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Chapter 5Choice
2
Economic Rationality
The principal behavioral postulate is that a decisionmaker chooses his most preferred alternative from those available to him.
The available choices constitute the choice set.
How is the most preferred bundle in the choice set located?
3
Rational Constrained Choice
x1
x2
4
Rational Constrained Choice
x1
x2
Affordablebundles
5
Rational Constrained Choice
x1
x2
Affordablebundles
More preferredbundles
6
Rational Constrained Choice
Affordablebundles
x1
x2
More preferredbundles
7
Rational Constrained Choice
x1
x2
x1*
x2*
(x1*,x2*) is the mostpreferred affordablebundle.
8
Rational Constrained Choice The most preferred affordable bundle
is called the consumer’s ordinary demand at the given prices and budget.
Ordinary demands will be denoted byx1*(p1,p2,m) and x2*(p1,p2,m).
9
Rational Constrained Choice
When x1* > 0 and x2* > 0 the demanded bundle is interior.
If buying (x1*,x2*) costs $m then the budget is exhausted.
10
Rational Constrained Choice
x1
x2
x1*
x2*
(x1*,x2*) is interior.(a) (x1*,x2*) exhausts thebudget; p1x1* + p2x2* = m.
11
Rational Constrained Choice
x1
x2
x1*
x2*
(x1*,x2*) is interior .(b) The slope of the indiff.curve at (x1*,x2*) equals the slope of the budget constraint.
12
Rational Constrained Choice (x1*,x2*) satisfies two conditions: (a) the budget is exhausted;
p1x1* + p2x2* = m (b) the slope of the budget
constraint, -p1/p2, and the slope of the indifference curve containing (x1*,x2*) are equal at (x1*,x2*).
13
Rational Constrained Choice Condition (b) can be written as:
That is,
Therefore, at the optimal choice, the ratio of marginal utilities of the two commodities must be equal to their price ratio.
2 1 1
1 2 2
dx MU pMRS
dx MU p
1 1
2 2
MU p
MU p
14
Mathematical Treatment
The consumer’s optimization problem can be formulated mathematically as:
One way to solve it is to use Lagrangian method:
15
Mathematical Treatment Assuming the utility function is
differentiable, and there exists an interior solution.
The first order conditions are:
16
Mathematical Treatment
So, for interior solution, we can solve for the optimal bundle using the two conditions:
1 1
2 2
1 1 2 2
,
.
MU p
MU p
p x p x m
17
Computing Ordinary Demands - a Cobb-Douglas Example. Suppose that the consumer has
Cobb-Douglas preferences.
Then
U x x x xa b( , )1 2 1 2
MUUx
ax xa b1
1112
MUUx
bx xa b2
21 2
1
18
Computing Ordinary Demands - a Cobb-Douglas Example. 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
112
1 21
2
1
//
.
ax
bx
pp
xbpap
x2
1
1
22
1
21
*
** *. (A)
19
Computing Ordinary Demands - a Cobb-Douglas Example. So now we have
Solving these two equations, we have
xbpap
x21
21
* * (A)
p x p x m1 1 2 2* * . (B)
xam
a b p11
*
( ).
x
bma b p2
2
*
( ).
20
Computing Ordinary Demands - a Cobb-Douglas Example. So we have discovered that the most
preferred affordable bundle for a consumer with Cobb-Douglas preferences
is
U x x x xa b( , )1 2 1 2
( , )( )
,( )
.* * ( )x xam
a b pbm
a b p1 21 2
21
A Cobb-Douglas Example
x1
x2
xam
a b p11
*
( )
x
bma b p
2
2
*
( )
U x x x xa b( , )1 2 1 2
22
A Cobb-Douglas Example Note that with the optimal bundle:
The expenditures on each good:
It is a property of Cobb-Douglas Utility function that expenditure share on a particular good is a constant.
( , )( )
,( )
.* * ( )x xam
a b pbm
a b p1 21 2
.)(
,)(
),( *22
*11
mba
bm
ba
axpxp
23
Special Cases
24
Special Cases
25
Some Notes
For interior optimum, tangency condition is only necessary but not sufficient.
There can be more than one optimum.
Strict convexity implies unique solution.
26
Corner Solution
What if x1* = 0?
Or if 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.
27
Examples of Corner Solutions -- the Perfect Substitutes Case
x1
x2
MRS = -1
28
Examples of Corner Solutions -- the Perfect Substitutes Case
x1
x2
MRS = -1
Slope = -p1/p2 with p1 > p2.
29
Examples of Corner Solutions -- the Perfect Substitutes Case
x1
x2
x1 0*
MRS = -1
Slope = -p1/p2 with p1 > p2.
2
*2 p
mx
30
Examples of Corner Solutions -- the Perfect Substitutes Case
x1
x2
x2 0*
MRS = -1
Slope = -p1/p2 with p1 < p2.
1
*1 p
mx
31
Examples of Corner Solutions -- the Perfect Substitutes Case So when U(x1,x2) = x1 + x2, the most
preferred affordable bundle is (x1*,x2*) where
and
0,),(
1
*2
*1 p
mxx if p1 < p2
2
*2
*1 ,0),(
p
mxx if p1 > p2
32
Examples of Corner Solutions -- the Perfect Substitutes Case
x1
x2
MRS = -1
Slope = -p1/p2 with p1 = p2.
1p
m
2p
m
33
Examples of Corner Solutions -- the Perfect Substitutes Case
x1
x2
All the bundles in the constraint are equally the most preferred affordable when p1 = p2.
1p
m
2p
m
34
Examples of Corner Solutions -- the Non-Convex Preferences Case
x1
x2Better
35
Examples of Corner Solutions -- the Non-Convex Preferences Case
x1
x2
36
Examples of Corner Solutions -- the Non-Convex Preferences Case
x1
x2Which is the most preferredaffordable bundle?
37
Examples of Corner Solutions -- the Non-Convex Preferences Case
x1
x2
The most preferredaffordable bundle
38
Examples of Corner Solutions -- the Non-Convex Preferences Case
x1
x2
The most preferredaffordable bundle
Notice that the “tangency solution” is not the most preferred affordable bundle.
39
Examples of ‘Kinky’ Solutions -- the Perfect Complements Case
x1
x2 U(x1,x2) = min{ax1,x2}
x2 = ax1
40
Examples of ‘Kinky’ Solutions -- the Perfect Complements Case
x1
x2
MRS = -
MRS = 0
MRS is undefined
U(x1,x2) = min{ax1,x2}
x2 = ax1
41
Examples of ‘Kinky’ Solutions -- the Perfect Complements Case
x1
x2U(x1,x2) = min{ax1,x2}
x2 = ax1
42
Examples of ‘Kinky’ Solutions -- the Perfect Complements Case
x1
x2 U(x1,x2) = min{ax1,x2}
x2 = ax1
Which is the mostpreferred affordable bundle?
43
Examples of ‘Kinky’ Solutions -- the Perfect Complements Case
x1
x2 U(x1,x2) = min{ax1,x2}
x2 = ax1
The most preferredaffordable bundle
44
Examples of ‘Kinky’ Solutions -- the Perfect Complements Case
x1
x2
U(x1,x2) = min{ax1,x2}
x2 = ax1
x1*
x2*
45
Examples of ‘Kinky’ Solutions -- the Perfect Complements Case
x1
x2 U(x1,x2) = min{ax1,x2}
x2 = ax1
x1*
x2*
(a) p1x1* + p2x2* = m(b) x2* = ax1*
46
Examples of ‘Kinky’ Solutions -- the Perfect Complements Case
(a) p1x1* + p2x2* = m; (b) x2* = ax1*.
Solving the two equations, we have
.app
amx;
appm
x21
*2
21
*1
47
Examples of ‘Kinky’ Solutions -- the Perfect Complements Case
x1
x2U(x1,x2) = min{ax1,x2}
x2 = ax1
xm
p ap11 2
*
x
amp ap
2
1 2
*