Rational Choice
CHOICE
1.
2.
Scarcity (income constraint)
Tastes (indifference map/utility
function)
ECONOMIC RATIONALITY
 The
principal behavioral postulate is
that a decision-maker chooses its
most preferred alternative from those
available to it.
 The available choices constitute the
choice set.
 How is the most preferred bundle in
the choice set located/found?
RATIONAL CONSTRAINED CHOICE
x2
More preferred
bundles
Affordable
bundles
x1
RATIONAL CONSTRAINED CHOICE
x2
x2*
x1*
x1
RATIONAL CONSTRAINED CHOICE
x2
(x1*,x2*) is the most
preferred affordable
bundle.
x2*
E
x1*
x1
RATIONAL CONSTRAINED CHOICE
At Equilibrium E
MRS=x2/x1
=
p1/p2
Slope of the
indifference curve
Slope of the budget
constraint
Individual’s
willingness to trade
Society’s willingness
to trade
RATIONAL CONSTRAINED CHOICE
 The
most preferred affordable bundle
is called the consumer’s ORDINARY
DEMAND at the given prices and
income.
 Ordinary demands will be denoted by
x1*(p1,p2,m) and x2*(p1,p2,m).
RATIONAL CONSTRAINED CHOICE
x2
The slope of the
indifference curve at
(x1*,x2*) equals the
slope of the budget
constraint.
x2*
x1*
x1
RATIONAL CONSTRAINED CHOICE
(x1*,x2*) satisfies two conditions:
 (i) the budget is exhausted, i.e.
p1x1* + p2x2* = m; and
 (ii) the slope of the budget constraint,
(-) p1/p2, and the slope of the
indifference curve containing (x1*,x2*)
are equal at (x1*,x2*).
COMPUTING DEMAND


1.
2.
How can this information be used
to locate (x1*,x2*) for given p1, p2
and m?
Two ways to do this
Use Lagrange multiplier method
Find MRS and substitute into the
Budget Constraint
COMPUTING DEMAND
Lagrange Multiplier Method
Suppose that the consumer has CobbDouglas preferences
a 1 a
1 2
U ( x1 , x2 )  x x
and a budget constraint given by
p1 x1  p2 x2  m
COMPUTING DEMAND
Lagrange Multiplier Method
Aim
max
a 1  a 
x1 x2
subject to
p1 x1  p2 x2  m
Set up the Lagrangian
L
 x1 , x 2 ,  
 x1a x 21  a    m  p1 x1  p2 x 2 
COMPUTING DEMAND
Lagrange Multiplier Method
Differentiate
L
 ax1a 1 x21 a   p1  0
x1
(1)
L
 (1  a ) x1a x2  a   p2  0
x2
(2)
L
 m  p1 x1  p2 x2  0

(3)
COMPUTING DEMAND
Lagrange Multiplier Method
From (1) and (2)
ax1a-1 x21 a  1  a x1a x2 a
λ

p1
p2
Then re-arranging
p1
ax2

1  a x1
p2
COMPUTING DEMAND
Lagrange Multiplier Method
Rearrange
ap2 x 2
p1 x1 
1  a 
Remember
p1 x1  p2 x2  m
COMPUTING DEMAND
Lagrange Multiplier Method
Substitute
ap2 x 2
p1 x1 
1  a 
p1 x1  p2 x2  m
COMPUTING DEMAND
Lagrange Multiplier Method
Solve x1* and x2*
*
x1
and
*
x2
am

p1

1  a m

p2
COMPUTING DEMAND
Method 2
Suppose that the consumer has
Cobb-Douglas preferences.
U( x1 , x 2 )  x1axb2
COMPUTING DEMAND
Method 2
 Suppose
that the consumer has
Cobb-Douglas preferences.
U( x1 , x 2 )  x1axb2
U
MU1 
 ax1a  1xb2
 x1
U
MU2 
 bx1axb2  1
 x2
COMPUTING DEMAND
Method 2
 So
the MRS is
a1 b
dx 2
 U/ x1
ax1 x 2
ax 2
MRS 



.
a b 1
dx1
 U/ x 2
bx1
bx1 x 2
COMPUTING DEMAND
Method 2
 So
the MRS is
a1 b
dx 2
 U/ x1
ax1 x 2
ax 2
MRS 



.
a b 1
dx1
 U/ x 2
bx1
bx1 x 2
 At
(x1*,x2*), MRS = -p1/p2 , i.e. the
slope of the budget constraint.
COMPUTING DEMAND
Method 2
 So
the MRS is
a1 b
dx 2
 U/ x1
ax1 x 2
ax 2
MRS 



.
dx1
 U/ x 2
bx1
bx1axb2  1
 At
(x1*,x2*), MRS = -p1/p2 so
ax*2
p1


*
p2
bx1

* bp1 *
x2 
x1 .
ap2
(A)
COMPUTING DEMAND
Method 2
 (x1*,x2*)
also exhausts the budget so
*
*
p1x1  p2x 2  m.
(B)
COMPUTING DEMAND
Method 2
 So
now we know that
* bp1 *
x2 
x1
ap 2
p1x*1  p2x*2  m.
(A)
(B)
COMPUTING DEMAND
Method 2
So now we know that
* bp1 *
x2 
x1
ap 2
Substitute
p1x*1  p2x*2  m.
(A)
(B)
COMPUTING DEMAND
Method 2
So now we know that
* bp1 *
x2 
x1
ap 2
Substitute
p1x*1  p2x*2  m.
and get
bp1 *
*
p1x1  p2
x1  m.
ap2
This simplifies to ….
(A)
(B)
COMPUTING DEMAND
Method 2
x*1 
am
.
( a  b)p1
COMPUTING DEMAND
Method 2
am

(a  b) p1
Substituting for x1* in
x1*
p1x*1  p2x*2  m
then gives
x2*
bm

(a  b) p2
COMPUTING DEMAND
Method 2
So we have discovered that the most
preferred affordable bundle for a consumer
with Cobb-Douglas preferences
U( x1 , x 2 )  x1axb2
is
( x1* , x2* ) 
(
am
bm
,
(a  b) p1 (a  b) p2
)
COMPUTING DEMAND
Method 2: Cobb-Douglas
x2
U( x1 , x 2 )  x1axb2
*
x2 
bm
( a  b)p 2
x*1 
am
( a  b)p1
x1
Rational Constrained Choice
 But
what if x1* = 0 or x2* = 0?
 If either x1* = 0 or x2* = 0 then the
ordinary demand (x1*,x2*) is at a
corner solution to the problem of
maximizing utility subject to a budget
constraint.
Examples of Corner Solutions:
Perfect Substitutes
x2
MRS = -1
x1
Examples of Corner Solutions:
Perfect Substitutes
x2
MRS = -1
Slope = -p1/p2 with p1 > p2.
x1
Examples of Corner Solutions:
Perfect Substitutes
x2
MRS = -1
Slope = -p1/p2 with p1 > p2.
x1
Examples of Corner Solutions:
Perfect Substitutes
x2
m
x 
p2
*
2
MRS = -1 (This is the indifference
curve)
Slope = -p1/p2 with p1 > p2.
x*1  0
x1
Examples of Corner Solutions:
Perfect Substitutes
x2
MRS = -1
ANOTHER
EXAMPLE
Slope = -p1/p2 with p1 < p2.
x*2  0
x1* 
m
p1
x1
Examples of Corner Solutions:
Perfect Substitutes
So when U(x1,x2) = x1 + x2, the most
preferred affordable bundle is (x1*,x2*)
where
m 
*
*
( x1 , x2 )   ,0  if p1 < p2
 p1

or
 m
( x , x )   0, 
 p2 
*
1
*
2
if p1 > p2.
Examples of Corner Solutions:
Perfect Substitutes
x2
m
p2
MRS = -1
Slope = -p1/p2 with p1 = p2.
The budget
constraint and
the utility curve
lie on each other
m
p1
x1
Examples of Corner Solutions:
Perfect Substitutes
x2
y
p2
All the bundles in the
constraint are equally the
most preferred affordable
when p1 = p2.
y
p1
x1
Examples of ‘Kinky’ Solutions:
Perfect Complements
X2 (gin)
U(x1,x2) = min(ax1,x2)
x2 = ax1 (a = .5)
X1 (tonic)
Examples of ‘Kinky’ Solutions:
Perfect Complements
x2
U(x1,x2) = min(ax1,x2)
x2 = ax1
MRS = 0
x1
Examples of ‘Kinky’ Solutions:
Perfect Complements
x2
U(x1,x2) = min(ax1,x2)
MRS = - 
x2 = ax1
MRS = 0
x1
Examples of ‘Kinky’ Solutions:
Perfect Complements
x2
U(x1,x2) = min(ax1,x2)
MRS = - 
MRS is undefined
x2 = ax1
MRS = 0
x1
Examples of ‘Kinky’ Solutions:
Perfect Complements
x2
U(x1,x2) = min(ax1,x2)
x2 = ax1
x1
Examples of ‘Kinky’ Solutions:
Perfect Complements
x2
U(x1,x2) = min(ax1,x2)
Which is the most
preferred affordable bundle?
x2 = ax1
x1
Examples of ‘Kinky’ Solutions:
Perfect Complements
x2
U(x1,x2) = min(ax1,x2)
The most preferred
affordable bundle
x2 = ax1
x1
Examples of ‘Kinky’ Solutions:
Perfect Complements
x2
U(x1,x2) = min(ax1,x2)
x2 = ax1
x2*
x1*
x1
Examples of ‘Kinky’ Solutions:
Perfect Complements
x2
U(x1,x2) = min(ax1,x2)
and p1x1* + p2x2* = m
x2 = ax1
x2*
x1*
x1
Examples of ‘Kinky’ Solutions:
Perfect Complements
x2
U(x1,x2) = min(ax1,x2)
(a) p1x1* + p2x2* = m
(b) x2* = ax1*
x2 = ax1
x2*
x1*
x1
Examples of ‘Kinky’ Solutions:
Perfect Complements
(a) p1x1* + p2x2* = m; (b) x2* = ax1*
Examples of ‘Kinky’ Solutions:
Perfect Complements
(a) p1x1* + p2x2* = m; (b) x2* = ax1*.
Substitution from (b) for x2* in
(a) gives p1x1* + p2ax1* = m
Examples of ‘Kinky’ Solutions:
Perfect Complements
(a) p1x1* + p2x2* = m; (b) x2* = ax1*.
Substitution from (b) for x2* in
(a) gives p1x1* + p2ax1* = m
which gives
m
*
x1 
p1  ap2
Examples of ‘Kinky’ Solutions:
Perfect Complements
(a) p1x1* + p2x2* = m; (b) x2* = ax1*.
Substitution from (b) for x2* in
(a) gives p1x1* + p2ax1* = m
which gives
x1*
m
am
*

; x2 
p1  ap2
p1  ap2
Examples of ‘Kinky’ Solutions:
Perfect Complements
(a) p1x1* + p2x2* = m; (b) x2* = ax1*.
Substitution from (b) for x2* in
(a) gives p1x1* + p2ax1* = m
which gives
m
*
*
am
x1 
; x2 
p1  ap2
p1  ap2
Examples of ‘Kinky’ Solutions:
Perfect Complements
x2
U(x1,x2) = min(ax1,x2)
*
x2 
x2 = ax1
am
p1  ap 2
x*1 
m
p1  ap 2
x1