### L03_Utility_Max_2014

```Utility Maximization
A Utility Function Mathematically
Representing Preferences
U
U=U(x, y)
U(A)
Y
U(B)
Utility functions have
U(A)  U(B) iff A
U(A)  U(B) iff A
B
B
A
Indifference curves
describe bundle-ordering
preferences
B
X
Consumption Opportunities: The Budget
Constraint
• Assume that an individual has I dollars
to allocate between good x and good y
pxx + pyy  M
y
M
py
The individual can afford
to choose only combinations
of x and y in the shaded
triangle
x
M
px
The Budget Constraint
•
M px
y

x
py py
• MC of consuming one more unit of x, the amount of y
that must be foregone.
• The slope of the budget line is this MC.
y
M
py
Slope 
p
y
 x
x
py
The slope is the change
in y for a one unit increase
in the consumption of x.
If Px = 10 and Py = 5, then
consuming one more x means
consuming two less y.
x
M
px
Maximizing Utility
• Keep buying x until the MB(x) = MC(x)
• Interaction of…
MB, MC
Not an indifference curve!
MC
MB
X
X*
– Preferences, diminishing MB because of diminishing MRS. MB
= MRS
• MB in terms of y willing to be given up
• In dollars, MB = py*MRS
– MC of x = px/py
• MC in terms of Y given up
• In dollars, MC = px.
Optimization Principle
• To maximize utility, given a fixed amount of income, an
individual will buy the goods and services:
– That exhaust total income
• Savings or borrowing is allowed (if we modify the budget
constraint to include a temporal component)
– So long as MB(x) ≥ MC(x), MB(y) ≥ MC(y), etc.
– Or, until MB(x) = MC(x), MB(y) = MC(y)
Intuition
• MRS is the maximum amount of y the person is willing to give up to consume
more x; the definition of MB.
• px/py tells us the number of units of y that must be given up to consume one
more x; the definition of MC.
y
At “A”, MRS>Px/Py (MB > MC),
At “B”, MRS<Px/Py, (MB < MC)
You are willing to pay more than
you have to, consumer surplus
increases.
Utility and consumer surplus
can be increased by consuming
less x.
A
Utility and consumer surplus
can be increased by consuming
more x.
Px = 10
Py = 5
Slope of budget line = -2
B
U0
x
Intuition
• At “C”, the MB = MC for the last unit of both
goods consumed.
• That is, at “C”, MRS = px/py, or
U x px

U y py
y
Ux U y

px p y
A
C
U1
B
U0
x
Optimization
• Unconstrained optimization is a lot easier to
solve than constrained optimization.
– Substitution: maximize the cross section of U
along the budget line
– Lagrange method
Substitution
• This turns the constrained
optimization into an
unconstrained problem.
• Find the equation for the
cross section of the U=U(x,y)
above the budget line and
maximize it -- i.e. find the top
of the parabola
U
y
y*
x*
x
Substitution
• Substitute and maximize
U  x y; M=p x x+p y y
dU  Mx 1

dx
py
 Mx 1
py
 Mx px x 1
x 


py
py

  1 px x


0
py
And substitute
again

p
 M
Ux 
 x
p
 y py
  1 px x


py
 Mp y
x

 1
x
  1 px p y
x
M
  1 px
 M 
M=p x 
 +p y y
   1 px 
M
py y  M 
  1
M   1   M
py y 
  1
y
M
  1 p y
Problem with this method
• It can get very mathematically complicated
very quickly.
• Even U=xαyβ gets very tricky to solve.
LaGrange Method
• LaGrange knew that unconstrained
optimization (like profit max) is relatively
simple compared to constrained optimization.
• Taking what he knew unconstrained
optimization he attempted to simplify the
constrained maximization problem by making
it mimic the unconstrained problem.
Unconstrained Optimization Example
max v  f ( x, y )  g ( x, y )
FOC
v x  f x ( x , y )  g x ( x, y )  0  f x  g x
v y  f y ( x, y )  g y ( x, y )  0  f y  g y
Maximizing v  f ( x, y )  g ( x, y ) means
fx  gx
fy  gy
• Profit maximization is an example of this. We maximize
the difference between two functions: π=R(q) – C(q).
Lagrange Method
• LaGrange wanted to find a way that was simpler
than constrained optimization and more
workable than simple substitution.
– He wanted to make constrained optimization take the
form of the simpler unconstrained problem.
max v  f ( x, y)  g ( x, y)
• First, let’s look at a simpler problem.
max v  f ( x)  g ( x)
Maximize Utility - Expenditure
• Maximize utility minus the cost of buying
bundles. Think about a one-good world.
U
U=U(x)
slope = Ux
Expenditure = E = pxx
max v  f ( x)  g ( x)
slope = Ex= px
max v  U ( x)  px x
x*
x
Problems:
• Ux not measured in \$ like E.
• E is not constrained, we can spend as much as we like.
Maximizing Utility - Expenditure
• First change the expenditure function by multiplying px
by λ. Now call that function EU.
U
U=U(x)
slope = Ux
Slope = EUx= λpx
EU=λpxx
max v  U ( x)  EU ( x)
max v  U ( x)   px x
x*
x
• We want λ to measure the marginal utility of \$1.
– In that case, units of x consumed would cost us utility and both
U(x) and EU(x) would be measured in the same units.
Maximizing Utility - Expenditure
• Problem, an infinite number of λ choices that will each solve this
with a different x*
EUx = λ1 px
U
U=U(x)
EUx= λ2 px
EUx = λ3 px
x*
x*
x*
x
• Now the slope of the expenditure function and expenditure are
measured in utils, not dollars. But we are not constraining x yet.
LaGrange Method
• So first subtract λM from the expenditure function to get
EL = λpxx - λM
U
U=U(x)
slope = Ux
EU = λpxx
EL = λpxx - λM
x*
-λM
slope = ELx= λpx
x
LaGrange Method
• We know we want to find the x* such that that distance between
U(x) and EL(x*) = U(x*). That is, where EL(x*) = 0
• So we maximize v = U(x) - 0
• Substitute λpxx – λM = 0 in for 0, to constrain x* to our budget.
U
U=U(x)
slope = Ux
max L  U ( x)  E ( x)
L
max L  U ( x)   ( px x  M )
U=U(x*)-0
EL = λpxx - λM
0
x*
-λI
slope =
ELx= λ px
x
LaGrange Method
• Our optimization becomes an unconstrained problem by including
the requirement that λpxx = λM.
• λ is chosen along with x to maximize utility so that λ = the marginal
utility of \$1. That is, λpx = Ux.
U
U=U(x)
slope = Ux
max L  U ( x)   ( px x  M )
U=U(x*)-0
EL = λ(pxx – M)
0
x*
-λM
slope = ELx= λ px
x
LaGrange Method
max L  U ( x)   ( px x  I ) or, equivalently
max L  U ( x)   ( I  px x)
FOC
Lx : U x   px  0
L : I  px x  0
L x is the condition that f x  g x when we
maximize v  f ( x) - g ( x)
L ensures that the solution satisfies  px x   I ,
i.e. that L( x*)  U ( x*)
Two Goods: Lagrange’s Manufactured
Plane
U
• To maximize utility, maximize the
height of the utility function
above the plane
EL = λpxx + λpyy – λM
• Such that
λpxx + λpyy – λM = 0
U = U(x,y)
y
When
x = 0 and
y = 0,
U = - λM
ELy= λ py
ELx= λ px
x
LaGrange Plane
EL=g(x,y)
EL= λpxx+ λpyy- λM
g’x=ELx= λ px
g’y=ELy= λ py
Lagrange Method
U
U = U(x,y)
max v  U ( x, y)  0, such that 0 = ( px x  py y  M )
max L  U ( x, y )   ( px x  p y y  M )
max L  U ( x, y )   ( M  px x  p y y)
y
x
UL=g(x,y) = 0
λ(pxx+pyy – M) = 0
Basic Demand Analysis
• Using Lagrangian to generate ordinary
(Marshallian) demand curves.
– FOCs necessary
– SOCs sufficient (check that they hold)
– Ordinary (Marshallian) demand curves
– Inverse demand curves
– Meaning of λ
– Indirect Utility
– Expenditure Function
– Comparative Statics General Results
Demand Functions using Lagrange’s
Method
• Set up and maximize:
L  (x, y)  U(x, y)  (M  p x x  p y y)
FOC: necessary conditions for a maximum
λ* chosen so that the constraint plane is
L x  U x  p x  0  U x  p x
L y  U y  p y  0  U y  p y
parallel to the utility function.
L  M  p x x  p y y  0  M  p x x  p y y
Solve to get two interesting results
Ux px
 , tangency condition
Uy py
Any x* and y* that maximizes utility
will also have to exhaust income.
Ux Uy

, bang for the buck the same for last unit
px
py
FOCs for an Optimum
• For utility to be maximized, it is necessary that the
indifference curve is tangent to the budget constraint (as
above). U x px
Uy

py
• But it is not sufficient, we also need a diminishing MRS.
y
Utility Maximized
y
Utility Maximized
y
FOCs
satisfied
FOCs satisfied
x
x
x
SOCs for an Optimum
• Sufficient condition for a maximum to exist
– If the MRS is non-increasing (utility function quasi concave) for all x, that is
sufficient for a maximum to exist – but it may not be unique.
– If the MRS is diminishing (utility function strictly quasi concave) for all x, that is
sufficient for a unique maximum to exist. Need this to satisfy second order
conditions for maximization.
SOCs do not hold
y
Utility Maximized
y
y
SOCs
satisfied
x
x
x
Expenditure Minimization: SOC
• The FOC ensure that the optimal consumption
bundle is at a tangency.
• The SOC ensure that the tangency is a minimum, and
not a maximum by ensuring that away from the
tangency, along the budget line, utility falls.
U*>U’
Y
U=U*
U=U’
X
Checking SOC:
utility function strictly quasi-concave
• The second order conditions will hold if the utility
function is strictly quasi-concave
– A function is strictly quasi-concave if its bordered
Hessian is negative definite. That is:
H 
0
Ux
U x U xx
0
Ux
Uy
 0 and H  U x U xx U xy  0
U y U yx U yy
• A function is strictly quasi-concave if:
1.
2.
-UxUx < 0
2UxUxyUy - Uy2Uxx - Ux2Uyy > 0
Checking SOC:
Constrained Maximization
• The second order condition for constrained maximization will
hold if the following bordered Hessian matrix is negative definite:
Note:
let L( x, y )  U ( x, y )   g ( x, y )
 L

H   L x
 L y

H 
0
L x
Lxx
Lyx
px
px U xx
Uy
Ux
, py 


So this Hessian and
The last only differ
by 
px 
L y 

Lxy 
Lyy 
0
px
py
 0 and H  px U xx U xy  0
p y U yx U yy
 px 2  0, and 2px p yU xy  px 2U yy  p y 2U xx  0
Ordinary Demand Curves
• And from the FOC:
From these three equations and unknowns:
Lx : U x   px
U
p
Ly : U y   p y
MRS 
x
Uy

x
py
, solve for y to get income consumption curve
L : M  px x  p y y
Solve to get the following:
x*  x( px , p y , M ), ordinary or Marshallian demand
y*  y ( px , p y , M ), ordinary or Marshallian demand
Ux U y
*   ( px , p y , M ), = = , marginal utility of \$1
px p y
Inverse Demand Curves
• Starting with the ordinary demand curves:
Solve to get the following inverse demand equations:
p*x  px ( x, p y , M )
Recognize that at the optimal bundle
p*x
 MRS
*
py
And p*x  p*y  MRS
So the inverse demand curve tells us the
(willingness to give up y for another x)  p y .
I.e. the dollar value of the y that the individual is
willing to give up for an x.
Utility and Indirect Utility
• Maximum Utility, a function of quantities
U *  U  x* , y * 
• Indirect Utility a function of price and income
Once we plug in
x*  x( px , p y , M ),
y*  y ( px , p y , M ) and get
utility as a function of only prices and income
V *  V  x( px , p y , M ), y ( px , p y , M ) 
V *  V  px , p y , M 
Optimization :
Expenditure Function
V *  V  px , p y , M 
• Solve for M:

M *  E px , p y ,U

• This equation determines the expenditure
needed to generate Ū, the expenditure
function:

E *  E px , p y ,U

Digression: Envelope Theorem
• Say we know that y = f(x; ω)
– We find y is maximized at x* = x(ω)
• So we know that y* = y(x*=x(ω),ω)).
• Now say we want to find out
*
*
*
dy*
d
*
dy
dy dy dx

 *
d d dx d
• So when ω changes, the optimal x changes, which
changes the y* function.
• Two methods to solve this…
Digression: Envelope Theorem
• Start with: y = f(x; ω) and calculate x* = x(ω)
• First option:
• y = f(x; ω), substitute in x* = x(ω) to get y* = y(x(ω); ω):
*
dy
x(), 

dy

 y* ()
d
d
*
• Second option, turn it around:
then substitute x* = x(ω)
f  x, 

 y (x; )


• First, take y
*
dy
 x(),   y* ()
into yω(x ; ω) to get dy 

d
d
*
•And these two answers are equivalent:
y* ()  y* ()
Envelope Result, 1st option to get
U*
 *
M
• Plug the optimal values into the LaGrangian
L*  U  x(p x , p y , M), y(p x , p y , M)   (p x , p y , M)  M  p x x(p x , p y , M)  p y y(p x , p y , M) 
Differentiate with respect to M
L*
x *
y*  * 
x *
y* 
* 
 Ux
 Uy
  1  p x
 py

   M  p x x *  p y y *
M
M
M  
M
M 
M 
L*
x *
y*
*
*
*
*
  Ux   px 
  Uy   py 
    M  p x x *  p y y *
M
M
M
M
And as we are at a maximum, the FOC get us:
L*
x *
y*
*
L*
*
  0
  0
   0

 *
M
M
M
M
M
U
As L=U (because M - p x x - p y y  0),
 (p x , p y , M)
M
In other words, when income rises by \$1, you gain \$1 worth of utility.
Optimization: Envelope Result
U*
 *x*
p x
• Plug the optimal values into the LaGrangian
L  U  x(p , p , M), y(p , p , M)   (p , p , M)  M  p x(p , p , M)  p y(p , p , M) 
*
x
y
x
y
x
y
x
x
y
y
x
y
Differentiate with respect to px
*
 *

x* *
y* 
*
* 
   p x
 x  py
   M  px x  py y  
p x
p x 
p x 
 
*
L*
x*
y* * *
*
*
*
* 
  Ux   px 
  Uy   py 
  x   M  px x  py y 
p x
p x
p x
p x
L*
x*
y*
 Ux
 Uy
p x
p x
p x
And as we are at a maximum, the FOC get us:
L*
x*
y* * *
* L*
  0
  0
  x  0

 * x
p x
p x
p x
p x p x
U x L*
U
U*
As L=U and as   ,
 x , means that
  x x(p x , p y , M)
p x p x
p x
px
In other words, when the price of x rises by \$1, you lose \$1 worth of utility for every x bought.
Optimization: Comparative Statics
• If we have a specified utility function and we
derive the equations for the demand
functions, the comparative statics are easy.
– Take the derivatives to calculate the changes in x
and y when prices or income change.
• However, what if all we know is U = U(x, y) and
we feel safe only assuming:
Ux > 0
Uy > 0
Uxx < 0
Uyy < 0
• Can we get anything from that?
Optimization: Comparative Statics
L  (x, y)  U(x, y)  (M  p x x  p y y)
FOC: necessary conditions for a maximum
L x  U x (x, y)  p x  0
L y  U y (x, y)  p y  0
L  M  p x x  p y y  0
And the equations for utility maximizing x, y, 
x*  x(p x , p y , M)
y*  y(p x , p y , M)
*  (p x , p y , M)
Comparative Statics:
Utility Maximizing x*, y*, λ*
Substitute equations for x*, y* and * into the FOC
(1) U x  x(p x , p y , M), y(p x , p y , M)   (p x , p y , M)p x  0
(2) U y  x(p x , p y , M), y(p x , p y , M)   (p x , p y , M)p y  0
(3) M  p x x(p x , p y , M)  p y y(p x , p y , M)  0
Whatever happens to prices or income, consumption
Comparative Statics: Effect of a change in M
Differentiate (1), (2), (3) w.r.t. M

y
x
0
 px
 U xy
M
M
M

y
x
0
 py
 U yy
U yx
M
M
M
y
x
0
 py
1  px
M
M
Rearrange
y
x

 1
 py
 px
0
M
M
M
y
x

0
 U xy
 U xx
p x
M
M
M
y
x

0
 U yy
 U yx
p y
M
M
M
U xx
Side note:
x
y
px
 py
1
M
M
Tells us that if income
increases by \$1, so will
total expenditure.
Comparative Statics: Effect of a change in M
Put in Matrix Notation
• Solve for Mx
0
p x
p x
U xx
p y
U yx

p y M 1
x
U xy •
 0
M
0
U yy y
M
Assuming H  2p x p y U xy  p 2x U yy  p 2y U xx  0
0
1 p y
p x
0
U xy
x  p y

M
0
U yy
H



?

p y U xy  p x U yy
()
X could be either normal or inferior.
 0

Comparative Statics: Effect of a change in I
Put in Matrix Notation
• Solve for
¶y
¶M
0
p x
p x
U xx
p y
U yx

 p y M 1
x
U xy •
 0
M
0
U yy y
M
Assuming H  2p x p y U xy  p 2x U yy  p 2y U xx  0
0
p x
p x
U xx
1
0
y  p y

M
U yx
0
H



?

p x U yx  p y U xx
()
Y could be either normal or inferior.
 0

Comparative Statics: Effect of a change in px
Differentiate (1), (2), (3) w.r.t. px
U xx
x
y

 U xy
 px
  0
p x
p x
p x
U yx
x
y

 U yy
 py
0
p x
p x
p x
p x
x
y
 x  py
0
p x
p x
Rearrange
  
x
y
0

p

p
x
x
y

p x
p x
 p x 

x
y
p x
 U xx
 U xy

p x
p x
p x
p y

x
y
 U yx
 U yy
0
p x
p x
p x
Comparative Statics: Effect of a change in px
Put in Matrix Notation
• Solve for px
x
0
p x
p y
p x
U xx
U xy 
p y
U yx
U yy

p x
x
x

p x
0
y
p x
Assuming H  2p x p y U xy  p 2x U yy  p 2y U xx  0
x

p x
0
x p y
p x
 U xy
p y
0 U yy
H
X could be giffen.


?

 



 x U xy p y  x p x U yy   p y 2
()
 0

Comparative Statics: Effect of a change in px
Put in Matrix Notation… AGAIN
• Solve for
¶y
¶ px
0
p x
p y
p x
U xx
U xy 
p y
U yx
U yy

p x
x
x

p x
0
y
p x
Assuming H  2p x p y U xy  p 2x U yy  p 2y U xx  0
0
p x
x
p x
U xx

 p y U yx
y

p x
H
0

 
 
?
 

p x p y   x p x U yx  x p y U xx 

0

()
X and y could be compliments or substitutes.
Comparative Statics:
Preview of income and substitution effects
Rearrange
these
x

p x


()





2
U
 p y  x  p y U xy  p x yy 




()
 
 

?
?

 
x p y U xy  p x U yy 

0;

M
()
x

p x

?

 x 
 p y 2  x 

M
()




 
 
?



p x p y  x  p x U yx  p y U xx 
y p x p y   x p x U yx  x p y U xx




p x
()
()


Income
effect
matters

 
 x U xy p y  x p x U yy   p y 2

Sub in
these

?



?
y p x U yx  p y U xx 

0

I
()

 
?
y 
p x p y  x 

 ; y 
 M 
p x
( )


Specific Utility Functions
• Cobb-Douglas
• CES
• Perfect Compliments
Cobb-Douglas: Utility Max
• Problem:
U(x, y)  x  y , s.t. M-p x x-p y y
• Set up the LaGrangian
L=x  y + (M-p x x-p y y)
• FOC
L x : x 1 y  p x  0

L y : x y
1
 p y  0
L : M-p x x-p y y=0
U x  x 1 y
U y  x  y1
x 1 y y
MRS   1 
x y
x
Cobb-Douglas: Demand
• FOC Imply, to maximize utility, these must hold.
yp y
xp x
x
; y
p x
p y
• Plug into the budget constraint to get the
ordinary (Marshallian) demand functions:
  M
  M
x*  
 ; y*  

     px
     py
• Note, demand only a function of own price
changes (one Cobb-Douglas weakness)
Cobb-Douglas: Demand
• Preferences are homothetic (only a function of
the ratio of y:x). When income rises, optimal
bundle along a ray from the origin.
– Expenditure a constant proportion of income
  
  
px x*  
 M; p y y*  
M
  
  
– Income elasticities are = 1
dx M 

eM 

dM x  (  )p x

M


 

 (  )p x

M

1
Cobb-Douglas: Indirect Utility
• Plug x* and y* into the utility function
U(x, y)  x  y
M
M
x* 
; y* 
(  )p x
(  )p y

 M    M
V
 
 (  )p x   (  )p y




Cobb-Douglas: Expenditure Function


 M   M 
V

 
 (  )p x   (  )p y 
• Solve for M, and then rename it E
M
(  )V

 
 px 


1

 

 py





E
(  )V

 
 px 


1

 

 py





CES: Utility Max
• Problem:
U(x, y)  x   y  , s.t. M - p x x - p y y
• Set up the Lagrangian
L  x   y   (M - p x x - p y y)
• FOC
L x : x
1
 p x  0
L y : y 1  p y  0
L : M - p x x - p y y  0
x
MRS   
 y
px  x 
 
py  y 
1
1
CES: Demand
• FOC Imply
 p x y 1 
x 
 p 
y


 1 
 1 


;
 p y x 1 
y
 p 
x


 1 
 1 


• Plug into the budget constraint and solve:
x
M

 py 

px  
 p
 x 

  


 1 

 1



; y
M
  





1


  px 

py   
 1
p 
 y 



CES: Indirect Utility
• Plug x* and y* into the utility function
U(x, y)  x   y 
M
x
;
  




 1 
p


px   y 
 1
 p

 x 



y

M

  px
py  
p
 y




  


 1 


 1




 


 


 


 
M
M

V
 
  
  
   p  1        1   

 px   y 
 1   p y   p x 

1
    p 

   p x 

y

     
 


CES: Expenditure Function
• Solve for M, then rename E.









M
V

  





 p y  1   

 px  
1 


   px 


 









M


  

 


 1 



 p  px

1

 y   p y 

  

 








1
V

  





 p y  1   

 px  
1 


   px 













1



2M
  

 


 1 



 p  px

1



 y   py 



 

 

CES: Expenditure Function (cont)
• Solve for M, then rename E.
V
M 








1
2

  





 p y  1   

 px  
1 


p
  x 


 
E









1


  

 


 1 



 p   px 
 1 
y 


p 

  y 

 
V

1









 
 

 
 

 
1
1
 2

 
  
  
  


 




    p y  1       p x  1   
1 
py   
 1
  px   p 





p
    y 

    x 

  














1

Perfect Compliments: Utility Max
• Problem:
U(x, y)  min(x, y), s.t. M-p x x-p y y
• No Lagrangian, just exhaust income such that:
x
y

• So plug this condition into the budget equation
– Essentially, we substitute the expansion path into the
budget line equation.
Perfect Compliments: Demand
• Demand equations
 y 
M=p x   +p y y
  
  

M=    p x +p y  y
  

M
y

  p x +p y

 x 
M=p x x+p y 

  

 
M=  p x +   p y  x
 

M
x

px +   py

Perfect Compliments: Indirect Utility




M
M


V(p x , p y , M)  min 
,


 
 p x +p y      p x +p y 
  


• Since utility from x = utility from y at utility
max:
M
M
V(p x , p y , M) 
or V(p x , p y , M) 


p x +p y  
  p x +p y


Perfect Compliments: Expenditure
Function
• Since the utility from consumption of each
must be equal,


  

p x +p y 
 p x +p y   



 





V
E
V, or E  














Bonus Topics
•
•
•
•
•
Money Metric Utility Function
Homogeneity
Corner Solutions (Kuhn-Tucker)
Lump-Sum Principle
MRS and MRT (Marginal Rate of
Transformation – slope of PPF)
Money Metric Utility Function
with Ū with U=U(x,y)
E  e  px , py , U 
E  e  p x , p y , U(x, y) 
• Now we know the minimum expenditure to get
the same utility as the bundle x’, y’.
U E  U E  p x , p y , x, y 
• That is, for an x’, y’, this function tells us the cost
of being at the tangency on the same indifference
curve. As all bundles on the same curve get the
same min. expenditure and prefernece ordering is
preserved, this is a utility function.
Evaluating Housing Policy
• Assume you find the poor generally expend
1/3 of income on housing.
• The government wants to double the quality
of housing the poor consume at the same 1/3
their income.
• How to evaluate?
Pre-Public Housing
If expenditure on
housing is generally 1/3
of income, assume
U=h1/3y2/3
Y
ph=\$1
px=\$1
M=\$1,000
667
M
2M
h* 
; y* 
3p x
3p y
333
667
H
Public Housing
ph=\$1
px=\$1
M=\$1,000
Qualified citizens get 667
housing units for 1/3 of income (\$333)
Y
V
E
1
3
 1   2 


 
3p
3p
 x  y
1
U 
E
3
h y
1
3
2
2
3
667
3
1  2
   
3  3
2
3
333
667
H
Public Housing: Money Metric Utility
V
E
1
3
 1   2 


 
 3p x   3p y 
1
U 
E
3
h y
1
3
2
ph=\$1
px=\$1
M=\$1,000
Qualified citizens get 667
housing units for 1/3 of income (\$333)
Y
2
3
3
1  2
   
3  3
2
3
667
UEPH=1,261
UE=1,000
333
667
The extra housing has a value to the poor of \$261. Depending on the cost to the
government of providing the housing, the program can be evaluated.
H
Homogeneity
• If all prices and income were doubled, the
optimal quantities demanded will not change
– the budget constraint is unchanged
xi* = xi(p1,p2,…,pn,M) = xi(tp1,tp2,…,tpn,tM)
• Individual demand functions are homogeneous
of degree zero in all prices and income
• To test, replace all prices and income with t*p
and t*M. The quantity demanded should be
unaffected (all the “t” should cancel out).
Corner Solutions
Y
Non-corner solutions:
Optimal bundle will be
where x > 0, y > 0 and
MRS = px/py
Two corner solutions:
Optimal bundle will be
where x =0
X
Corner Solutions
Y
A
X
• At “A”, MRS = px/py, but the optimal quantity of X = 0
Corner Solutions
Y
B
X
• At “B”, the tangency condition holds where x* < 0.
Given the budget line, the optimal feasible x is where
x = 0 and MRS < px/py
Corner Solution
• To develop these conditions as part of a
constraints:  =  2 (we could add  =  2 if
we wanted to be really thorough).
• Lagrangian:
– L = (, ) + (M −   −  ) + ( −  2 )
– Requiring x = s2 is simply a way of ensuring that x
≥ 0.
Corner Solutions
• FOC
L x  U x  p x    0
L y  U y  p y  0
L  M  p x x  p y y  0
L  x  s  0
2
Ls  2s  0
Corner Solutions
• Kuhn-Tucker Condition
Ls  2s  0
• This tell us that either:
• μ =0 (the optimum is at a tangency)
• s = 0 (the optimum is where x = 0)
• μ =0 and s = 0 (the optimum is at a tangency where x
= 0)
Corner Solutions
Y
If s > 0 and μ = 0
L x  U x  p x  0
L y  U y  p y  0
Ux px

Uy py
• The usual assumption is that the optimal
px
bundle will be where x>0, y>0 and MRS 
py
X
Corner Solutions
If s = 0 and μ = 0
Y
L x  U x  p x  0
A
L y  U y  p y  0
Ux px

Uy py
X
•
At “A”, MRS 
px
py
, but the optimal quantity of x = 0
Corner Solutions
If s = 0 and μ > 0
Y
L x  U x  p x    0
L y  U y  p y  0
Ux   px

Uy
py
B
Ux Ux   px
MRS 


Uy
Uy
py
B’
px
MRS  , at B
py
X
• At B’, the tangency condition holds where x* < 0.
px
• The optimum is where x=0 and MRS 
py
Kuhn-Tucker Example
• Utility: U=xy+20y, M = 40, px = \$4 and py = 1.
L  xy  20y  (40  4x  y)
L x : y  4; L y : x  20  
L : 40  4x  y
y
Gets  4x  80  y, x   20
4
Solve  x  5, y  60
Oops.
Looks Like
• Tangency where x=-5, y=60
Y
y
MRS =
4
x  20
X
• Tangency where x=-5, y=60
Y
MRS =
40
2
0  20
px 4
60
MRS =


5  20 p y 1
Slope =
px 4

py 1
X
Kuhn-Tucker Set-up
• Utility: U=xy+20y, M = 40, px = \$4 and py = 1
L  xy  20y  (40  4x  y)  (x  s 2 )
L x : y  4    0
L y : x  20    0
L : 40  4x  y  0
L : x  s 2  0
Ls : 2s  0
 Kuhn-Tucker Condition
Kuhn-Tucker Result
Kuhn-Tucker: 2s  0, so  or s or both = 0
y   px
Use L x and L y :

x  20 p y
px
y
If μ=0, optimum is a tangency where

x  20 p y
If s  0, optimal x  0.
If   0, and s  0 optimum at corner and:
px
y
y

 MRS 
p y x  20
x  20
If   0, and s  0 optimum at corner but tangency where x  0.
If   0, and s  0 optimum at interior.
Kuhn-Tucker Result
• In this example, at x = 0, y = 40,   0 :
y
At corner, this condition holds:
=4
x  20
So   4x  80  y and   40
y
40
At the corner, MRS =

2
x  20 0  20
If p x  2, then the optimum is at x  0, y  40,
  0 and s  0.
• Optimum where x=0, y=40
Y
At optimum:
px
y
y

 MRS =
p y x  20
x  20
p x 40  40
40

 MRS =
py
20
20
px
y
y

 4  MRS = 2 
p y x  20
x  20
X
Solving Kuhn-Tucker
• If you find that the optimal bundle is not on the
budget constraint, check all corners for a
maximum utility.
Lump Sum Tax
• Taxing a good vs taxing income
– Tax on x only
 px   x  py y  M
M  p  
which is y= - x
x
py
py
x * and y * are optimal bundles so
 p x    x *  p y y*  M , and R* = x *
– Lump sum tax (income tax)
p x x  p y y  M  x *
M  x * p x
y
 x
py
py
Lump Sum Tax
• Difference in the budget lines: sales tax
Without a tax,
M px x *
y* 

py
py
With the unit tax on x, at any x*,
y* 
M p x x * x *


py
py
py
M p x x *  M x * p x x * 
At any x*, y*-y 

 


py
py
p y 
 p y p y
*

y*-y* 
x *
py
Lump Sum Principle
• A tax on x rotates the budget line to have:
Y
M
py
y*
x *
py
yτ*
slope  
px  
py
x*
M
px  
M
px
X
Lump Sum Tax
• Difference in the budget lines: income tax
Without a tax,
M px x *
y* 

py
py
With an income tax R*= x *
y*R 
M x * p x x *


py py
py
M p x x *  M x * p x x * 
At any x*, y*-y 

 


py
py
p
p
p
 y
y
y 

*
R
x *
y*-y 
py
*
R
Lump Sum Principle
• Tax paid = x*τ. Alternatively, an income tax of that same
amount would shift the budget line so that the
consumer can just afford the same bundle they chose
under the tax on x.
Y
x *
py
M
py
M  x *
py
x *
py
y*
x*
M
px  
M  x *
px
M
px
X
Lump Sum Principle
• When
U  xy.5
2M
M
x
; y
3p x
3p y
• Indirect Utility function is:
.5
2M  M 
V(p x , p y , M) 


3p x  3p y 
Lump Sum Principle
• Set, I=60, py = 2, px=1
2M
M
x
 40; y 
 10
3p x
3p y
• Utility is:
.5
2M  M 
V(p x , p y , M) 

  126.49
3p x  3p y 
Lump Sum Principle
• With a \$1 tax on x,
x
2M
M
 20; y 
 10
3  p x  1
3p y
• Utility is:
.5
 M 
2M
V  V(p x , p y , M) 

  63.24
3  p x  1 Px  3p y 
*
• And tax revenue is \$20.
Lump Sum Principle
• With a \$20 income tax,
x* 
2(M  20)
M  20
=26.667; y* 
 6.667
3Px
3Py
• Utility is:
.5
2(M  20)  M  20 
V  V(p x , p y , M) 

  68.85
3Px
 3Py 
*
• 68.85 > 63.24
The Lump Sum Principle: Perfect
Compliments
• If the utility function is U = Min(x,4y) then plug
x=4y into the budget equation to get:
M
x 
p x  0.25p y
*
M
y 
4p x  p y
*
• The indirect utility function is then:

 M 
M
V(p x , p y , M)  Min 
,4
 p x  0.25p y  4p x  p y  



The Lump Sum Principle: Perfect
Compliments
• Set, I=60, py = 2, px=1
M
x 
 40
p x  0.25p y
*
M
y 
 10
4p x  p y
*
• Utility is:

 M 
M
V(p x , p y , M)  Min 
, 4
 40


 p x  0.25p y

4p

p
x
y




The Lump Sum Principle: Perfect
Compliments
• With a \$1 tax on x,
M
x 
 24
(p x  1)  0.25p y
*
M
y 
6
4(p x  1)  p y
*
• Utility is:



M
M
V(p x , p y , I)  Min 
,4
 24


 (p x  1)  0.25p y

4(p

1)

p
x
y




The Lump Sum Principle: Perfect
Compliments
• With a \$24 income tax,
M - 24
x 
 24
p x  0.25p y
*
M - 24
y 
6
4p x  p y
*
• Utility is:
 M - 24
 M - 24  
V(p x , p y , M)  Min 
,4
 24


 p x  0.25p y

4p

p
x
y




• Since preferences do not allow substitution, the
consumer makes the same choice either way and
the tax does not affect utility.
MRS=MRT
• Using Varian’s example about milk and butter
– B* and M* may be different for all consumers.
– However, the MRS at the tangency is the same for ALL
consumers, no matter the income or preferences.
Milk
Pb=3; Pm=1
MRS=3
MRS=3
MRS=3
Butter
Marginal Rate of Transformation
(a.k.a Rate of Product Transformation)
• Varian is sort of implicitly assuming the PPF is
linear. So there is a constant trade of in producing
butter or milk as resources are reallocated.
Milk
MRS=3
MRT=3
Social Indifference curve
Butter
Marginal Rate of Transformation
• With a more realistic PPF, the MRT rises as
more butter and less milk is produced.
MRT 
Milk
MRT=3
Cb q m

Cm q b
Assuming competative firms producing b and m
C
p
MRT  b  b
Cm p m
or
Cb Cm

, marginal cost  p the same at
pb pm
the margin for all goods.
Butter
MRS=MRT
• In the long run (π=0), the cost of producing
butter must be 3 times the cost of producing
milk. That is, the tradeoffs in consumption =
Milk
MRT 
Cb p b U b


 MRS
Cm p m U m
px/py=3
Butter
Back to Varian’s treatment
• A new technology that allows you to produce
butter with 4 gallons of milk is not going to be
a winner as everyone would choose the
cheaper butter previously offered.
Milk
MRT=4
Pb=4; Pm=1
Butter
• However, a new technology that allows you to
produce 1 pound of butter with 2 gallons of
milk IS going to be a winner!
Milk
MRS=MRT=2
Pb=2; Pm=1
Butter
MRS=MRT
• Here is what improved butter making
technology does with a more standard PPF.
Milk
px/py=3
px/py=2
Butter
```