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Rational Choice
CHOICE
1. Scarcity (income constraint)
2. 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
Affordablebundles
x1
x2
More preferredbundles
RATIONAL CONSTRAINED CHOICE
x1
x2
x1*
x2*
RATIONAL CONSTRAINED CHOICE
x1
x2
x1*
x2*
(x1*,x2*) is the mostpreferred affordablebundle.
E
RATIONAL CONSTRAINED CHOICE
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
At Equilibrium E
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 byx1*(p1,p2,m) and x2*(p1,p2,m).
RATIONAL CONSTRAINED CHOICE
x1
x2
x1*
x2*
The slope of the indifference curve at (x1*,x2*) equals the slope of the budgetconstraint.
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
How can this information be used to locate (x1*,x2*) for given p1, p2 and m?
Two ways to do this
1. Use Lagrange multiplier method
2. Find MRS and substitute into the Budget Constraint
COMPUTING DEMAND Lagrange Multiplier Method
Suppose that the consumer has Cobb-Douglas preferences
and a budget constraint given by
mxpxp 2211
aaxxxxU 12121 ),(
COMPUTING DEMAND Lagrange Multiplier Method
Aim
Set up the Lagrangian
mxpxpxx aa 2211
121 tosubject max
22111
21,,
21
xpxpmxxL aa
xx
COMPUTING DEMAND Lagrange Multiplier Method
Differentiate
(3) 0
(2) 0)1(
(1) 0
2211
2212
11
21
11
xpxpmL
pxxax
L
pxaxx
L
aa
aa
COMPUTING DEMAND Lagrange Multiplier Method
From (1) and (2)
Then re-arranging
2
21
1
12
11 1
p
xxa
p
xaxλ
aaaa-
1
2
2
1
1 xa
ax
p
p
COMPUTING DEMAND Lagrange Multiplier Method
Rearrange
Remember
axap
xp
1
2211
mxpxp 2211
COMPUTING DEMAND Lagrange Multiplier Method
Substitute
axap
xp
1
2211
mxpxp 2211
COMPUTING DEMAND Lagrange Multiplier Method
Solve x1* and x2*
and 1
*1 p
amx
2
*2
1
p
max
COMPUTING DEMAND Method 2
Suppose that the consumer has Cobb-Douglas preferences.
U x x x xa b( , )1 2 1 2
COMPUTING DEMAND Method 2
Suppose that the consumer has Cobb-Douglas preferences.
U x x x xa b( , )1 2 1 2
MUUx
ax xa b1
11
12
MUUx
bx xa b2
21 2
1
COMPUTING DEMAND Method 2
So the MRS is
MRSdxdx
U xU x
ax x
bx x
axbx
a b
a b
2
1
1
2
11
2
1 21
2
1
//
.
COMPUTING DEMAND Method 2
So the MRS is
At (x1*,x2*), MRS = -p1/p2 , i.e. the slope of the budget constraint.
MRSdxdx
U xU x
ax x
bx x
axbx
a b
a b
2
1
1
2
11
2
1 21
2
1
//
.
COMPUTING DEMAND Method 2
So the MRS is
At (x1*,x2*), MRS = -p1/p2 so
MRSdxdx
U xU x
ax x
bx x
axbx
a b
a b
2
1
1
2
11
2
1 21
2
1
//
.
ax
bx
pp
xbpap
x2
1
1
22
1
21
*
** *. (A)
COMPUTING DEMAND Method 2
(x1*,x2*) also exhausts the budget so
p x p x m1 1 2 2* * . (B)
COMPUTING DEMAND Method 2
So now we know that
xbpap
x21
21
* * (A)
p x p x m1 1 2 2* * . (B)
COMPUTING DEMAND Method 2
So now we know that
xbpap
x21
21
* * (A)
p x p x m1 1 2 2* * . (B)
Substitute
COMPUTING DEMAND Method 2
So now we know that
xbpap
x21
21
* * (A)
p x p x m1 1 2 2* * . (B)
p x pbpap
x m1 1 21
21
* * .
Substitute
and get
This simplifies to ….
COMPUTING DEMAND Method 2
xam
a b p11
*
( ).
COMPUTING DEMAND Method 2
2
*2 )( pba
bmx
Substituting for x1* in p x p x m1 1 2 2
* *
then gives
1
*1 )( pba
amx
COMPUTING DEMAND Method 2
So we have discovered that the mostpreferred affordable bundle for a consumerwith Cobb-Douglas preferences
U x x x xa b( , )1 2 1 2
is )(21
*2
*1 )(
,)(
),(pba
mb
pba
maxx
COMPUTING DEMAND Method 2: Cobb-Douglas
x1
x2
xam
a b p11
*
( )
x
bma b p
2
2
*
( )
U x x x xa b( , )1 2 1 2
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
x1
x2
MRS = -1
Examples of Corner Solutions: Perfect Substitutes
x1
x2
MRS = -1
Slope = -p1/p2 with p1 > p2.
Examples of Corner Solutions: Perfect Substitutes
x1
x2
MRS = -1
Slope = -p1/p2 with p1 > p2.
Examples of Corner Solutions: Perfect Substitutes
x1
x2
2
*2 p
mx
x1 0*
MRS = -1 (This is the indifference curve)
Slope = -p1/p2 with p1 > p2.
Examples of Corner Solutions: Perfect Substitutes
x1
x2
1
*1 p
mx
x2 0*
MRS = -1
Slope = -p1/p2 with p1 < p2.
ANOTHER EXAMPLE
Examples of Corner Solutions: Perfect Substitutes
So when U(x1,x2) = x1 + x2, the mostpreferred affordable bundle is (x1*,x2*)where
0,),(
1
*2
*1 p
mxx
or
2
*2
*1 ,0),(
p
mxx
if p1 < p2
if p1 > p2.
Examples of Corner Solutions: Perfect Substitutes
x1
x2
MRS = -1
Slope = -p1/p2 with p1 = p2.
1p
m
2p
m
The budget constraint and the utility curve lie on each other
Examples of Corner Solutions: Perfect Substitutes
x1
x2
All the bundles in the constraint are equally the most preferred affordable when p1 = p2.
yp2
yp1
Examples of ‘Kinky’ Solutions: Perfect Complements
X1 (tonic)
X2 (gin) U(x1,x2) = min(ax1,x2)
x2 = ax1 (a = .5)
Examples of ‘Kinky’ Solutions: Perfect Complements
x1
x2
MRS = 0
U(x1,x2) = min(ax1,x2)
x2 = ax1
Examples of ‘Kinky’ Solutions: Perfect Complements
x1
x2
MRS = -
MRS = 0
U(x1,x2) = min(ax1,x2)
x2 = ax1
Examples of ‘Kinky’ Solutions: Perfect Complements
x1
x2
MRS = -
MRS = 0
MRS is undefined
U(x1,x2) = min(ax1,x2)
x2 = ax1
Examples of ‘Kinky’ Solutions: Perfect Complements
x1
x2U(x1,x2) = min(ax1,x2)
x2 = ax1
Examples of ‘Kinky’ Solutions: Perfect Complements
x1
x2U(x1,x2) = min(ax1,x2)
x2 = ax1
Which is the mostpreferred affordable bundle?
Examples of ‘Kinky’ Solutions: Perfect Complements
x1
x2U(x1,x2) = min(ax1,x2)
x2 = ax1
The most preferredaffordable bundle
Examples of ‘Kinky’ Solutions: Perfect Complements
x1
x2U(x1,x2) = min(ax1,x2)
x2 = ax1
x1*
x2*
Examples of ‘Kinky’ Solutions: Perfect Complements
x1
x2U(x1,x2) = min(ax1,x2)
x2 = ax1
x1*
x2*
and p1x1* + p2x2* = m
Examples of ‘Kinky’ Solutions: Perfect Complements
x1
x2U(x1,x2) = min(ax1,x2)
x2 = ax1
x1*
x2*
(a) p1x1* + p2x2* = m(b) x2* = ax1*
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* = mwhich gives
21
*1 app
mx
Examples of ‘Kinky’ Solutions: Perfect Complements
(a) p1x1* + p2x2* = m; (b) x2* = ax1*.
Substitution from (b) for x2* in (a) gives p1x1* + p2ax1* = mwhich gives
21
*2
21
*1 ;
app
amx
app
mx
Examples of ‘Kinky’ Solutions: Perfect Complements
(a) p1x1* + p2x2* = m; (b) x2* = ax1*.
Substitution from (b) for x2* in (a) gives p1x1* + p2ax1* = mwhich gives
21
*2
21
*1 ;
app
amx
app
mx
Examples of ‘Kinky’ Solutions: Perfect Complements
x1
x2U(x1,x2) = min(ax1,x2)
x2 = ax1
xm
p ap11 2
*
x
amp ap
2
1 2
*