - g(x_1,x_2)
+ \lambda\ (
)
\displaystyle{\partial \mathcal{L} \over \partial x_1} =

FIRST ORDER CONDITIONS

\displaystyle{\partial \mathcal{L} \over \partial x_2} =
\displaystyle{\partial \mathcal{L} \over \partial \lambda} =
\displaystyle{\partial f \over \partial x_1}
\displaystyle{\partial f \over \partial x_2}
k - g(x_1,x_2)
=0
- \lambda\ \times
\displaystyle{\partial g \over \partial x_1}
\displaystyle{\partial g \over \partial x_2}
- \lambda\ \times
=0
=0
\displaystyle{\Rightarrow \lambda\ = {{\partial f /\partial x_1} \over {\partial g/\partial x_1}}}
\displaystyle{\Rightarrow \lambda\ = {{\partial f /\partial x_2} \over {\partial g/\partial x_2}}}
\displaystyle{\Rightarrow {{\partial f /\partial x_1} \over {\partial f/\partial x_2}} = {{\partial g /\partial x_1} \over {\partial g/\partial x_2}}}

TANGENCY
CONDITION

CONSTRAINT

## Example: Fence Problem

You have 40 feet of fence and want to enclose the maximum possible area.

## Example: Fence Problem

You have 40 feet of fence and want to enclose the maximum possible area.

f(x_1,x_2)
\text{s.t. }
g(x_1,x_2) = k
k - g(x_1,x_2) = 0
\mathcal{L}(L,W,\lambda)=
\displaystyle{\max_{x_1,x_2}}
L \times W
40 - 2L - 2W
+ \lambda\ (
)

OBJECTIVE

FUNCTION

CONSTRAINT

L \times W
40 - 2L - 2W = 0
2L + 2W = 40
\mathcal{L}(L,W,\lambda)=
L \times W
40 - 2L - 2W
+ \lambda\ (
)
\displaystyle{\partial \mathcal{L} \over \partial L} =

FIRST ORDER CONDITIONS

\displaystyle{\partial \mathcal{L} \over \partial W} =
\displaystyle{\partial \mathcal{L} \over \partial \lambda} =
W
L
40 - 2L - 2W
=0
- \lambda\ \times
2
2
- \lambda\ \times
=0
=0
\displaystyle{\Rightarrow \lambda\ = {W \over 2}}
\displaystyle{\Rightarrow \lambda\ = {L \over 2}}
W = L

TANGENCY
CONDITION

CONSTRAINT

2L + 2W = 40
W = L

TANGENCY
CONDITION

CONSTRAINT

2L + 2W = 40

## Meaning of the Lagrange multiplier

Suppose you have $$F$$ feet of fence instead of 40.

\displaystyle{\partial \mathcal{L} \over \partial L} =
\displaystyle{\partial \mathcal{L} \over \partial W} =
\displaystyle{\partial \mathcal{L} \over \partial \lambda} =
W
L
F - 2L - 2W
=0
- \lambda\ \times
2
2
- \lambda\ \times
=0
=0
\displaystyle{\Rightarrow \lambda\ = {W \over 2}}
\displaystyle{\Rightarrow \lambda\ = {L \over 2}}
W = L

TANGENCY
CONDITION

CONSTRAINT

2L + 2W = F

SOLUTIONS

L^* = {F \over 4}
W^* = {F \over 4}
\lambda^* = {F \over 8}

## Meaning of the Lagrange multiplier

Suppose you have $$F$$ feet of fence instead of 40.

SOLUTIONS

L^* = {F \over 4}
W^* = {F \over 4}
\lambda^* = {F \over 8}

Maximum enclosable area as a function of F:

A^*(F) = L^*(F) \times W^*(F) = {F \over 4} \times {F \over 4} = {F^2 \over 16}

## Meaning of the Lagrange multiplier

Suppose you have $$F$$ feet of fence instead of 40.

\mathcal{L}(L,W,\lambda)=
L \times W
40 - 2L - 2W
+ \lambda\ (
)

# Verbal Analysis: MRS, MRT, and the “Gravitational Pull" towards Optimality

## Fish vs. Coconuts

• Can spend your time catching fish (good 1)
or collecting coconuts (good 2)
• What is your optimal division of labor
between the two?
• Intuitively: if you're optimizing, you
couldn't reallocate your time in a way
that would make you better off.
• The last hour devoted to fish must
bring you the same amount of utility
as the last hour devoted to coconuts

Marginal Rate of Transformation (MRT)

• The  number of coconuts you need to give up in order to get another fish
• Opportunity cost of fish in terms of coconuts

Marginal Rate of Substitution (MRS)

• The number of coconuts you are willing to give up in order to get another fish
• Willingness to "pay" for fish in terms of coconuts

Both of these are measured in
coconuts per fish

(units of good 2/units of good 1)

Marginal Rate of Transformation (MRT)

• The  number of coconuts you need to give up in order to get another fish
• Opportunity cost of fish in terms of coconuts

Marginal Rate of Substitution (MRS)

• The number of coconuts you are willing to give up in order to get another fish
• Willingness to "pay" for fish in terms of coconuts

Opportunity cost of marginal fish produced is less than the number of coconuts
you'd be willing to "pay" for a fish.

Opportunity cost of marginal fish produced is more than the number of coconuts
you'd be willing to "pay" for a fish.

Better to spend less time fishing
and more time making coconuts.

Better to spend more time fishing
and less time collecting coconuts.

MRS
>
MRT
MRS
<
MRT

Better to produce
more good 1
and less good 2.

MRS
>
MRT
MRS
<
MRT

## “Gravitational Pull" Towards Optimality

Better to produce
more good 2
and less good 1.

These forces are always true.

In certain circumstances, optimality occurs where MRS = MRT.

# Graphical Analysis: PPFs and Indifference Curves

The story so far, in two graphs

Production Possibilities Frontier
Resources, Production Functions → Stuff

Indifference Curves
Stuff → Happiness (utility)

Both of these graphs are in the same "Good 1 - Good 2" space

Better to produce
more good 1
and less good 2.

MRS
>
MRT

Better to produce
less good 1
and more good 2.

MRS
<
MRT

# Mathematical Analysis: Lagrange Multipliers

We've just seen that, at least under certain circumstances, the optimal bundle is
"the point along the PPF where MRS = MRT."

CONDITION 1:
CONSTRAINT CONDITION

CONDITION 2:
TANGENCY
CONDITION

This is just an application of the Lagrange method!

(see other deck for worked examples)

## Next Time

Examine cases where the optimal bundle is not characterized by a tangency condition.

New concepts:
corner solutions and kinks.

By Chris Makler

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