Homework 1 EC422 (2/2549)
Q1. (from exercise 5.1 in Dixit) The CD cost function.
Let make it simpler, suppose we have two inputs, x1, and x2. So, the production
a
b
function is y = A x 1 x 2 . Let w1, and w2 are input prices for input 1 and 2,
respectively. Find the cost function.
Our problem is to minimize cost for a given level of output.
Min w1x1 + w2x 2
a
b
st. A x 1 x 2 = y
Assuming the interior solutions for x1 and x1, we could form the FOC as
...
From the FOC conditions, you can solve for 3 unknowns.
The conditional factor demand for input 1:
x 1(w1, w2, y ) = A
- 1/ ( a + b )
and for input 2:
x 2 (w1, w2, y ) = A
- 1/ ( a + b )
So, the cost function
éa w2 ùb / ( a + b ) 1/ ( a + b )
ê
ú
y
êb w1 ú
ë
û
éa w2 ù- a / ( a + b ) 1/ ( a + b )
ê
ú
y
êb w1 ú
ë
û
c(w1, w2, y ) = w1x 1(w1, w2, y ) + w2x 2(w1, w2, y )
éé ùb / ( a + b ) éa ùb / ( a + b ) ù a
b
1/ ( a + b )
- 1/ ( a + b ) ê a
(a + b )
(a + b )
ú
ê
ú
ê
ú
= A
+
w
w
y
2
êêb ú
ú 1
êëb ú
û
êëë û
ú
û
For simplicity, if we assume that A=1, and a + b = 1 then the cost function is
reduced to
c(w1, w2, y ) = Kw1a w2b y , where K = a - a (1 - a )a - 1
If a + b < 1 , Find the profit function.
Note that the condition implies that the production function exhibits decreasing
returns to scale technology.
Our problem is to maximize the profit given technology by selection inputs and
output levels. For simplicity, let A=1.
Max py - w1x 1 - w2x 2
a
b
y = x1 x 2
st.
we can substitute y into the objective function, and choose only inputs.
The FOC with respect to x1 and x2 are,
pa x 1a - 1x 2b - w1 = 0.............(1)
pb x 1a x 2b - 1 - w2 = 0.............(2)
and (1) / (2) we have
1
a
b
æx 2 ÷
ö
çç ÷ =
÷
çèx 1 ÷
ø
æ ö
æb w1 ö
÷
ççw1 ÷
ç
÷
÷
or
x
=
x
2
1ç
÷
÷, or w2x 2 = ( b / a )w1x 1
÷
çèw2 ÷
ç
ø
èa w2 ø
Substitute x2 into (1), we get
1 
  

1
1  

1
1  
  
  p
 w2  
By symmetry, we know that
x 1( p, w1, w2 ) =  
 w1 
1 
  
  
  p
 w1  
And x1 and x2 into the production function to get the output supply function as
x 2 ( p, w1, w2 ) =  
 w2 
y ( p, w1, w2 ) = x 1( p, w1, w2 )a x 2( p, w1, w2 )b


 
   1    1   1  
 
p
 
 w1 
 w2 
Therefore the profit function is
p ( p, w1, w2 ) = py ( p, w1, w2 ) - w1x 1( p, w1, w2 ) - w2x 2( p, w1, w2 )


   1    1   1  
 
p
(1     ).
 
w
w
 1
 2
It is obvious that the profit be positive when a + b < 1 .
And the profit is negative when a + b > 1
Also, you can check that the profit function is homogenous of degree 1 in (w1,
w2, p).
1
2
Q2. (From exercise 5.2 in Dixit) The CES Expenditure Function
1
 x    y  
Suppose the direct utility function is U(x,y)= 
where x and y are the quantities of the two goods, and ,
and are given
constants, with ,
positive, and < 1.
(a) Find the Expenditure function, where p, and q are the prices of the goods,
Min
E = px + qy
1
 x    y  
s.t.
U(x,y) = 
=u.
First, we form the Lagrangian function


1
L  px  qy    x    y    u
Assuming the interior solutions in both goods, the first-order conditions are
1
1
L
1
 p   ( )  x  1  x    y     0

FOC x
1
1
 p   x  1  x    y     0
…………………. (1)
1
L
1
 q   ( )  y  1  x    y     0
y

1
 q   y
 1
 x   y 


1

1
0
…………………..(2)
L
 u   x    y     0

1
(1) / (2) ;
p  x
  
q   y
…………………..(3)
 1
1
 p    1
x  y  
q 
or
…………………. (4)
So, by eliminating
these conditions reduced to the two equations (4) and (3)
in two unknowns.
r

.
  1 This will g reatly reduce the mess.
At this point, we can define
Substitute (4) into (3), we can solve for y as
u
y* 
1
1



 1
  p      


  q  


 ....................................Ans
and substitute y* into (4) we will get
3
1
 p    1
u
x*     
1
1
q 



 1
  p      
  q  


 .............................Ans
Ratio of cost minimizing quantities x*/y*
1
1
x*  p    1  q 1 
    
y*  q  
 p
1 r
q
 
 p
Let
a
1
 1
1
x *  q 1 

 
y*  p 
1
    1
 
 
1 r
 
 
 
....................................Ans
  r 1 , b  
1
 1
1
1 r
    1  q 
    
 
 p
  r 1.
a
b
..............................Ans
Note that we just solve for the Hicksian demand functions. We can find the
expenditure function by substituting x*, y* into the objective function.
Then E = px* + qy* ;
1
 p    1
u
u
E  p   
q
1
1
1
1
q 






 1
 1
  p      
  p      




  q  

  q  





r
 p  a 
qu
    1
E

1
 q b 
r
1
 





p
a
      1
 q  b 


1
qu  ap r  bq r 
E 1 

bq r 
 
1


r


  1 r 1 1 1 

 

r
r
 
r 1
- Note that 
1
1
u
u
p r q r 1r
r
r r
E  1 1  ap r  bq r  r 
ap

bq

u
(
 )


1
b
a
 r
r
(ab)
 b
Hence,
Take ln of x*/y*;
ln( x / y)  ln(a / b)  ln(q / p)1r
4
 ln( x / y )
 1 r
 ln(q / p)
r

..........................................Ans

  1 therefore to ensure that r<1, we need   1
Since
1
 0   1
 1
.
5
1
or
Q3. Stone-Geary utility function
Max
U  ( x1   1 )1 ( x2   2 ) 2
s.t.
I  p1 x1  p2 x2
; 1   2  1
First of all, let A  x1   1 , B  x2   2
L  A1 B2    I  p1 ( A   1 )  p2 ( B   2 ) 
L
 1 A 2 B 2   p1  0
FOC A
……………………………. (1)
L
  2 A1 B 1   p2  0
B
……………………………. (2)
L
 I  p1 ( A   1 )  p2 ( B   2 )  0

…………………. (3)
1 B p1


A
p2
2
(1) / (2) ;
 2 p1
 A
1 p2
Sub. B into (3);
B
……………………………. (4)
I  p1 ( A   1 )  p2 (
I  p1 1  p1 A 
 2 p1
  A 2)
1 p2
2
 p  A  p2 2
1 1
1
( I  p1 1  p2 2 )
p1
Sub. A into (4);
A
2
( I  p1 1  p2 2 )
p2
Find x1 and x2 , x1  A   1 , x2  B   2 hence,
B
x1 
1
p1
( I  p1 1  p2 2 )   1
2
( I  p1 1  p2 2 )   2
p2
Sub. A and B to find indirect utility function, V(p1, p2, I);
x2 
V (
1
p1
( I  p1 1  p2 2 ))1 (
1
2
p2
( I  p1 1  p2 2 )) 2
2
   
V   1   2  ( I  p1 1  p2 2 )
 p1   p2 
Note that if you define I’= ( I  p1 1  p2 2 ) and use this in the maximization


A  1 ( I '), and B  2 ( I ').
p1
p2
problem. We know that we could solve for
6
Use the definition of A, and B, we can rewrite these as
p1 x1  p1 1 
1
p1
p2 x2  p2 2 
( I  p1 1  p2 2 )
2
( I  p1 1  p2 2 )
p2
This is called Linear Expenditure system.
7