Solutions Manual to
Exercises in
Mathematical Tools
For Economics
1
2
Exercises for Chapter 1
1.1
1.
(i)
 4 2 
 6 15 
A B  
 ,  3A  
.
 4 8
 18 3 
 0 1  4 2   4 8 
C  A  B  


.
 5 8  4 8   52 54 
1
 6
 2 7 
 4 8
CA  
 , CB  
 , CA  CB  
.
 58 17 
 6 71
 52 54 
 4 2  0 1  10 12 


.
 4 8  5 8   40 68 
 A  BC  
(ii)
 2 5  2 3  14 29 
AB  


.
1 2 7  10 25 
6
 2 3  2 5   22 7 
BA  



1  38 17 
 2 7  6
14 29  0 1   145 218 
ABC  



10 25  5 8   125 210 
 15 26  2 5   186 49 
BCA  



1  394 121
 35 54  6
 0 1 14 29   10 25 
CAB  


.
 5 8 10 25  150 55 
Clearly AB and BA are not the same.
(iii) trAB  14  25  39
trBA  22 17  39
trABC  145  210  65
trBCA  186 121  65
trCAB  10  55  65.
2.
(i)
 1 4 2
1 3 7
  16 12 41
AB  
  2 5 1  
.
29
12
58
5 0 8



 3 1 6 
As B is 3  3 and A is 2  3 , BA does not exist.
 16 29 

(ii)  AB    12 12  .
 41 58 


 1 2 3  1 5   16 29 


 

BA   4 5 1 3 0    12 12  .
 2 1 6  7 8   41 58 


 

 1 5
1 3 7
  59 61 
(iii) AA  
 3 0  
,
61
89
5 0 8



7 8
which is clearly symmetric.
3.
(i)
 2 3 5  1 3 5   0 0 0 


 

AB   1 4 5  1 3 5    0 0 0  .
 1 3 4  1 3 5   0 0 0 


 

 1 3 5  2 3 5   0 0 0 


 

BA   1 3 5  1 4 5    0 0 0  .
 1 3 5  1 3 4   0 0 0 


 

Both AB and BA equal the 3  3 null matrix.
4
(ii)
 2 3 5  2 2 4   2 3 5 


 

AC   1 4 5  1 3 4    1 4 5   A. .
 1 3 4  1 2 3   1 3 4 


 

 2 2 4  2 3 5   2 2 4 


 

CA   1 3 4  1 4 5    1 3 4   C. .
 1 2 3  1 3 4   1 2 3 


 

 2 3 5  2 3 5   2 3 5 
 

(iii) A  AA   1 4 5 
 1 4 5    1 4 5   A.
 1 3 4  1 3 4   1 3 4 


 

2
 2 2 4  2 2 4 



C  CC   1 3 4  1 3 4   C. .
 1 2 3  1 2 3 



2
(iv) (a)
AACB
 AAB from ii
 AO from i
 O.
(b)
 A  B
2
  A  B  A  B 
 A2  AB  BA  B 2
 A2  B2
4.
from i.
A square matrix A is symmetric idempotent if A  A and A2  A.
(i)
A  (i i) / n  (i)i / n  A.
B  (I  A)  I  A  B.
AA  i ii2 i  A as ii  n.
n
BB   I  A   I  2 A  A2  B.
2
5
trA  tr i i   tr i i  1.
n
n
trB  tr(I  A)  trI  trA  n  1.
(ii)
Ax  i ix / n.
 x1 
Now ix  (1 . . . 1)    x1  . . . + xn .
x 
 n
x
n
 
Hence Ax  i x    where x   xi / n , so Ax is an n1 vector where elements are x. Now
i 1
x
 
 x1  x 

 so Bx is an
Bx   I  A  x  x  Ax  
n1 vector where jth element is x j  x .
,
x x
 n

1.2
1.
A  6, B  18  4  22,
AB 
A 
B 
2 4
4 74
1 5
0
6
2 1
4
9
 6
 18  4  22.
4 1 6
2.
 148  16  132.
4 1
6
A7 2 9

1 0 3 .
r2  r2  2r1
3 0 8
3 0 8
6
Expanding using the second column we have
A  1 1
1 2
1 3
3
8
 1(8  9)  1.
Similarly,
4 7 3
4 1 3
A  1 2 0  1
6 9 8
1
3.
A
0 0  1.
6 3 8
2 3 2.5
0 0 0

r2  r2  r3 1 2 3
3 1 1
7
3
.
2
Expanding using the second row we have
A  7(1)
2 4
1
2 3
1
2 3 7 1
3 1 1
0
0 0
2 3  0.
3 1 1
2 1 3 0
B
3 2 1 0

 0.
c4 c4 3c2 1 3 0 0
2 4 1 0
1.3
1.
 4 2   4 1
1
1  4 1 .
AdjA  

 
 and A  4  2  6. Hence A  6 
 2 1
 1 1  2 1
 8 5   8 0 
1
1  8 0 .
AdjB  

 
 and B  16. Hence B   16 
 5 2 
 0 2   5 2 
 1 1 2 0   7 8 
AB  


.
 2 4  5 8   16 32 
7
8
8
 32 16   32
1
1  32
Adj( AB)  
.
 
 and AB  96 so ( AB)   96 

16

7
8

7

16

7



 

2
2.
A
1 0
2
1
 6, so A is nonsingular.

6 2 6  6(1) 23

13

6
3
r3 r3  r2 13 6 0
2




AdjA   





2 6
3 9
1
6 6

4 9
0
2
3 9
0
4 9
1 0

2 6
2 0
6 6
6 2 

4 3 
 36 78 10 
2 1  

  9

18
2 .

4 3  

  6 12 2 
2 1
6 2 
Hence
6
 36 9


A1   1  78 18 12  .
6

 10 2 2 
0 1 0
B
1 0

0 1 1  1(1)31
 1, so B is nonsingular.
1 1
r1 r1  r3
1 4 3
r2  r2  r3
 3

 4
 3
AdjB   
 4

 3
 3

4
3
1 4

1 3
3
1 3
3
1 3
3
4

1 3
1 4
1 3 

1 4
1 1  7 3 3 
 7
1 3 
  3 0 1    1 0  1  .

 

1 4  
  1 1 0 
3

1
0
 

 
1 3
1 3 
 7 3 3 


Hence B   1 0 1  .
 1 1 0 


1
8
3.
2 4 5

( A I)   0 3 0
1 0 1

1 0 1

 0 3 0
r1 r3
2 4 5

1 0 0

0 1 0
0 0 1 
0 0 1

0 1 0
1 0 0 
1 0 1


0 3 0
r3 r3 2r1 
0 4 3
1 0 1

 0 1 0
r2  1 r2  0 4 3
3 
1

0 1 0
1 0 2 
0 0
1

0 1 0
3

1 0 2 
0 0

1 0 1


0 1 0
r3 r3  4r2 
0 0 3



1 0 0

 0 1 0
r1 r1  r3 
0 0 1



1 0 1

 0 1 0
r4  1 r4  0 0 1
3 


1 0 1

 0 1 0
r4  1 r4  0 0 1
3 

0

1

0

2 

0
1
3
1 4
3
0
1
3
4
9
1
3
4
9
0
1
3
0
0
1
3
1 4
9
0
0
0
1
3
1 4
3
9
0
5 
3

0 

2
3

1 

0 

2
 
3



0 

2 
3 
1
9

1 0 0

 0 1 0
r1 r1  r3 
0 0 1


1
3
4
9
1
3
4
9
0
1
3
5 
3

0 

2
3
 3 4 15 

Hence A1  1  0
3 0 .
9

 3 4 6 
1 2 1
 B I    0 1 1
2 1 4

1 0 0

0 1 0
0 0 1 
1 2 1

0
1 0
0 0 1
1 0 0


 0 1 1
c2 c2  2c1 
2 3 2
c  c  c 
3
2
1
2 1

0 1 0 
0 0 1
1 0 0

 0 1 1
c2 c2 
2 3 2
1
1 0 0


0 1 0
c3 c3 c2 
 2 3 1
2 3 

0 1 1
0 0
1
1 0 0


0 1 0
c1 c1  2c3 
0 0 1
c2 c2 3c3 
5 7 3 

2
2
1
2
3
1
1 0 0

 0 1 0
c3 c3 
0 0 1
1
5 7
2
2
3

2 1
3 1
 5 7 3 


Hence B   2 2 1 .
 2
3 1

1
10
1 1 0
4.
A

1
r2  r2  r3 2
A   1
3 3
1 0 . Expanding this determinant using the third column we have
1 1
1 1
1




Adj A =  





1
2 1
1 1
 2, so A is nonsingular.

3 1
2 1
1 0
1 0
1 1
2 1
1 0
2
1

1 0
3 1
3 2 

2 1
 1 1 1

1 1
   1 1 3  .


2 1  


1

1
5

 
1 1 
3 2 
 1 1 1
Hence A  1  1 1 1 .
2

 1 3 5 
1
1.4
1.
(i)
A set of m n1 vectors a1, . . . ,am are linearly dependent if there exists scalars 1 , . . . ,m
not all of which are zero such that 1 a1  . . .  mam  0.
(ii) Consider
1 x1  2 x2  3 x3  0
which is the following set of equations in 1 , 2 , and 3 :
21  22  143  0
71  42  273  0
1  2  33  0
From the third equation 1  2  33 and substituting back into the other two equations
gives 2  23 , so if we take 3  1, 2  2 , and 1  5 we get 5 x1  2 x2  x3  0 .
Thus the vectors are linearly dependent.
11
(iii) Consider
1 x1  2 x2  3 x3  0
which gives
91  22  43  0
31  0
1  22  83  0.
The second equation gives 1  0 so we are left with 22  43 and 22  83 which are true only if
2  3  0. Hence the vectors are linearly dependent.
2.
(i)
11
 1 3 2


0
1 11 33 

A

r2  r2  2r1  0 1 5 15 

r3 r3  r1  0 5
1 3 
r4  r4  r1
11
 1 3 2


0
1 11 33 


6 18 
r3 r3  r2  0 0

r4  r4 5r2  0 0 54 162 
11
 1 3 2


0
1 11 33 


.
r4 r4 9r1  0 0 6 18 


0
0 0 0
Hence r ( A)  3.
(ii)
B
1 1 2
2
1
0 2
0
0
1

r2  r2  r1 0
r3 r3  r1 1
r4  r4  r1
1
0 0 1
1 2
 1(1)
1 2
0
0
1
1
1 0 1
12
1
2
0
1 1
 1.

0
1 1  1(1)11
2 1
r3r3 r1
0 2 1
Hence r ( B)  4.
3.
Expanding A by the third column we have
A  (1)31
2 1 x
1
0
 (1  x)(1)3 3
5 x
2
2
1 x
 (1  x)  (1  x) (5  x)(1  x)  4
 (1  x) x ( x  6).
Therefore A  0 if x  1, or x  0, or x  6 but is nonzero for all other values of x.
So if x does not take the value 0, 1, or 6 then r ( A)  3. If x takes any of these values consider the minor
5 x 1
2
0
 2
for all values of x. Thus we have r ( A)  3 for all x other than 0, 1, 6,
0, 1, or 6.
4.
r ( A)  2 for x taking the value
(i)


1

1
1 
1
N   X  X X  X    X    X X   X   X  X X   X   X  X X  X   N .




M    I  N   I   N   M .
Hence both N and M are symmetric.
Consider
NN  X ( X X )1 X X ( X X )1 X   N
13
and
MM  ( I  N )( I  N )  I  2 N  N 2  M .
So M and N are idempotent.
Now
NX  X ( X X )1 X X  X
so
MX  X  NX  O.
(ii)
r ( N )  trN  tr( X X ) 1 X X  trI K  K .
r (M )  tr( I  N )  trI n  trN  n  K .
1.5
1.
(i)
 0
 1
8
A B  
 0
 3
 8
 
7
 0 7    0 7 0
 2
 
5
 8 5    8 5 16

7
 0 7    0 21 0
 4
   24 15 32
5
8
5

 
  1
 0
 3
B A
  1
 8

  3
2
 1
 7
4
 3
2
 1
 5
4
 3
 0

7
( A  B)  
 0

 14
8
14 

10 
.
28 

20 
2    0 0 7 14 

4    0 0 21 28 
.

2    8 16 5 10 

 
4    24 32 15 20 
0 24 

5 21 15 
.
16 0 32 

10 28 20 
14
 0
 1
7
A  B  
 0
 2
 7
 
8  0 8   0
 3

5   7 5    7


 0
8
 0 8
 4
   14
5
 7 5  
8
0 24 

5 21 15 
.
16 0 32 

10 28 20 
tr( A  D)  0  5  0  20  15
trA trB  ( 1  4)(0  5)  15.
(ii)
 1 2  0 7  3 4   16 3  3 4   54 61 
ABC  


 


,
 3 4  8 5  2 1  32 41 2 1 178 87 
so
 54 


178 

vecABC 
.
 61 


 87 
  1
 3
3
(C   A)vecB  
  1
 4
 3
 
2
 1
 2
4
3
2
 1
 1
4
3
2 

4 
vecB
2

4  
 3 6 2 4   0 

 
9 12 6 8   8 


 4 8 1 2   7 

 
 12 16 3 4   5 
 54 


178 

.
 61 


 87 
 1 2  0 7   16 3 
(iii) AB  



 3 4  8 5   32 41
15
so
trAB  16  41  57.
0
 
8
(vecA)vecB  (1 2 3 4)    57.
7
 
5
(iv) trABC  54  87  141.
C is 2  2 and vecB is 4 1 so the required identity matrix is 2  2 and we have
 1
 3
0
(vecA)(C   I )vecB  (vecA) 
 1
 4
 0
 
  1 2 3 4 
  9 22 5 0 
0 

1 
vecB
0

1  
2 0

0 2
vecB
1 0 

0 1
0
1
 2
1
0
0
1
 1
1
0
3

0
4

0
0
3
0
4
0
 
8
7
 
5
 176  35  141.
2.
 a1 
 b1 
 a1 




Let a    , b    . Then ab     b1
a 
b 
a 
 n
 n
 n
  a1 
  
b  a   b1  
 a 
  n
 a1  b1

a  b  
 a b
 n 1
 a1  
 
bm     ab
a 
 n 
bm  

  ab.
bm  
16
 a1b1

bm   
a b
 n 1
a1bm 

.
anbm 
 a1b1 




 anb1 


vec ab  
  vec (b  a ).
 a1bm 




a b 
 n n
 b1  a1  
  
  
  an  


ba  
  vec ab.
b a 
 m  1 
  
   
  an  
Exercises for Chapter 2
2.2
1. (i)
In matrix notation we have
Ax  0
1

A1


0 2

2 1
1 1 
2 
1 0



0 2
1 
r2' r2 r1 

r3  r3   r1  0 1 1  2 
where
2 
1 0



0 2
1  .

r3'  r3  0.5r2 

 0 0 1.5  2 
Thus r ( A)  3 for all values of  other than   0.75 , in which case we only have the trivial solution.
(ii). If   0.75 , r ( A)  2  3 so we have an infinite number of solutions. If this is the case the
original system has the same solutions as
x1  2 x3  0
2 x2  x3  0.
17
Let x3   , any real number. Then x2   / 2 and x1  2 , so our general solution is
 2 


x    / 2 .
  


*
2.
In matrix notation we have
Ax  0
 1 2 2 3 
where A   1 0 2 13  .
3 5 4 0 


As A is 3  4 we must have r ( A)  4, the number of variables. Hence if a nontrivial solution exists, an
infinite number of solutions exist.
 1 2 2 3 


A

0 2 4 16 

r2  r2  r1 
0 1 2 9 
r  r 3r 
3
3
1
 1 2 2 3 


  0 1 2 8 
r2  1 r2  0 1 2 9 
2
 1 2 2 3 


  0 1 2 8  .
r3 r3  r2 
1
0 0 0
Hence r ( A)  3 and the original system has the same solution as
x1  2 x2  2 x3  3x4  0
x2  2 x3  8 x4  0
x4  0
 x2  2 x3 , x1  2 x3 .
Let x3   , where  is any real number. Then x2  2 and x1  2 so
 2 


2  
x*  
 


 0
is the general solution.
18
3.
Rewrite the system as
x  3z  ky  0
2x  4z  3y  0
3x  22  ky  0
which can be represented in matrix notation as
Ax  0
1 3 k
where A   2 4 3 
 3 2 k 


3
k
3
k
1
1




0 10 3  2k

0 10
3  2k  .




11
r2 r2  2r1
2k  r3 r3 10 r2  0
0 3.3  0.2k 
 0 11
r3 r3 3r1
If k  33/ 2 then r ( A)  3 and the only solution is the trivial solution. Thus for k  33/ 2 we have a
nontrivial solution and for this case our system has the same solution as
x1  3x2  33 / 2 x3  0
10 x2  30 x3  0.
So x2  3x3 and x1  15x3 / 2 , and our general solution is
 15 /12
x*   3 





for any real number .
2.3
1. In matrix notation our system is Ax  b
1 1 1
 2


 
where A   2 1 2  and b   5  .
 4 3 4
c


 
2
2
1 1 1 2
1 1 1
1 1 1






Then ( A b)   2 1 2 5 

0 1 0
1   0 1 0
1

 4 3 4 c  r2  r2  2r1  0 1 0 c  8  r2  r2  0 1 0 c  8 

 r3  r3  4r1 



19
2
1 1 1


 0 1 0
1 .
r3 r3  r2 

0 0 0 c 9
If c  9 then r ( A)  2 but r ( A b)  3 and the equations are inconsistent. For c  9 our system has the
same solution as
x1  x2  x3  2
x2  1.
A particular solution to this system would be x3  0, x2  1, x1  3. The homogeneous equations are
x1  x2  x3  0
x2  0
which have a general solution of the form x3   , x2  0, x1   for any real number . Hence the
general solution to our nonhomogeneous equation is
3  


x   1  .
  


*
2.
In matrix notation we have
3 3  x   9 
1

   
 3 17 8  y    49  ,
3
  
4
1

 z   c 
or Ax  b where
3 2 9 
3 2
9
3 2
9
1
1
1






( A b)   3 17 8 49 

0 26 14
22 

0 13 7
11 


r   r  3r
1
 3 4
1 c  r2r 2 3r 1  0 13 7 c  27  r2   2 r2  0 13 7 c  27 

3
3
1
9
 1 3 2


  0 13 7
11 .
r3 r3  r2 

 0 0 0 c  38 
Clearly r ( A)  2  3, the number of variables. Hence no unique solution exists as this requires that
r ( A b)  r ( A)  3.
20
If c  38 then r ( A b)  r ( A) and the equations are inconsistent. No solution exists.
(a)
(b)
If c  38 then r ( A b)  r ( A)  2 which is less than the number of variables so an infinite number
of solutions exist. For this case our equations have the same solution as
x  3y  2z  9
13 y  7 z  11.
Let y  0, then z  11/ 7 and x  85/ 7 so a particular solution to these nonhomogeneous equations is
 85 
1  0 .
7  
 11
The corresponding homogeneous equations have the same solution as
x  3y  2z  0
13 y  7 z  0.
Let z   , where  is any real number. Then y  7 /13 and x  5 /13 so the general solution to the
homogeneous equations can be written as
 5
 
7 .
13  
 13 
 
Adding, we have the general solution to the nonhomogeneous equations is
 85 / 7  5 /13 


7 /13  .

 11/ 7   


3.
(i)
In matrix notation our system of equations is
 x1 
 1 4 17 4     38 

  x2  

2
12
46
10

  x    98  .
 3 18 69 17   3  153 

 x


 4
21
Consider
 1 4 17 4 38 


( A b)   2 12 46 10 98  .
 3 18 69 17 153 


If the equations are consistent
r ( A b)  r ( A).
But r ( A)  min(3, 4)  4, where 4 is the number of variables, so the equations will then have an infinite
number of solutions.
(ii)
 1 4 17 4 38 
 1 4 17 4 38 
 1 4 17 4 38 






( A b)

0 4 12 2 22    0 2 6 1 11

0 2 6 1 11  .


r2  r2  2r1 
 r2  1 r2  0 6 18 5 39  r3  r3 3r2  0 0 0 2 6 
0
6
18
5
39





2 
r  r 3r
3
3
1
Hence r ( A b)  r ( A)  3 and the equations are consistent.
2 16 
1 0 5
 1 0 5 0 10 




( A b)

0 2 6
1 11

0 2 6 0 8 .



r1 r1  2r2  0 0 0
1 3  r1 r1  2r3  0 0 0 1 3 

1
r
2  r2  r3
r3 r3
2
It follows that the nonhomogeneous equations have the same solution as
x1  5 x3  10
2 x2  6 x3  8
x4  3.
Let x3  0, then x4  4 and x1  10, so a particular solution to this system is
 10 
 
 4 .
 0
 
 3
22
The corresponding homogeneous system has the same solution as
x1  5 x3  0
2 x2  6 x3  0
x4  0.
Letting x3   , where  is any real number we have x2  3 and x1  5 so the general solution to
these homogeneous equations can be written as
 5 


 3  .
 


 0
Adding we see that the general solution to the nonhomogeneous equations can be written as
 10  5 


4  3 
*

x 
.
  


 3 
2.4
1. Writing our system as Ax  b we have
 1 2 2 
 1 2 2 




(i) A   0 2
1
 0 2
1 .
r  r  r
1 0
1 3 3 1  0 2 3 

Thus A  1 1
11
2 1
2 3
 8, and the equations have a unique solution. Now
 2
1

1
 0
 2 2
1
1
A  
8 0
1

 2 2
 2
1


0
1
1
1
1 2
1

1
1 2
0
1
0 2 

1 0
 2 1 2 
1 2  1
  2 3 2 


1 0  8 

6

1
2



1 2
0 2 
23
so the unique solution is
 2 2 6  1 
 42 



 
x  1  1 3 1 4   1  5  .
8
  8  22 
 2 2 2  8 
 
*
(ii)
 1 1 1
 1 1 1




A   1 2 3

1 2 3  .

 2 2 2  r3 r3  2r1  0 0 0 




Thus A  0 and no unique solution exists. Moreover the first and the third equations are inconsistent,
so no solution exists.
(iii)
 1 1 2 
 1 1 2 




A   2 2 4 

0 0 0,

 3 3 6  r2  r2  2r1  3 2 6 




so A  0, and no unique solution exists. Consider
 1 1 2 3 
 1 1 2 3 
 1 1 2 3 




 A b    2 2 4 6    3 2 6 9    0 1 0 0  ,
 3 2 6 9  r2  r3  2 2 4 6  r2  r2 3r1  0 0 0 0 



 r  r  2 r 

3
3
1
thus r ( Ab)  r ( A)  2 and we have an infinite number of solutions. Our system has the same solution as
x  y  22  3
 y  0.
Let z   , any real number. Then the general solution can be written as
 3  2 


x   0 .
  


*
2.
Consider a system of n linear equations in n variables which we write as
Ax  b.
24
If A is nonsingular then the solution for xi can be written as
a11
b1
a1n
x 
*
i
/ A,
an1
bn
ann
where the numerator is obtained by replacing the ith column A by the vector b and taking the determinant.
By Cramer’s rule,
1 2 2
x1 
x2 
x3 
4 2
1
8 0
A
1
3 0 3

4 2
1
8 0
A
1
1 1 2
1 1 2
0 4
1
0 4
1
1 8
A
1
0 7
A
3

1 2 1
1
2
0 2 4
0
2 4
2

3 3
8
1
 42 / 8.
A
4 1

7 3
 5/8
A
1
2 4
1 0 8 0 2 7 2 7


 22 / 8.
A
A
A
Exercises for Chapter 3
3.2
1. (i)
The endogenous variables are the variables of interest to us in our economic analysis. The
whole purpose of building the model in the first place is to get some insight into what determines
the values of these variables. The exogenous variables are variables whose values are taken as
final for the purposes of our analysis. There are noneconomic variables, economic variables
determined by noneconomic forces, or economic variables determined by economic forces other
than those at play in the model. Definitional equations represent relationships between the
variables of the model which are true by definition. Behavioral equations purport to tell us
something of the behavior of some economic entities. The structural form is the original form of
the model that comes to us from economists. The reduced form is obtained by solving for the
endogenous variables in terms of the exogenous variables. The solutions are called the
equilibrium values of the endogenous variables. A model is complete when the number of linear
equations in the model equals the number of endogenous variables and the model has a unique
solution.
25
(ii) Comparative static analysis concerns itself with how the equilibrium values of the endogenous
variables change when we change the given values of the exogenous variables. If we write the
structural form as
Ax  b
where x is a vector containing the endogenous variables and b is a vector containing the
endogenous variables or linear combinations of the exogenous variables, then the reduced form
is given by
x  A1b
and all our comparative static results are summarized by
x  A1b
where b is the vector of changes in the values of the exogenous variables. Often in our
economic analysis we are not interested in finding the complete reduced form. Instead, all that
we are interested in is the equilibrium value of one of the endogenous variables and our
comparative static analysis is concerned with how this equilibrium value changes when there are
changes in the values of the exogenous variables. In this case we can use Cramer’s rule to solve
for the equilibrium value of the endogenous variable in question.
2.
(i)
Isolating the endogenous variables on the left hand side of the equations we have
Y C  I G
bY  C  a
dY  I  c
or in matrix notation
 1 1 1 Y   G 

   
 b 1 0  C    a  ,
 d
  
0 1

 I   c 
which we write as
Ax  b.
Now we have three equations in three endogenous variables and
1 1 1
A  1 b
d
0 1   1 1
1 2
0
1
1  b 1
d
26
1
 1 b  d,
which we assume is nonzero so the model is complete. The reduced form is given by
x *  A1b
 1
0

1
 0
 1 1
1


1 b  d  0 1

 1 1

1 0


b
0
b
d
1
d
1 1
d

1

1
d
1 1
b
1
b
0
1 

0
G 
1   
 a
0   
 c
1 
1 
b
d   G 
1
1 G 
1
1

 

 
1
1

1 1 d
d a
b  a  .
 b 1 d
1  b  d 
1

b

d
d

b 1  b   c 
d 1  b 
1

 c 
(ii) We go to the first column of A1 and the elements of this column gives us the results. So
1
Y * 
1 b  d
b
C * 
1 b  d
d
I * 
.
1 b  d
(iii) Proceding as before we have
Y  C I G
 bY  C  a  bT
 dY  I  c
 G 
so all that changes is that the vector b is now  a  bT  whereas before it was
 c 


In our answer for (i) we replace a by a  bT .
If both exogenous variables increase by unit amounts then
1
 
b   b 
 0
 
27
G 
 
 a .
c
 
and
1
1 1 
1

 
1
x 
b 1 d
b  b  .

1 b  d 

b 1  b 
b
 0 
*
That is our results are obtained by adding to the first column of A1 , b times the second column of A1 .
For example
C* 
3.
b  b(1  d )
 bd .
1 b  d
1 b  d
Isolating the endogenous variables on the left hand side we have
Q  βP    βT
Q  bP  a  cR
which in matrix notation is
1    Q   α + βT 

   
.
 1  b  P   a +cR 
Using Cramer’s rule, the equilibrium values of the endogenous variables are given by
  βT
Q* 

a  cR b b   βT     a  cR 

1 
 b
1 b
and
1   βT
P* 
(i)
1
a  cR
a  cR    βT

.
 b
   b
T
28
We have
Q*  
bT   ve   ve   ve   ve   ve


  ve.
 b
 ve
  ve     ve 
Hence the increase in T leads to a decrease in the equilibrium quantity.
Similarly
P* 
 T   ve   ve   ve 

  ve,
 b
 ve
so the increase in T leads to a decrease in the equilibrium price.
(ii)
R
Similarly
βcR  ve   ve  ve 
Q* 

 ve,
 b
ve
so the decrease in rainfall leads to a decrease in the equilibrium quantity, and
P* 
cR   ve   ve 

  ve,
 b
 ve
so the decrease in rainfall leads to an increase in equilibrium price.
4.
Substituting the consumption and investments functions into the definitional equation, then isolating
the endogenous variables o the left hand side gives
Y (1   )  δr      G
τY + λr  M .
Using Cramer’s rule, the equilibrium values of our endogenous variables are given by
29
    G 
M
      G   δM
Y* 

1   
 (1   )  


1      G

r* 
M
1 




(1   ) M   (    G )
.
 (1   )  
If G and M change by G and M respectively we have the resultant change in r * is given by
1    M  G .
r * 
 1     
We want r * to be zero which implies that
1    M  G  0.
So the required decrease in M is given by
G
M 
.
1   
Exercises for Chapter 4
4.1
 1 0.5  x1 
.
0  x2 
(i)
 x1 x2   0.5
(ii)
 x y
 13 16  x 
  .
16 17  y 
 3 1 1.5  x1 

 
(iii)  x1 x2 x3   1 1 2  x2  .
1.5 2

3 

 x3 
30
 3 1 0  x1 

(iv)  x1 x2 x3   1 1 2 
 x2  .
 0 2 8  x 

 3 
 42 1.5 4  x1 
 
(v)  x1 x2 x3  1.5 2
3  x2  .
 4

3 7 

 x3 
4.2
1. (i)
Consider
A  I 
1 
1
1 1  
 1     1      2 
2
so the matrix has the eigenvalues 1  0 and 2  2.
(ii) Consider
5
A  I 
2
1
0  1 1
2 1 
1
2
31
1
1  0
0 1 
 1    1
3 3
5
2
2 1 
   1  1     5   1     4   1     5  6   2  5   1        6 
so the matrix has eigenvalues 1  1, 2  0 and 3  6.
(iii) Consider
2
A  I 
1
1
0  1 1
1 2
1 1 
1
0 1 
1
0
1 1 
 1    1
2 2
2
1 1 
  1     1     2   1     1  1     2  3   2  2 
 1        3 ,
so the matrix has eigenvalues 1  1, 2  0 and 3  3.
(iv) Consider
3  
A  I 
2
2 3  
0
0
0    5    1
3 3
3  
0 5  
2
   5     3     4     5      2  6  5 


   5      5    1 ,
so the eigenvalues are 1  2  5 and 3  1.
31
1
2
2 3  
2.
Consider
A  I 
a1  
b
b a2  
  a1    a2     b2  a1a2    a1  a2    2  b 2
so the equation which defines the eigenvalues is the quadratic equation
 2    a1  a2   a1a2  b2  0.
This equation has equal roots if and only if
2
 a1  a2   4  a1a2  b2   0.
That is  a1  a2   4b 2  0, but clearly this is impossible, for b  0.
2
4.4
1. (i)
The vectors x1 , ... , xm are orthogonal if xi x j  0 for i  j. They are orthonormal if they are
orthogonal and xi xi  1 for i  1,..., n.
(ii) Consider
 1
 
yx   y1 y2 y3   1  y1  y2  3 y3 .
 3
 
For y and x to be orthogonal we want this expression to equal zero. Hence three vectors orthogonal to x
would be
 1   1  3 
     
 1  ,  1 ,  0  .
 0   0   1
     
(iii) Consider
 1
 
y x1  1  y1 y2 y3  1  1  y1  y2  y3  .
3
3
 1
 
For y to be orthogonal to x1 we want this expression to be zero. Hence
 1 
 1 
 
 
 1 and  0 
 0
 1
 
 
would be orthogonal to x1 . Normalizing these vectors we have that
32
 1
 1
 
 
1
1
x2 
1 and x3 
0

2 
2  
 0
 1
meet the requirement.
2.
(i)
Consider
A  I 
2
4
4 2
  2     16  4  4   2  16   2  4  12,
2
so the matrix has two eigenvalues 1  6 and 2  2.
Eigenvector for 1  6
Consider
 4 4   x1   0 
     .
 4 4   x2   0 
 A  1I  x  
Both the equations give
x1  x2 .
We want our eigenvector to be normalized, that is
x12  x22  1
so
2 x22  1
and we can take
x2  1 .
2
Thus an eigenvector corresponding to 1 that meets the requirements is
x1 
1  1
 .
2  1
Eigenvector for 2  2
Consider
 4 4   x1   0 
     .
 4 4   x2   0 
 A  2 I  x  
33
Both these equations give
x1   x2
so a normalized vector corresponding to 2 would be
x2 
1  1
 .
2  1
Clearly x1 and x2 are orthonormal.
(ii) Consider
A  I 
4
2
2 1 
  4   1     4      5 ,
so the matrix has eigenvalues 1  0 and 2  5.
Eigenvector for 1  0
Consider
 4 2   x1   0 
     .
 2 1   x2   0 
 A  1I  x  Ax  
Both these equations give x2  2 x1. We want our eigenvector to be normalized, so we require that
5 x12  1 and we can take x1  1/ 5.
Our eigenvector associated with 1 would then be
x1 
1 1
 .
5  2 
Eigenvector for 2  5
Consider
 1 2   x1   0 
     .
 2 4   x2   0 
 A  2 I  x  
Both equations give
x1  2 x2
34
so the required eigenvector is
 2
x2  1   .
2 1
(iii) From exercise 1. (iii) of 4.2 we know the eigenvalues for this matrix are 1  1, 2  0, and
3  3.
Eigenvector for 1  1
Consider
1 1 1  x1   0 
  
 A  1I  x  1 0 0 
 x2    0 
1 0 0  x   0 

 3   
which render the equations
x1  x2  x3  0
x1  0,
so x3   x2 . For the vector to be normalized we want
x12  x22  x32  1
which implies that 2 x22  1 so take x2  1/ 2.
Our eigenvector associated with 1 would then be
 0
 
1
x1 
1 .
2  
 1 
Eigenvector for 2  0
Consider
 2 1 1  x1   0 
  
 A  2 I  x   1 1 0 
 x2    0 
 1 0 1  x   0 

 3   
which gives
2 x1  x2  x3  0
x1  x2  0
x1  x3  0.
35
Thus x1   x3   x2 . The normalized requirement implies that 3 x12  1 so we can take x1  1 and the
3
required eigenvector would be
 1
 
x2  1  1 .
3 
 1
Eigenvector for 3  3
Consider
 1 1 1  x1   0 
  
 A  3 I  x   1 2 0 
 x2    0  .
 0 0 2  x   0 

 3   
That is
 x1  x2  x3  0
x1  2 x2  0
x1  2 x2  0,
so x1  2 x2  2 x3 . For the vector to be normalized we require 6 x22  1 so we take x2 
eigenvector associated with 3 would be
 2
1  
x3 
1 .
6  
1
4.5
1. (i)
Consider
1 1 1 1 1   1 0  ;
 

2 1 1 
 1 1  0 1 
so the matrix is orthogonal.
(ii) Consider
 0
2

1 3  2
6
 3  2

2  0
3  3  1 0 0

 

1 2  2  2    0 1 0 


2
1
1  0 0 1 
1


which establishes the result.
36
1
and the
6
(iii) Consider

0
5 5  0
6 2 6   1 0 0 


 

1 
6 2 5 2  5 2 5
 5   0 1 0 ;
30 


2
1  0 0 1 
 2 6  5 1 5



so the matrix is orthogonal.
2.
Now
 1  2  1
QQ  1 

3 2
1

  2
2  1 0

.
1   0 1 
 1
QAQ  1 
3 2

2  1

1 
  2
and
 2  2

1 
 2
0 1
 0
 1

3  3 2 3    2

0 0

.
 0 3
3.
2

1 
2

1 
From exercise 1 of 4.2 we have that the eigenvalues of this matrix are 1  2  5 and 3  1.
Consider the following set of equations
 2 2 0  x1   0 
  
 A  1I  x   2 2 0 
 x2    0 
 0 0 0  x   0 

 3   
which imply that x1   x2 and x3 can take any value.
Taking x3  0 the normalization condition requires that x12  x22  1. That is 2 x12  1 so we can take
x1  1/ 2.
This gives the following eigenvector
 1
 
1
x1 
1 .
2  
 0
Taking x3  1 the normalization condition requires that x12  x22  1  1. So x1  x2  0. A second
eigenvector associated with the repeated roots that meet our requirements would be
0
 
x2   0  .
1
 
37
Consider now
 2 2 0  x1   0 
  
 A  3 I  x   2 2 0 
 x2    0 
 0 0 4  x   0 

 3   
which imply that
x1  x2 , x3  0.
The normalization condition requires then that
2 x12  1
so we take x1  1 and our eigenvector is
2
1
 
x3  1  1  .
2 
0
A Q that meets our requirement is
 1/ 2 0 1/ 2 


Q   1/ 2 0 1/ 2  .


0 1
0



The main diagonal elements of QAQ are the eigenvalues 5, 5, and 1.
4.
(i)
(a) From exercise 1. (i) of 4.2 the eigenvalues of this matrix are 1  0 and 2  2.
Consider the equations
 1 1  x1   0 
( A  1I ) x  Ax  
     ,
 1 1  x2   0 
which renders x1  x2 . The normalized condition requires x12  x22  1 so we take x1  1/ 2 and
 1
x1  1  
2  1
as our eigenvector.
The equations
 1 1  x1   0 
( A  2 I ) x  
    
 1 1  x2   0 
38
give x1   x2 so we take
 1
x2  1  
2  1
as our eigenvector. Then
 1 1
Q 1 
.
2 1 1
The main diagonal elements are the eigenvalues 0 and 2.
(b) From exercise 1. (ii) of 4.2 the eigenvalues of this matrix are 1  1, 2  0, 3  6 , so first we
consider
 4 2 1  x1   0 

   
( A  1I ) x   2 0 0  x2    0 
 1 0 0  x   0 

 3   
which gives x1  0 and x3  2 x2 . The normalization requirement is then 5 x22  1 so we take x2  1
5
and
 0
 
1
x1 
1
5  
 2 
as the eigenvector associated with 1.
Next consider
 5 2 1  x1   0 

   
( A  2 I ) x  Ax   2 1 0  x2    0 
 1 0 1  x   0 

 3   
which gives
5 x1  2 x2  x3  0
2 x1  x2  0
x1  x3  0.
So we have x1   x3   x2 / 2.
The normalization requires then that 3 x22  1 so we take x2  2 and as our eigenvector,
3
2
 1
 
1
x2 
2 .
6  
 1
39
Lastly,
1 x1   0 
 1 2

   
( A  3 I ) x   2 5 0  x2    0 
 1 0 5  x   0 

 3   
which gives
x1  5 x2  0
2 x1  5 x2  0
so x1  5 x3  2.5 x2 .
Normalization requires that x12  x22  x32  1 so 30 x32  1 and we take
x3  1/ 30 and as our eigenvector
5
 
x3  1  2  .
30  
1
An orthogonal Q that meets our requirements is
 0
 5 5


Q 1  6
2 5 2 .
30 

 2 6

5
1


The main diagonal elements of QAQ are the eigenvalues 1, 0, 6.
(c) From exercise 2. (iii) of 4.4 we have that a set of orthonormal vectors for this matrix are
 0
 1
 2




 
x1  1  1 , x2  1  1 , x3  1  1
2 
3 
6 
 1
 1
 1
so
 0
2

Q 1  3  2
6
 3  2

2

1 .

1
The main diagonal elements are the eigenvalues.
(ii) (a) As the eigenvalues are 0 and 2 we know the matrix is positive semidefinite and xAx  0
for all x .
(b) As the eigenvalues are 0, 1, and 6 the matrix is positive semidefinite and xAx  0 for all x.
(c) Again the eigenvalues are 0, 1, and 3 so the matrix is positive semidefinite and xAx  0
for all x.
40
4.6
1. (i)
From exercise 1 of 4.5 a set of orthonormal eigenvectors for this matrix is
1
1 
1  2 .

 and


3  2
2  1
so we consider the transformation
y  Qx

 1 
where Q   1 
 3  2



That is
1  2   .
 
2  1  
2 x2
y1  1 x1 
3
3
y2  x1  1 x2 .
2
(ii) From exercise 2 of 4.5 a transformation that meets our requirements is
 1  1 0
 2

2


y 0
0
1  x.
 1

1
0

2
 2

That is
y1  1 x1  1 x2
2
2
y2  x3
y3  1 x1  1 x2 .
2
2
2.
In matrix notation the quadratic form can be written as
 1 2  x 
 x y
  ,
 2 1  y 
so we consider
A  I 
1 
2
2 1 
 1     4
2
    3   1
and equating this determinant to zero gives eigenvalues 1  3 and 2  1.
41
Consider
 2 2   x1 
   0
 2 2  x2 
 A  1I  x  
which gives x1  x2 so a normalized eigenvector would be
x1 
1  1
 .
2  1
Consider
 2 2   x1   0 
    
 2 2   x2   0 
 A  2 I   
so x1   x2 and a normalized eigenvector would be
 1
x2  1   .
2  1
The required transformation is
 x  1 1 1  v1 
 

  .
2 1 1 v2 
 y
The quadratic form is indefinite as one of the eigenvalues is positive whereas the other is negative.
3.
If A is an n  n positive definite matrix then all the eigenvalues of A are positive so there exists an
orthogonal matrix Q such that
0
 1


QAQ  

0


n

where 1 ,..., n are all positive. Let
 1

D 

 0
0 

.

n 
Then as Q is orthogonal
A  Q D D Q
so let
P  D Q .
42
4.7
1. Consider
A 
 i i 
n
 i in  A
and
AA  i ii i / n2  A as i i  n, so A is symmetric and idempotent.
Now
B   I  A   I  A  B
as A is symmetric and
BB   I  A I  A)   I  2 A  A2  B
as A is idempotent, so B is also symmetric idempotent.
AB  A  I  A  A  A2  0
as A is idempotent.
As A and B are symmetric idempotent their ranks are equal to their traces so
r ( A)  trA  trii / n  1
and
r ( B)  trB  tr  I  A  trI  trA  n 1.
As the eigenvalues of a symmetric idempotent matrix are 1 or 0 and as the rank of a symmetric matrix is
equal to the number of nonzero eigenvalues, A has n1 eigenvalues of 0 and 1 eigenvalue of 1 whereas B
has n1 eigenvalues of 1 and 1 eigenvalue of 0. It follows that both A and B are positive semi definite
matrices (clearly all symmetric idempotent matrices are). Moreover as the determinant of a symmetric
matrix is the product of the eigenvalues both matrices have zero determinants.
2.
Consider


1

1









N  X X X  X X  X X 
X  N


and
NN  X  X X  X X  X X  X   N
1
1
so N is symmetric idempotent. Now
M    I  N   I  N
43
as N is symmetric and
MM   I  N   I  N   I  2 N  N 2  M
as N is idempotent, so M is also symmetric idempotent. Also
MN   I  N  N  N  N 2  0
as N is idempotent. Finally
1
r ( N )  trN  tr  X X  X X  trI K  K
and
r (M )  trM  trI n  K  n  K .
The matrix N has K eigenvalues of 1 and nK eigenvalues of 0 whereas M has nK eigenvalues of 1 and K
eigenvalues of 0. Both matrices have determinants of 0.
3.
(i)
From exercise 1. (iii) of 4.2 the eigenvalues of this matrix are 1, 0 and 3, so clearly the trace of
the matrix is the sum of the eigenvalues. Now
2 1 1 1 1 0
3 3 1 1
1 1 0  1 1 0  1 1
 0,
1 1
1 0 1 1 0 1
so the determinant is the product of the eigenvalues. The rank is 2.
(ii) From exercise 1. (ii) of 4.2 the eigenvalues are 1, 0, and 6 so clearly the trace is equal to the sum
of the eigenvalues. Now
5 2 1
4 2 0
2 1 0  2 1 0  1 1
1 0 1
1 0 1
3 3
4 2
2 1
0
so the determinant is the product of the eigenvalues. The rank is 2.
(iii) From exercise 4. (i) of 4.5 the eigenvalues are 1, 0, and 3, so clearly the trace is the sum of the
eigenvalues. The determinant is clearly zero, the product of the eigenvalues and the rank is 2.
4.8
1. (i)
The leading principal minors are
3 2 0
3 2
3 2
3,
 5, 2 3 0  5(1)33
 25.
2 3
2 3
0 0 5
These alternate in sign, the first being negative. Thus the matrix is negative definite.
44
(ii) The leading principal minors are
3,
3
3
1
1
 2, 1 1
1 1
0
0
0 2
2  1 1
2 8
0
6
2  1(1) 21
2 8
2
6
2 8
 4.
These alternate in sign, the first being negative. Thus the matrix is negative definite.
2.
The principal minors are
2, 1, 1,
1 0
0 1
2 1 1
 1,
2 1
1 1
 1,
2 1
1 1
1
1 1 0
1 1 0  1 1 0  0.
1 0 1
1 0 1
The principal minors are all nonnegative with one being zero. Thus the matrix is positive semidefinite.
3.
The first order principal minors are 1, 3, and 8 but a second order principal minor is
1 2
2 3
 1. This
is enough to establish that the matrix is indefinite.
Exercises for Chapter 5
5.2
1. Let f  x1,..., xn   f ( x) be a differentiable function of many variables. Then the gradient vector of
f(x) is
 f1 ( x ) 
.
f ( x )  

 f ( x) 
 n

That is it is the vector of first order partial derivatives of the function.
The Hessian matrix of the function is
f1n ( x ) 
 f11 ( x )
.
H  x   

 f ( x)

f
(
x
)
nn
 n1

That is it is the matrix of all second order partial derivatives of f(x).
(i)
so
y1  3x12 x22 , y2  x13 x2
 3x12 x22 
y ( x )   3  .
 2 x1 x2 
y11  6 x1 x22 , y12  6 x12 x2 , y22  2 x13
45
so the Hessian matrix is
 6 x1 x22
H ( x)   2
 6 x1 x2
(ii)
so
6 x12 x2 
.
2 x13 
y1  5 x14  6 x1 x2 , y2  3x12  2 x2
 5x14  6 x1 x2 
y ( x )  
.
2
 3x1  x2 
y11  20 x13  6 x2 , y12  6 x1 , y22  2
so
 20 x13  6 x2
H ( x)  
 6 x1
6 x1 
.
2 
(iii) y1  1/ x2 , y2   x1 / x22 so
 1/ x2 
y ( x )  
.
2
  x1 / x2 
y11  0, y12  1/ x22 , y22  2 x1 / x23 ,

0 1/ x22 
H  x  
.
2
3

1/
x
2
x
/
x
2
1
2

(iv) Write y   x1  x2  x1  x2  . Then
1
y1   x1  x2    x1  x2  x1  x2   2 x2  x1  x2  ,
1
2
2
y2    x1  x2    x1  x2  x1  x2   2 x1  x1  x2  ,
1
2
2
so
 2 x2  x1  x2 2 
.
y  x   
 2 x  x  x 2 
1
1
2


y11  4 x2  x1  x2  ,
3
y12  2  x1  x2   4 x2  x1  x2   2  x1  x2  x1  x2  ,
2
3
3
y22  4 x1  x1  x2  ,
3
so
H ( x) 
4 x2 2  x1  x2  

1
.
3 
4 x1 
 x1  x2   2  x1  x2 
46
(v)
y1  5  x12  x22  2 x1  10 x1  x12  2 x22  ,
4
4
y2  5  x12  2 x22   4 xx   20 x2  x12  2 x22  ,
4
4
so
4  10 x1 
y  x    x12  2 x22  
.
 20 x2 
y11  10  x12  2 x22   40 x1  x12  2 x22   2 x1 
4
3
  x12  2 x22   90 x12  20 x22 
3
y12  40 x1  x12  2 x22   8 x2   320 x1 x2  x12  2 x22 
3
3
y22  20  x12  2 x22   80 x2  x12  2 x22   4 x2 
4
3
  x12  2 x22   360 x12  20 x22  ,
3
so
2
2
3  90 x  20 x
1
2
H ( x )   x12  2 x22  
 320 x1 x2
(vi) y1 
320 x1 x2 
.
360 x22  20 x12 
2 x13
3x24
,
y

,
x12  x23 2 x12  x23
so
y  x  
 2 x13 
1

.
x12  x23  3x24 
Write y1  2 x13  x12  x23  so
1
y11  6 x14  x12  x23   2 x13  x12  x23 
1
 2 x14  x12  3x23  x12  x23 
y12  2 x13  x12  x23 
 6 x13 x24
2
 2 x 
3
1
2
 3 x 
x  x  .
2
1
2
4
2
3 2
2
Write y2  3x24  x12  x23  so
1
y22  12 x25  x12  x23   3x24  x12  x23 
1
 3x25  4 x12  x23  x12  x23 
2
47
2
 3x 
4
2
so
 2 x14  x12  3x23 

6 x13 x24
1

.
H  x 
2
3x25  4 x12  x23  
 x12  x23   6 x13 xx4
(vii) y1  e 2 x1 3 x2 , y2  3e 2 x1 3 x2
so
2
y  x   e2 x1 3 x2   .
 3
y11  4e2 x1 3 x2 , y12  6e2 x1 3 x2 , y22  9e2 x1 3 x2 ,
so
4 6
H  x   e2 x1 3 x2 
.
6 9
(viii) y1  6 x1 x2  7 x2 , y2  3x12  7 x1 / 2 x2
so
 6 x1 x2  7 x2 
y  x    2

 3x  7 x / 2 x 
1
1
2


3
y11  6, y12  6 x1  7 / 2 x2 , y22  7 x1 / 4 x2 2
so

6
H  x  
 6x  7 / 2 x
2
 1
2.
6 x1  7 / 2 x2 
.
3
2

7 x1 / 4 x2

y
x
y
x
 2 1 2,
 2 2 2,
 x1 x1  x2  x2 x1  x2
x22  x12
 2y
2
2 1
2
2 2

x

x

x
x

x
2
x

 1 2  1 1 2  1 2 2 2 .
 x12
x  x 
2
2
By symmetry
x12  x22
 2y

,
 x22  x 2  x 2 2
2
2
48
so clearly,
 2y  2y

 0.
 x12  x22
5.3
1.
(i)
(a)
AB
x2
x1
A  B is convex.
A  B  E2
x2
x1
A  B  E 2 is convex.
49
(b)
AB
x2
x1
A  B is convex
AB
x2
x1
A  B is not convex
50
(c)
AB
x2
x1
A  B is convex.
AB
x2
x1
A  B is convex.
(ii) (a) Let u  X , v  X  cu  z,and cv  z
Consider
c  u  1    v   cu  1    cv
  z  1    z
 z,
so u  1    v  X .
51
0   1
(b)
Let u, v  X  Au  b and Av  b and consider
A   u  1    v    Au  1    Av
0   1
 b  1    b
b
so u  1    v  X .
(c)
Similar to (b)
(d)
Let u and v  X  ui  0, vi  0 and consider u  1    v for 0    1. The
ith element of this point is
ui  1    vi   0  1    0  0,
so this point also belongs to the non negative orthant.
2.
Suppose S had two points say u and v. Then all points u  1    v  S for 0    1. But this
contradicts S having a finite number of elements. So S has no or one point only.
3. Suppose  s1, t1   S  T  s1  S and t1  T and  s2 , t2   S  T  s2  S and t2  T.
Consider the point
  s1 , t1   1    s2 , t2     s1 , 1    s2 , t1  1    t2  0    1
As S is convex  s1  1    s2  S and as T is convex t1  1    t2  T.
Graph of S  T
t
t2
T
t1
s1
S
52
s2
s
4.
Let u, v  X 1  X 2  u, v  X 1 and u, v  X 2
 u  1    v  X 1 and u  1    v  X 2
 u  1    v  X1  X 2
so X 1  X 2 is convex.
X 1  X 2 need not be convex as the following counter example shows:
x1
X1
x2
X2
X 1  x  E 2 / x1  0, x1  2, x2  0, x2  2
X 2  x  E 2 / x12  x22  1.
5.
Let y  f  x1 ,..., xn   f  x be a function of many variables. This function is homogeneous of
degree r if
f   x1 ,...,  xn    r f  x1,..., xn  .
Euler’s theorem
If y  f  x  is homogeneous of degree r and differentiable then
x1
f
f
 . . .  xn
 rf  x  .
 x1
 xn
53
(i)
Consider
1
1
f   x,  y     x  2   y  2  3   x    y    f  x , y  ,
1
2
and the function is homogeneous of degree 1. Now
1  12 12
1 12  12
1
f x  x y  6 xy , f y  x y  3x 2 y 2
2
2
so
xf x  yf y 
1 12 12
1 1 1
x y  6 xy 1  x 2 y 2  3x 2 y 1  f ( x, y )
2
2
and Euler’s theorem holds.
(ii) Consider
3
1
f   x,  y     x  4   y  4  6 x   f  x, y 
so the function is homogeneous of degree 1. Now
3 1 1
1 3 3
f x  x 4 y 4  6, f y  x 4 y 4
4
4
so
3 43 14
1 43 14
xf x  yf y  x y  6 x  x y  f  x, y  ,
4
4
and Euler’s theorem holds.
  x     y   3  f x, y
f   x,  y  
 
2
2
 x    y 
2
(iii)
2
so the function is homogeneous of degree 0.
f x  2 x  x 2  y 2   x 2  x 2  y 2  2 x  2 xy 2  x 2  y 2 
1
2
2
f y  2 y  x 2  y 2   y 2  x 2  y 2  2 y  2 yx 2  x 2  y 2  .
1
Thus
2
xf x  yf y   2 x 2 y 2  2 y 2 x 2  x 2  y 2   0,
2
so Euler’s theorem holds.
(iv) f   x,  y   3   x    y   2   x    y   3   x    y    6 f  x, y 
so the function is homogeneous of degree 6.
f x  15x 4 y  4 xy 4  9 x 2 y 3 , f y  3x5  8x 2 y 3  9 x 3 y 2 ,
5
2
4
54
3
3
so
xf x  yf y  15x5 y  4 x 2 y 4  9 x 3 y 3  3 yx5  8x 2 y 4  9 x 3 y 3  6 f  x, y 
and Euler’s theorem holds.
6.
(i)
For the Cobb-Douglas function


Q   K ,  L   A   K    L     Q  K , L 
so the function is homogeneous of degree    .
For the CES production function

Q   K ,  L   A a1   K   a2   L 


1
 
1
 Q  K , L        Q  K , L  ,
so the function is homogeneous of degree 1.
(ii) For the Cobb-Douglas function
QK   AK  1L , QL   AK  L 1
so
KQK  LQK   AK  L   AK  L      Q  K , L  ,
and Euler’s theorem holds.
For the CES production function
1
1
1
1
1
1
QK  A  a1K   a2 L   a1 K  1 , QL  A  a1K   a2 L   a2  L 1


so
1
KQK  LQL  A  a1K   a2 L  
1
a K

1
 a2 L   Q  K , L 
and Euler’s theorem holds.
(iii) For the Cobb-Douglas function
 1

QK   K ,  L    A   K    L 
     1QK  K , L 
so the marginal product QK is homogeneous of degree     1 . Similarly QL is homogeneous of
degree     1 .
For the CES production function

QK   K ,  L   A a1   K   a2   L 



1 

  oQK  K , L  ,
55
a1   K 
 1
 QK  K , L     
1 

  1
so the marginal product QK is homogeneous of degree 0.
7.
As f1  x1, x2  is homogeneous of degree r1 we have by Euler’s theorem that
x1 f11  x2 f12   r  1 f1
so
x12 f11  x1x2 f12   r  1 x1 f1. (1)
Similarly as f 2  x1, x2  is homogeneous of degree r1 we have
x2 x1 f12  x22 f 22   r  1 x2 f 2 (2)
Adding (1) and (2) gives
x12 f11  2 x1 x2 f12  x22 f 22   r  1 x1 f1  x2 f 2   r  r  1 f ,
by Euler’s theorem.
8.
(i)
A set X  Rn is a convex set if u, v  X  u  1    v  X , 0    1.
A function f(x) is convex if  f  u   1    f  v   f   u  1    v  for all points u, v in the
domain of the function.
f1  x2 , f 2  x1 , f11  0, f12  1, f 22  0
so the Hessian matrix is
0 1
H ( x)  
.
1 0
Principal minors are 0, 0, 1, so the function is neither convex nor concave.
(ii)
f x  4 x3  6 xy 2 , f y  4 y 3  6 x 2 y
f xx  12 x 2  6 y 2 , f xy  12 xy, f yy  12 y 2  6 x 2
so the Hessian matrix is
12 x 2  6 x 2
12 xy 
H ( x, y )  
.
2
2
12
xy
12
y

6
y


The principal minors are 12 x 2  6 y 2 ,12 y 2  6 x 2 and
12 x
2
 6 x 2 12 y 2  6 x 2   144 x 2 y 2  72 x 4  72 y 4  36 x 2 y 2
so these principal minors are  0 on R 2 . Thus the Hessian matrix is positive semidefinite and the
function is convex.
56
(iii) f x  6 x  2 y  3,
so the Hessian matrix is
 6 2 
H 
.
 2 2 
f y  2 x  2 y  4,
f xx  6, f xy  2 f yy  2
The principal minors are 6, 2, 8 so the matrix is negative semidefinite and the function is concave.
(iv) f x  e x  y  e x  y  3 , f y  e x  y  e x  y  1 ,
2
2
x y
x y
x y
x y
f xx  e  e , f xy  e  e , f yy  e x y  e x y
so the Hessian matrix is
 e x y  e x y
H ( x, y )   x  y
x y
e  e
e x y  e x y 
.
e x y  e x y 
The principal minors of this matrix are e x  y  e x  y and
e
x y
 e x  y    e x  y  e x  y   4e2 x which are all greater than zero on R2 and thus H is
2
2
positive semidefinite. The function is (strictly) convex.
(v) f x  2  2 x  2 y ,
so the Hessian matrix is
 2 2 
H 
.
 2 2 
f y  1  2 x  2 y ,
f xx  2, f xy  2, f yy  2,
The first order principal minors are 2 and 2 and the second order principal minor is 0 so the matrix is
negative semidefinite and the function is concave.
(vi) f x  1  e x  e x y , f y  1  e x y ,
Thus the Hessian matrix is
 e x  e x  y e x  y 
H ( x, y )  
.
e x  y e x  y 

f xx  e x  e x y ,
f xy  e x y ,
f yy  e x y .
The first order principal minors are e x  e x  y and  e x  y which are both  0 on R 2 and the second
order principal minor is
e x  y  e x  e x  y   e 2( x  y )  e x e x  y
which is  0 on R 2 thus the Hessian matrix is negative definite on R2 and the function is concave.
9. f x  2 x  y  3x 2 , f y  2 y  x,
so the Hessian matrix is
 2  6 x 1
H  x, y   
.
1 2 

f xx  2  6 x,
57
f xy  1,
f yy  2
For the function to be concave we want this matrix to be negative semidefinite. Thus we require all the first
order principal minors to be  0,
that is, 2  6 x  0  x  1 .
3
We also require that the second-order principal minor be  0,
that is, 4  12 x  1  0  x  7 /12.
Hence the largest convex domain in E 2 for which the function is concave is
 x 

D    x  7 /12  .
 y 

10. (i) f x  6 x  2 y  3,
so the Hessian matrix is
 6 2 
H 
.
2

2


f y  2 x  2 y  4,
f xx  6, f xy  2, f yy  2
Leading principal minors are 6 and H  8, so the matrix is negative definite on R2 and the function is
strictly concave.
(ii)
f x  4 x3  2 xy 2  3, f y  2 x 2 y  4 y 3  8,
f xx  12 x 2  2 y 2 ,
f xy  4 xy,
so the Hessian matrix is
12 x 2  2 y 2
H  x, y   
4 xy

f yy  2 x 2  2 y 2

.
2 x  12 y 
4 xy
2
2
The first order leading principal minor is 12 x 2  2 y 2 which is  0 on the set. The second order leading
principal minor is
12 x 2  2 y 2  2 x 2  12 y 2   16 x 2 y 2  24 x 4  132 x 2 y 2  24 y 4
which is also  0 on the set. Hence the Hessian matrix is positive definite on the set and the function is
strictly convex on the set.
(iii)
3
1
1
3


fx  1 x 4 y 4 , fx  1 x4 y 4 ,
4
4
7 1
3 3
f xx   3 x 4 y 4 , f xy  1 x 4 y 4 ,
16
16
1
so the Hessian matrix is
 3x  74 y 14
H  x, y   1 
16   43  43
 x y
7

f yy   3 x 4 y 4
16
3 3
x 4 y 4 
.
1 7 
4
4
3x y 
58
5
1

The first order leading principal minor is  3 x 4 y 4 which is  0 on the positive orthant whereas the
16
second order principal minor is
3
3
3
3
3
3
9 x2 y2  1 x2 y2  1 x2 y2
256
256
32
which is  0 on the positive orthant so the Hessian matrix is negative definite on the positive orthant and
the function is strictly concave on this set.
(iv) f x  3e x ,
f y  20 y 3 ,
f xx  3e x ,
f xy  0,
so the Hessian matrix is
 3e x

H  x, y , z    0
 0

f z  1/ z,
f xz  0,
0
60 y 2
0
f yy  60 y 2 , f yz  0,
f 22  1/ z 2 ,
0 

0 .
1/ z 2 
The first-order principal minor is 3e x which is positive, on the positive orthant, the second order principal
minor is 180e x y 2 which is positive on the positive orthant, and the third order principal minor is
180e x y 2 / z 2 which is also positive on the set. So the Hessian matrix is positive definite on the positive
orthant and the function is strictly convex on that set.
11. QK  aAK a 1Lb , QL  bAK a Lb1 ,
QKK  a  a  1 AK a2 Lb , QKL  abAK a1Lb1, QLL  b b  1 AK a Lb2
so the Hessian matrix is
 a(a  1) AK a 2 Lb
abAK a 1Lb1 
H  K, L  

abAK a 1Lb1 b(b  1) AK a Lb2 

(i)
For the function to be concave we want the first-order principal minors to be  0 and the second
order principal minor to be  0. Consider the first order principal minor
a(a  1) AK 2 Lb
which is  0 on the positive orthant if Aa (a  1)  0. With A  0 this requires that a  0 and a  1.
Similarly we require b  0 and b  1.
The second-order principal minor is
 AK

a 1 b1 2
L
ab(a  1)(b  1)  a 2b2    AK a 1Lb1  ab(1  a  b).
2
With A  0, a  0, b  0 for this to be  0 on the positive orthant implies that a  b  1. Moreover
a  b  1 with a  0 and b  0 implies that both a and b are  1. Hence the function is concave on the
positive orthant if A  0, a  0, b  0 and a  b  1.
59
(ii) For the function to be strictly concave, the first order leading principal minor of the Hessian matrix
should be  0 which requires that
Aa (a  1)  0
so with A  0 we need a  0 and a  1. The second order leading principal minor must also be  0 which
requires that ab 1   a  b   0. With a  0 this implies that b  0 and a  b  1 as required.
12. As f  x  and g  x  are convex functions
f   u  1    v    f  u   1    f  v 
g   u  1    v    g  u   1    g  v  ,
for 0    1.
Let h  x   af  x   bg  x  and consider
h   u  1    v   af   u  1    v   bg   u  1    v 
 a   f  u   1    f  v    b   g  u   1    g  v  
  h  u   1    h  v  ,
so h(x) is convex.
13. Let x1 and x2 belong to S  so f  x1   k and f  x2   k.
We wish to prove that x1  1    x2  S  for 0    1. Now as f(x) is convex for 0    1,
f   x1  1    x2    f  x1   1    f  x2 
  k  1    k
 k.
If f(x) is a concave function then S   x / f  x   k  is a convex set.
5.4
1. Consider a nonlinear equation
(5.1)
F  y, x1,..., xn   0.
The implicit function theorem addresses the question when it is possible to solve this equation for y as a
differentiable function of x1 ,..., xn .
In nonlinear economic models equilibrium conditions often give rise to an equation like (5.1). In this
equation the y is the endogenous variable and x1 ,..., xn are the exogeneous variables. Before any
comparative static analysis can be conducted we need to know the condition required to ensure that we
can solve the equilibrium equation for y as a differentiable function of x1 ,..., xn .
60
2.
(i)
Clearly the point satisfies the equation and the function on the left hand side has continuous
partial derivatives. Consider
Fy  4 x 2  14 xy  10
at the point in question, so an implicit function exists in the neighbourhood of this point.
Differentiating both sides of the equation with respect to x, remembering that y can now be regarded as a
function of x gives
dy
dy
3x2  8xy  4 x 2
 7 y 2  14 xy  0
dx
dx
so


dy
4 x2  14 xy  7 y 2  3x 2  8xy
dx
and
dy 7 y 2  3x 2  8 xy

.
dx
4 x 2  14 xy
(ii) The point satisfies the equation and the function on the left hand side has continuous partial
derivatives. Moreover
Fy  8 x  4 y 3  1
at the point, so an implicit function exists in the neighbourhood of this point. Differentiating both sides of
the equation with respect to x, remembering y can be treated now as a function of x gives
dy
dy
2 x  8 y  8x  4 y3
0
dx
dx
so


dy
4 y3  8x    2 x  8 y
dx
and
3.
dy
2x  8 y
 3
.
dx
4 y  8x
The point satisfies the equation and the function of the equation has continuous partial derivatives.
Consider
Fy  2 x2  2 y  2
at the point. So such an implicit function exists. Differentiating both sides of the equation with respect to
x1 and x2 gives
y
y
1  3 x2  2 x 2
 2y
0
 x1
 x1
61
so
y
2  x2  y    1  3x2 
 x1
and
1  3 x2
y

 2
 x1
2  x2  y 
at the point in question.
Similarly
3x1  2 y  2 x2
giving
y
y
 2 x2  2 y
0
 x2
 x2
3 x  2 y  2 x2
y
 1
5
 x2
2
2  x2  y 
at the point.
4.
Write the equilibrium condition as
Y  C Y   I Y   M Y   G  X  0.
We assume the functions have continuous derivatives. Consider
FY  1  C  I   M .
For a solution to exist we require
I  C  I   M   0.
Differentiating our equilibrium condition with respect to G, regarding Y as a function of G and X gives
 Y  C  Y  I   Y  M   Y  1  0
G
G
G
G
so
5.
Y 
1
.
 G 1  C  I   M 
The point satisfies the equations and the functions have continuous partial derivatives. The Jacobian
determinant is
5 y14 1
J 
 15 y14 y22  2 y1  2
2
2 y2 3 y2
at the point so the implicit functions exist.
62
Differentiating both sides of the equations with respect to x, treating y1 and y2 as functions of x, we have
dy dy
5 y14 1  2  1
dx dx
dy
2 y1 1  3 y22 dy2 dx  2 x.
dx
By Cramer’s rule
1
1
2 y1
3 y22
 3 y22  2 x 
2 x 3 y22
dy


dx1 5 y14
1
15 y14 y22  2 y1 
and
5 y14
1
2 y1 3 y22
dy

 15 y14 y22  2 y1  15 y14 y22  2 y1  .
4
2
dx2 15 y1 y2  2 y1 
At the point in question
dy
dy
 1 and
 1.
dx1
dx2
6.
(i)
Write our equilibrium equations as
F 1 Y , r , G , M   Y  C Y   I  r   G  0
F 2 Y , r , G , M   L  r , Y   M  0.
If we assume that all functions have continuous derivatives, then it is possible to solve for Y and r as
functions of G and M at points where
F 1 Fr1 1  C  I 
J  Y2

LY
Lr
FY Fr2
is nonzero, that is, the condition we require is
Lr 1  C  I LY  0.
(ii) Differentiating both sides of equilibrium equations with respect to M, remembering that now we
regard Y and r as functions of G and M we have
 Y  C  Y  I   r  0
M
M
M
Lr  r  LY  Y  1
M
M
63
which in matrix notation is
 Y
1  C  I     M
 L
Lr    r
Y

 M

   0 .
 1
  

Using Cramer’s rule we have
0 I 
 Y  1 Lr
I

.
 M 1  C   I  Lr  L  C   I LY
LY
Lr
and
1  C 0
LY 1
r 
1  C

.
 M  Lr  I  C   I LY  Lr 1  C   I LY
Using the apriori information we have on the signs of derivatives we obtain
Y 
 ve
  ve   ve.
 M   ve   ve     ve   ve   ve
So M and equilibrium Y move in the same direction,
 r   ve   ve.
 M  ve
So M and equilibrium r move in opposite directions.
5.5
1. (i)
For a function f(x) of a single variable the Taylor’s approximation of order three is
f   x0 
f   x0 
2
3
f  x   f  x0   f   x0  x  x0  
x  x0  
x  x0 


2!
3!
f   x0  2 f   x0  3
 f  x0   f   x0  dx 
dx 
dx
2!
3!
where
1
f  x0   x02 ,
1

f   x0   1 x0 2 ,
2
3
f   x0    1 x0 2 ,
4
Evaluating our approximation at x0  4 and dx  0.05 we have
2
3
f (4.05)  4.05  2  1  0.05  1  0.05  1  0.05
4
64
512
 2  0.0125  0.000039  0.0000002
 2.012461.
64
5

f   x0   3 x0 2 .
8
(ii) (a)
f  x    x  1 2
f  0  1
1
1
f   x   1  x  1 2
2
3
f   x   1  x  1 2
4
5
f   x  3  x  1 2
8
f  0  1
2
f   0   1
4
f   0  3
8
so
1 dx 3.
 x  1 2  1  12 dx  81 dx 2  16
1
For dx  0.2 we have
1.2  2  1  12  0.2   18  0.2 
1
(b)
2
3
 1  0.2   1  0.1  0.005  0.0005  1.0955
16
f  x   f   x   f   x   f   x   e x
so all these render 1 at x  0, thus
e x  1  dx  1 dx 2  1 dx 3
2
3!
and
2
3
e0.2  1  0.2  1  0.2   1  0.2   1  0.2  0.02  0.00133  1.22133
2
6
(c)
f  x   log x
f 1  0
f  x  1
x
f  1  1
f   x   12
x
2
f   x  3
x
f  1  1
f  1 2
so
log x  0  dx  1 dx 2  1 dx 3.
2
3
For dx=0.2 we have
2
3
log1.2  0.2  1  0.2   1  0.2   0.2  0.02  0.00266  0.18266.
2
3
65
f  x   f  x0   f  x0  dx  1 dxH  x0  dx,
2
where f  x  is the gradient vector of f(x), H  x  is the Hessian matrix of f(x).
2.
(i)
(ii) For this Cobb-Douglas function
1 1
3 3
f 1,1  1, f1  1 x1 4 x24 , f 2  3 x14 x2 4 ,
4
4
so
1
4
f 1,1    .
 3
 
4
Moreover
7
3

f11   3 x1 4 x24 ,
16
3
1


f12  3 x1 4 xx 4 ,
16
1
5

f 22   3 x14 x2 4 ,
16
so
 1 1 
H 1,1  3 
.
16  1 1 
Thus
f  x1, x2   1  1 dx1  3 dx2  3 dx12  3 dx1dx2  3 dx2 2 .
4
4
32
16
32
In moving from the point x1o  1, x2o  1 to the point x1  1.1 and x2  0.9 dx1  0.1 and dx2  0.1.
Thus we have
3
1
3  0.12  3  0.1 0.12  3  0.12
1.1 4  0.9  4  1  14  0.1  43  0.1  32
16
32
 0.94625.
3
3
1
1


(iii) dy  f1dx1  f 2dx2  1 x1 4 x24 dx1  3 x14 x2 4 dx2
4
4
o
o
At the point x1  1, x2  1 with dx1  0.1 and dx2  0.1 we have
dy  1  0.1  3  0.1  0.05.
4
4
3.
The total derivative is defined as
dy  y  y dx2


.
dx1  x1  x2 dx1
66
The two concepts differ when x2 is itself a function of x1 . If this is the case dy / dx1 is an approximation
for the rate of change in y when x1 changes by a small amount, taking into account as it does the direct
 y dx2
effect of this change given by  y /  x1 and the indirect effect through x2 given by
.
 x1 dx1
dx2 1
y
y
 3x12 ,
 14 x2 ,

so
 x1
 x2
dx1 x1
dy
 3x12  14 x2 / x1.
dx1
4.
We know that
dy  f1dx1  f 2dx2 .
Taking the differential of both sides remembering that f1 and f 2 are functions of x1 and x2 we have
d 2 y  d  dy    f11dx1  f12dx2  dx1   f 21dx1  f 22dx2  dx2
 f11  dx1   2 f12dx1dx2  f 22dx22
2
 dxH  x  dx.
Exercises for Chapter 6
6.1
1. (i)
Critical points where
f1  3x12  9 x2  0
f 2  3x22  9 x1  0.
From the second equation we have x1  x22 / 3. Substituting in the first equation gives
x24
 9 x2  0  x2  x23  27   0
3
so x2  0 or x2  3. The equation then has two critical points
0
 3
x1*    or x2*    .
0
 3 
Now
f11  6 x1, f12  0, f 22  6 x2
so the Hessian matrix is
9
 6x
H  x   1
.
9

6
x

2
67
and
0 9
H  x1*   
.
9 0
The first order principal minors of this matrix are 0, 0 and the second order principal minor is -81, so x1* is a
saddle point. Also
9
18
H  x2*   
.
 9 18 
The first order leading principal minors of this matrix is 18 and the second order leading principal minor is
233 so the critical point is strict local minimum.
(ii) Critical points where
f x  3x 2  y  1  0
fy  x  2 y  0
so x  2 y and
12 y 2  y  1  0   3y  1 4 y  1  0  y  1 or y   1 .
3
4
The function then has two critical points
 2
 1
 3
 2
*
*
x1  
 , x2   1 
1
 


 4
 3
Now
f xx  6 x, f xy  1, f yy  2
so the Hessian matrix is
 6 x 1
H  x  
.
 1 2
Now
4 1
H  x1*   

1 2
whose leading principal minors are 4 and 7 so H  x1*  is positive definite and x1* is a strict local minimum.
Also
 3 1
H  x2*   
,
 1 2
whose first order principal minors are -3 and 2 so H  x2*  is indefinite and the point is a saddle point.
68
(iii) Critical points where
f x  4 x3  2 x  6 y  0
f y  6 x  6 y  0
thus x  y and
4 x3  4 x  0  x  x 2  1  0
so x  0 or x  1 or x 1. The function then has three critical points
0
 1
 1
x1*    , x2*    , x3*    .
0
 1
 1
The Hessian matrix is
12 x 2  2 6 
H  x  

6 6 

so
 2 6 
H  x1*   
.
 6 6 
First order principal minors are 2, 6, whereas H  x1*   24 so this matrix is indefinite and x1* is a
saddle point. Also
 14 6 
H  x2*   
,
 6 6 
whose leading principal minors are 14 and 48 so this matrix is positive definite and x 2* is a strict local
minimum. Similarly x3* is a strict local minimum.
(iv) Critical points where
f1  2 x1  3x2  0
f 2  6 x2  3x1  4 x3  0
f3  4 x2  12 x3  0,
giving x2  3x3 , x1  3 x2  14 x3 which is only true if x1  x2  x3  0. Thus the function has one
2
3
critical point
0
 
*
x   0 .
0
 
69
The Hessian matrix is
 2 3 0 


H   3 6 4  .
 0 4 12 


The leading principal minors of this matrix are
2 3 0
2 3
3
2 2 2
2,
 3, H  3
6 4  4  1
4
3 6
9 14
9 14 0
so the matrix is positive definite and the critical point is a strict local minimum.
(v) Critical points where
f1  x3  2 x1  0
f 2  1  x3  2 x2  0
f3  x1  x2  6 x3  0,
so x3  2 x1 , x2  11x1 , and x1  1/ 20. Thus the function has one critical point
 1
 
1
x   11 .
20  
 2 
*
The Hessian matrix is
2 0 1


H  0 2 1
1 1 6


which has leading principal minors
0 2 11
2 0
11
3 1 2
2,
 4, H  0 2
1  1 1
 20.
0 2
2
1
1 1
6
All the leading principal minors are positive so H is positive definite and x * is a strict local minimum.
(vi) (d) The first order conditions are
fx  x  6 y  6  0
f y  6 x  2 y  3z  17  0
f z  3 y  8 x  2  0
70
and the first of these equations gives x    3  3 y  so substituting into the other equations gives
16 y  3 z  1
3 y  8 z  2,
Using Cramer’s rule to solve these equations gives
1 3
2 8
y
 14 /137
16 3
3 8
16
1
3 2
 29 /137
137
x  3  42 /137  369 /137,
z
and this is the one critical point the function has.
The Hessian matrix is
 2 6 0


H   6 2 3  ,
 0 3 8 


which has leading principal minors of 2 and
2 6
6 2
 32, so H is indefinite and the critical point is a
saddle point.
2.
Critical points where
f x  y 2  3x 2 y  y  y  y  3x 2  1  0
(1)
f y  2 xy  x 3  x  x  2 y  x 2  1  0 .
(2)
Clearly y  0 and x  0 define critical points. When y  0 we have from (2) that x  x 2  1  0 which gives
x  0, x  1, or x  1. When x  0 from (1) we get y  y  1  0 which gives y  0 or y  1.
Alternatively critical points are defined by
y  3x 2  1  0
2 y  x2  1  0
which yields 5 x 2  1 so x  1 or  1 with y  2 in both cases.
5
5
5
71
All in all we have six critical points
1 5 
 1 5 
0
0
1
 1
*
x1*    , x2*    , x3*    , x4*    , x5*  
 , x6  
.
0
1
 0
0
 25 
 25 
The Hessian matrix is

6 xy
2 y  3 x 2  1
H  x  
.
2
2x
 2 y  3x  1

 0 1
*
H  x1*   
 which is indefinite so x1 is a saddle point.
 1 0 
0 1
*
H  x2*   
 which is indefinite so x 2 is a saddle point.
1 0
 0 2
*
H  x3*   
 which is indefinite so x 3 is a saddle point.
2
0


0 2 
*
H  x4*   
 which is indefinite so x 4 is a saddle point.
 2 2 
 12
5 5
H  x5*   
 2
 5
2
5
.
2 

5
The leading principal minors of this matrix are 12 / 5 5 and 4/5 so H  x5*  is positive definite and x 5* is a
strict local minimum.
 12
5 5
H  x6*   
 2
 5
2
5
.
2


5
The first leading principal minor is 12 / 5 5 and the second leading principal minor is 4/5 so H  x6*  is
negative definite and x 6* is a strict local maximum.
72
6.2
1. (i)
Total profit is given by
1
 1

  p  L2  K 2   wL  rK .


(ii) The critical point is obtained from
1
 L  1 pL 2  w  0
2
1
 K  1 pK 2  r  0,
2
so we have one critical point
L*  p 2 / 4w2 , K *  p 2 / 4r 2 .
(iii) For a global maximum we require that the objective function  is concave on the nonnegative
orthant. The Hessian matrix of this function is
  32

pL
0 
1

H  L, K   
.
4 
3 
2 
pK 
 0
The first order principal minors of this matrix are
3
3
 1 pL 2 and  1 pK 2
4
4
which are both  0 on the nonnegative orthant and the second order principal minor is
3
3
H  1 p2 L 2 K 2
16
which is  0 in the nonnegative orthant so H(L,K) is negative semidefinite on the domain of our function,
and thus  is concave on its domain. Any local maximum will then be a global maximum.
(iv) L*   p,  w     p 
2
4   w   L*  p, w  , so the demand for labour is homogeneous of
2
degree 0. Similarly for K * .
*
p2
p2
p2
(v)  L   3  L*   3 w. Similarly K *   3 r.
w
2w
2w
2r
(vi) Substituting L* and K * back into the profit function gives the maximum profit function
*1 
p2 p2
 *1
M  p, w, r   p  L 2  K 2   wL*  rL* 

4 w 4r


73
 M  p  p  M   p  p   p
 2 w 2r 
 p 2 w 2r


 M   p  M   p  w
w
4 w2
4 w2
2
2
 M   p  M   p  r .
r
4r 2
4r 2
2
2.
(i)
2
Total profit is given by
1
1
  4 pL4 K 2  wL  rK .
(ii)
3
1
1
 L  pL 4 K 2  w  0,  K  2 pL4 K
 K  w  K  2wL .
2L r
r
1
2
 r  0,
Substituting into the first order conditions gives
1
1
2 pL 4 w 2
 L* 
4 p4
8 p4
*
,
K

.
r 2 w2
r 3w
(iii) For a global maximum want  to be concave on its domain, the nonnegative orthant. The
Hessian matrix is
 3  74 12 1  34  12 
L K 
 L K
H  L, K   p  4 3 1 2 1 3  .

 1 L 4 K  2
 L4 K 2 
 2

The first order principal minors are
1
1
7
3
 3 pL 4 K 2 and  L4 K 2
4
which are  0 on the domain of  . The second order principal minor is
3
H  1 p 2 L 2 K 1
2
which is  0 on the domain. Hence the Hessian matrix is negative definite,  is concave on its domain,
and any local maximum is a global maximum.
74
4  p
(iv) L   p,  r ,  w  
*
4
  r    w
2
2
 L*  p, r , w  , so L* is homogeneous of degree zero.
Similarly for K .
*
*
8 p4
8 p4
(v)  L   3  L*   3 w
w
rw
rw
4
4
*
 K   24 p  K *   24 p r.
r
r 4w
r 4w
(vi) The maximum total profit is formed by substituting L* and K * into . Thus the firm’s profit
function is
*1
M  p, r, w  4 pL 4 K
*1
2
 wL*  rK *  4 p 4 / r 2 w
 M  16 p 3 / r 2 w  M  16 p 3p / r 2 w
p
 M  4 p 4 / r 2 w2  M  4 p 4w / r 2 w2
w
 M  8 p 4 / r 3 w  M  8 p 4r / r 3w.
r
6.3
1. The Lagrangian function is
Z   x1  1 x2  2   12  2x1  4x2 .
The first order conditions are
Z   12  2 x1  4 x2  0
Z1  x2  2  2  0
Z 2  x1  1  4  0
 x1  2 x2  3
3
9
11
 x2  , x1  ,   .
4
2
8
The bordered Hessian matrix is
0 2 4


H   2 0 1 .
4 1 0


8 0 4
H  2 0
4
1  1 1
1 0
3 2
8 4
2
1
 16,
so we have a local maximum.
75
2.
(i)
The Lagrangian function is Z  x1x2   Y  p1x1  p2 x2  .
(ii) The first order conditions are
Z   Y  p1 x1  p2 x2  0
Z1  x2   p1  0
Z 2  x1   p2  0.
(iii) The border Hessian matrix is
 0 p1 p2 


H   p1 0
1
p

 2 1 0
H   1
1 2
p1
p1
p2
1
0
  1
1 3
p2
p1
p2
0
1
 2 p1 p2  0,
so the second order condition holds for maximization.
(iv) The second two of the first order conditions imply that
p
x2  1 x1
p2
so substituting into the first of these gives
x1*  Y / 2 p1 , x2*  Y / 2 p2 .
Also x1*   p1, Y   Y / 2p1  x1*  p1,Y  so x1* is homogeneous of degree 0. Similarly for x2* .
Suppose prices and income increase so they are  times their original values. All that changes in our
problem is the constraint which now becomes
 p1x1   p2 x 2  Y .
But this is the original constraint so nothing changes in the problem. Thus we would expect this result.
(v) M  p1, p2 ,Y   x1* x2*  Y 2 / 4 p1 p2
2
2
(a)  M   Y  M   Y2 p1
 p1
4 p1 p2
4 p1 p2
(b)  M  Y2  M  Y Y .
 Y 2 p1 p2
2 p1 p2
76
(vi) From the first order conditions we obtain
Y
M
 *  x1* / p2 

.
2 p1 p2  Y
3. (i) The problem is,
Minimize wL  rK
1
1
subject to 4 K 4 L4  Qo ,
so the Lagrangian function is
1 1


Z  wL  rK    Qo  4K 4 L4  .


(ii) The first order conditions are
1
4
1
4
Z  Qo  4 K L  0
1

3
Z L  w   4 L 4  0

3
1
Z K  r   K 4 L4  0.
(iii) The deriviatives needed to form the bordered Hessian matrix are
1

3

3
1
gL  K 4 L 4 ,
g K  K 4 L4
1
7

3
Z LL   K 4 L 4 ,
4
7
1


3
4
Z KK    K L 4 ,
4
3
3


1
Z LK   K 4 L 4
4
so the bordered Hessian matrix is
H  L, K ,   

7
4
1
L K
4

7
4

0
4 K 2 L 4 KL2 
 2

2
KL 
 4 K L 3 K
 4 KL2  KL 3 L2 


.
As our problem is a minimization problem we require H  0 at the critical point obtained from the first
order conditions.
(iv) From the last two first order conditions we have
w  K  K  wL .
r L
r
Substituting into the first of these conditions gives
 
Q 4 w
r
o
1
4
Q 
L 0 L  o 
 4 
1
2
*
2
 
r
w
1
2
Q 
K  o 
 4 
77
*
2
 
1
w 2.
r
(v) The cost function is
M  w, r , Qo   wL*  rK *  Qo2  rw  2 / 8.
1
1
(vi)  M  Qo  rw  2 / 4.
 Qo
(vii) From the first order completions
*
3
1
*
4
4
1
 Qo  rw  2 / 4   M /  Qo
  rK L
*
4. (i) The problem is,
Minimize p1 x1  p2 x2
subject to x1a x2b  uo ,
and the Lagrangian function associated with this problem is
Z  p1 x1  p2 x2    uo  x1a x2b 
The first order conditions are
Z  uo  x1a x2b  0
Z1  p1  a x
(1)
a 1 b
1
2
x 0
Z 2  p2  b x x
a
1
(ii)
b 1
2
(2)
0
(3)
Z11  a  a  1  x1a  2 x2b
Z12   ab x1a 1 x2b 1
Z 22  b  b  1  x1a x2b  2
so the border Hessian matrix is
 0
a2 b2 
H  x1 x2  ax1 x22
 bx12 x2

ax1 x22
a  a  1  x
2
2
ab x1 x2


ab x1 x2  .
b  b  1  x12 
bx12 x2
(iii) As this is a minimization problem we require H  0 at the critical point obtained from the first
conditions.
(iv) From (2) and (3)
ax2 p1
bp x

 x2  1 1 .
bx1 p2
ap2
78
Substituting into (1) gives
 bp 
uo  x1a  b  1   0.
 ap2 
Solving for x1 and using the fact that a  b  1 gives
b
 ap 
x1  uo  2  .
 bp1 
By symmetry,
a
 bp 
x2  uo  1  .
 ap2 
(v) Substituting x1 and x2 into the objective function gives
b
a
 ap 
 bp 
M  p1uo  2   p2uo  1 
 bp1 
 ap2 
 p1a p2buo a1 ab  b  p2b p1auo a  ab1 b
 p1a p2buo a  ab  b  a  b 
 a  ab  b p1a p2buo ,
as required.
(iv)  M  aa  abb p1a 1 p2buo ,
 p1
thus for small changes
M a bb  p p1b p2buo p1.
6.4
The set of feasible solutions is a hyperplane as it is the set of points which satisfy a linear constraint so this
set is a closed convex set. Consider the Hessian matrix of the objective function which is
 1
0
 x13

H  x1 , x2   

 0  13 

x2 

The first leading principal minor is  13  0 for all x  0 and the second leading principal minor is
x1
1  0 for all x  0 so this matrix is negative definite and the objective function is strictly concave.
x13 x23
Thus there will be a unique local maximum which is also a global maximum.
79
1.
(i)
The Lagrangian function is
1 1
Z     Y  p1 x1  p2 x2  .
x1 x2
(ii) The first order conditions are
Z   Y  p1 x1  p2 x2  0
Z1 
1
  p1  0
x12
Z2 
1
  p2  0.
x22
(iii) The bordered Hessian matrix of this problem is


 0
p1
p2 


H  x1 , x2    p1  23
0


x1


0  2 3
 p2
x2 

H  p1  1

2 1
p1
p2
0
2 / x
3
2
 p2  1
p1
3 1
p2
2 / x
3
1
0
2 p12 2 p22
 3  0,
x23
x1
for all positive x1 and x2 . Thus the second order condition holds.
(iv) From the first order conditions we have
x22 p1
p1

 x2 
x.
2
p2
p2 1
x1
Substituting gives
p1
x 0
p2 1
Y  p1 x1  p2
 x1* 
p 
Y

Y
p1 p2
1
x2* 
Now
p2 
p1 p2


.


x1*   p1 ,  p2 , Y   Y /  p1   2 p1 p2  x1*  p1 , p2,Y  ,
80
so x1* is homogeneous of degree zero. Similarly for x2* .
M  p1 , p2 , Y    1*  1*  
(v)
x1 x2

(a)  M   p1 2
 p1
1
(b)  M 
Y



p1 
 /Y.
2
p2

1
p1  p2 / Y  M   p1 2
 /Y
2
p1 
p2
2
 M 

p1 

p1  p2 p1 / Y
p2
 Y / Y .
2
(vi) From the first order conditions
1
* 

p1 x1*2

p1  p2
Y
2

2

M
.
Y
Exercises for Chapter 7
7.2
1. Total profit is
1
 12

  p  L  K 2   wL  rK .


Let M be the maximum profit function. By the Envelope Theorem
(i)
∂ M 

  L*   p 2 / 4w2
∂w ∂wL
*
(ii)
 M 

  K *   p 2 / 4r 2
r r K
*
(iii)
1
1
*
*
 M 
p
p
2

 L  K 2 

 p  p L ,K
2w 2r
*
*
which are the same results we got in exercise 1 of 6.2.
2.
Total profit is
1
1
  4 pL4 K 2  wL  rK .
81

2
Let M be the maximum profit function. By the Envelope Theorem
(i)
 M 
4 p4

  L*   2 2
w wL
r w
*
(ii)
 M 
8 p4

 K *   3
r r K
r w
*
1
1
*
*
 M 
4
(iii)

 4 L K 2  16 p3 / r 3 w,
 p  p L ,K
*
*
which are the same results we got in exercise 2 of 6.2.
7.3
1. (i)
The Lagrangian function is
Z  x1 x2   Y  p1x1  p2 x2  .
Let M be the indirect utility function.
Then by the Envelope Theorem
M Z

 p1  p1
M Z

 p2  p2
M Z

Y Y
  * x1*  
Y2
,
4 p12 p2
  * x2*  
Y2
,
4 p12 p2
 * , x1*
 * , x2*
 * 
*
Y
,
2 p1 p2
which are the same results we got in exercise 2 of 6.3.
(ii) The Lagrangian function is
1 1
Z      Y  p1 x1  p2 x2  .
x1 x2
If M is the indirect utility function, then by the Envelope Theorem
p  p2
M Z

  * x1*   1
,
 p1  p1 * , x*
p
Y
1
1
82
M Z

 p2  p2
M Z

Y Y
p1  p2
  * x2*  
p2 Y
 * , x2*
 *


*
p1  p2

,
2
Y2
,
which are the same results as we obtained for exercise 6.4.
2.
The Lagrangian function is
1 1


Z  wL  rK    Qo  4K 4 L4  .


If M is the cost function then by the Envelope Theorem
1
M Z
 Q   r 2

 L*   o    ,
w wL
 4   w
2
*
M Z

r r
K*
M Z

 Qo  Qo
3.
1
2
 Q   w 2
 K  o    ,
 4  r
*
1
 * 
*
Qo  rw  2
.
4
The Lagrangian function is
Z  p1 x1  p2 x2    uo  x1a x2b  .
Let M be the minimum value function. Then by the Envelope Theorem
M Z

 p2  p2
M Z

 uo  uo
a
 bp 
 x  uo  1  ,
 ap2 
*
2
x2*
   a  a b b p1a p2b .

83
7.4
1. (i)
We write the Lagrangian function as
Z  U  x1 , x2     p2 x2  p2 x2  Y  .
Then the first order conditions are
Z1  U1   p1  0
Z 2  U 2   p2  0
Z   p1 x1  p2 x2  Y  0.
(ii) The bordered Hessian matrix is
 0 p1 p2 


H  x1 , x2    p1 U11 U12 
p U
U 22 
21
 2
and the second order condition is
H 0
at the critical point obtained from the first order conditions.
(iii) We have
p1dx1  x1dp1  p2 dx2  x2 dp2  dY  0
U11dx1  U12 dx2   dp1  p1d   0
U 21dx1  U 22 dx2   dp2  p2 d   0.
Isolating the differentials of the exogenous variables on the right hand side we have
p1dx1  p2 dx2  dY  x1dp1  x2 dp2
U11dx1  U12 dx2  p1d    dp1
U 21dx1  U 22 dx2  p2 d    dp2
or in matrix notation
 d    dY  x1dp1  x2 dp2 

 

H  dx1   
 dp1
.
 dx  

 dp2
 2 

(1)
84
(iv) In (1) let dp1  dY  0
Then we obtain
 d     x2 dp2 

 

H  dx1    0 
 dx    dp 
2 
 2 
so using Cramer’s rule to solve for dx1 we get
dx1 
0
 x2 dp2
p2
p1
0
U12
p2
 dp2
U 22
H
Expanding the determinant in the numerator using the second column gives
H
dx1   x2 dp2 H12   dp2 32 ,
H
where H12 and H 32 are the cofactors of the (1, 2) and (3, 2) elements of H .
Dividing both sides of this equation by dp2 , and remembering that the ratio of differentials is a derivative
we have
H
 x1
H
  x2 12   32 .
(2)
 p2
H
H
(v) Let dp1  dp2  0 in (1) to obtain
 d    dY 

  
H  dx1    0 
 dx   0 
 2  
and using Cramer’s rule to solve for dx1 we have
dx1 
0
dY
p2
p1
0
U12
p2
0 U 22
.
dYH12
H 
H
Again dividing through both sides of this equation by dY gives
H
  x1 
 12 .


H
  Y Prices
constant
85
(3)
(vi) If utility remains constant then
dU  U1dx1  U 2 dx2  0
or
U1
dx1  dx2  0.
U2
But from the first order conditions
U1 p1

U 2 p2
so holding utility constant requires
p1dx1  p2 dx2  0.
But taking the differential of both sides of the budget constraint gives
p1dx1  x2 dp1  p2 dx2  x2 dp2  dY
or
p1dx1  p2 dx2  dY  x1dp1  x2 dp2 .
Thus holding utility constant requires that
dY  x1dp1  x2 dp2  0 .
(4)
(vii) Suppose that Y and p2 change in such a way that utility remains constant and suppose further
p1 does not change.
Then we must have
dY  x1dp1  x2 dp2  0
dp1  0.
Substitute these conditions in (1) gives
 d   0 

 

H  dx1    0  ,
 dx    dp 
2
 2 
and by Cramer’s rule
0
0
p2
p1
0
U12
p  dp2 U 22
dx1  2
dp H
H   2 32
H
86
so
  x1 
 H 32



H
  p2 Utility
constant
.
(5)
Substituting (3) and (5) into (2) gives Slutsky’s equation
 x1   x1 
 x 

 x2  1 

 p1   p1  Utility
  Y Prices
constant
constant.
(vii) The substitution effect of a change in p 2 on good 1 is
  x1 
 H 32

.


H
  p2  Utility
constant
Clearly the substitution effect of a change in p1 on good 2 is
  x2 
 H 23

.


H
  p1  Utility
constant
But as the bordered Hessian matrix H is symmetric, H 23  H 23 and the substitution effects are the same.
2.
(i)
Multiply both sides of (7.7) by p2 / xx we have
  x1  p2   x1 
p2
p2
 x 
.

 x2  1 



  Y Prices x1
  p2  x1   p2  Utility x1
constant
constant
Now we can write
p2 p2 x2   x1 
Y
 x 
x2  1 

  21


Y   Y Prices x1
  Y Prices x1
constant
constant
where
 2  the proportion of income Y spent on good 2
1  income elasticity of demand for good 1.
Thus in terms of elasticities we can write Slutsky’s equation (7.7) as
e12  12   22
where e12 is the cross elasticity of demand.
87
x 
p1   x2 
p1
(ii) 11  12   1 


  p1  Utility x1   p1  Utility x2
constant
constant


 p1H 22  pH32  .
H x1
But the term in the bracket is obtained by multiplying elements of the first column of H by the
corresponding cofactors of the second column of H so by our theorem this term is zero and thus
11  12  0.
7.5
1. (i)
The Marshallian demand functions are the optimal point of the following problem:
max imize
x1 x2
p1 x1  p2 x2  Y
subject to
The Lagrangian function associated with this problem is
Z  x1 x2   Y  p1x1  p2 x2 
and the first order condition are
Z1   x1 1 x2   p1  0
Z 2   x2 1 x   p2  0
Z   Y  p1 x1  p2 x2  0.
From the first two equations we obtain
 x2 p1
x p

 x2  1 1 .
 x1 p2
 p2
Substituting into the third equation gives
Y
.
x1* 
    p1
From symmetry
x2* 
Y
.
    p2
These are the Marshallian demand functions (we assume the second order conditions hold).
(ii) The indirect utility function is the maximum value function

V  p, Y   x x
*
1
*
2

 
      Y 
    

 p1   p2      
88
.
(iii) From the consistency properties
V  p, e   u,
so


 
      e 
    

 p1   p2      
1
  p   p      
 u  e      u  1   2   .
       
(iv) By Shephard’s Lemma the Hicksian demand function x1 is given by x1 
  p   p   
x1  u  1   2  
       
  p   p   
 u  1   2  
       
1  
 
1  
 

 1


e
, so
 p1
p  p 
u 2   1 
    
p  p  
u 2   1 
      p1
1
1
  p   p       
 
1
2
 u     
u
p1
       

  p2   

 .
  p1 
By symmetry
x2  u
2.
(i)
1
 

  p1   

 .
  p2 
The Hicksian demand functions are the optimal point of the following problem:
min imize
p1 x1  p2 x2
x
subject to

1
1
 x2    u.
The Lagrangian function associated with this problem is
1


  
L  p1 x1  p2 x2    u   x1  x2  


and the first order conditions are
L1  p1    x1  x2



1 
L2  p2    x1  x2 
L  u   x1  x2



1


x1 1  0
1 

x2 1  0
 0.
89
From the first two equations we have
p1  x1 
 
p2  x2 
 1
1
 p   1
 x1   1  x2 .
 p2 
Substituting into the third equation we have
1

 
    x  p  p  

1
2  1
2 

   p1 

u  x2    1   

  p


p2



2 


 
 
1
where    /    1
1

x2  p1  p2  
1
p2 1
so
1
x2 
up21
p

1

 p2
1




,
and
1
x1 
 1
up1
p

1

 p2
1
.
(ii) The expenditure function is the minimum value function of this problem that is
1
u  p1  p2 

 
e  p, u   p1 x1  p2 x2 

u
p

p
 1 2 .
1

 
 p1  p2 
By the consistency properties
e  p,V   Y
so
V  p1  p2


1
 
 y  V  y  p1  p2


1
 
90
(iii) By Roy’s Identity the Marshallian demand function is given by
V

p1
x1* 
V
y
1   p11
 1
    y  p1  p2 

 
1

 p1  p2  

yp1 1
.
p1  p2
x2* 
yp2 1
.
p1  p2
By symmetry
3.
(i)
The conditional demand functions for inputs are the optimal point of the following problem
w1 x1  w2 x2
min imize


1
y   x1  x2  .
subject to

The Lagrangian function is
1


  
Z  w1 x1  w2 x2    y   x1  x2   ,


and the first order conditions are
Z1  w1 
1 
 
x1  x2   ρx1 1  0


 w1    x1  x2 
Z 2  w2    x1  x2 
1

1 

x1 1  0
x2 1  0
1
Z   y   x1  x2    0.
The first two conditions give
w1  x1 
 
w2  x2 
 1

 1
 x1
w 
w

 1 x2 1  x1   1  x 2 , where  
.
w2

 w2 
Substituting into the third condition gives
  w  w
y   1  2
  w2
1

  
w2 y
x

x

.

 2
2
1

 
 
 w1  w2 
91
By symmetry

x1 
w1 y
w

1

 w2

1
.

(ii) The cost function is the minimum value function of this problem, that is,
1
w1  w2 


 
c  w, y   w1 x1  w2 x2  y

y
w

w
 1 2 .
1

 
 w1  w2 
(iii) Consider
1
 c  y  w  w   w 1  x .
1
2
1
1
 w1
Similarly
c
 x2 , so Shephard’s Lemma holds.
 w2
4.
The conditional demand functions are the optimal point of the following problem
(i)
min imize
w1 x1  w2 x
subject to
100 x12 x24  y.
1
1
The Lagrangian function associated with this problem is
1 1
Z  w1 x1  w2 x2    y  100 x12 x24  ,


and the first order conditions are
1
1
1
3
1
1
Z1  w1  50 λx1 2 x24  0
Z 2  w2  25 λx12 x2 4  0
Z   y  100 λx12 x24  0.
From the first two equations
w1 2 x2

w2
x1
so
x1 
2 w2 x2
.
w1
92
Substituting into the third equation gives
1
 2w  2 3
y  100  2  x24
 w1 
so
2
4
 y  3  w1  3
x2  

 
 100   2w2 
and
2
4
1
4
2w  y  3  w1  3  y  3  2w2  3
x2  2 
 
 .
 
 
w1  100   2w 2   100   w1 
(ii) The cost function is the minimum value function for this problem that is
c  w , y   w1 x1  w2 x2
1
4
2
4
 y  3  w1  3
 y  3  w1  3
 w1 

w
 
 
2


 100   2w2 
 100   2 w2 
4
1
 y  3 23
3.
 3
w
2
w


2
2  100  1
(iii) Marginal cost is
3
 c   y  3 w 23  2w2  .
 y  100  1 50
1
1
Now from the first order conditions
1
1
w x2 x 4
 1 1 2
50
2
1
1
 y  3  w1  6  y  3  w1 
 w1 
 
 
 

 100   2w2   100   2w2 
 c.
y
1
6
(iv) The factor demand functions are the optimal point of the following problem
1
max imize
1
  py  w1 x1  w2 x2  100 px12 x24  w1 x1  w2 x2 .
93
The first order conditions are
1
1
1
3
1  50 px1 2 x24  w1  0
 2  25 px12 x2 4  w2  0,
2w x
2 x2 w1

 x1  2 x
x1
w2
w1

so from the first condition
 2w x 
50 p  2 2 
 w1 

1
2
1
4
2
x  w1
thus
1

x2
1
4
1
w  2w  2  2w1w2  2
 1  2 
50 p  w1 
50 p
and
 50 p 

4
*
2
 50 p 

4
*
1
x
x
4 w13 w22
2 w13 w2
.
The profit function is the maximum value function of this problem that is
4
4
4
1
1
*
*
50 p   50 p   50 p 

*
*
*
2
4
  w, p   100 px1 x2  w1 x1  w2 x2  2


w1 w2
2w12 w2 4w12 w2
 50 p 

4
4w12 w2
.
(v) By Hotelling’s lemma the firm’s supply function is given by
*  50 p  50 504 p3
y  w, p  

 2 .
p
w12 w2
w1 w2
3
*
94
Exercises for Chapter 8
8.2
1. (i) The ordinates for the subintervals are
y1  0
y2  1
n
y3  2
n
yn 1  1
so
SR  1  y2 
n
 yn1   12 1  2 
n
 
 n   1 1  1  1 as n  
2
n
2
(ii) The ordinates for the subintervals are
y1  0

 2
n
y2  1
n
y3
2
2
yn 1  1
so
S R  1  y2   yn 1   13 1  22 
n
n
 1 as n  .
3
 n 2   1  2  3  12 
6
n n 
(iii) The ordinates for the subintervals are
y1  0

 2
n
y2  1
n
y3
3
3
yn 1  1
so
S R  1  y2 
n
2
2
 yn 1   14 1  2   n   1 4 n 2  n  1
n
4n
 1 1  2  12   1 as n  .
4 n n  4
95
8.3
1. (i) This function has a maximum at x  0 and cuts the x axis at x  4 and x  4.
So we want
I
2.

4
4
16  x dx  16 x  x / 3
2
4
 256 3.
3
4
Required areas are given by the following definite integrals:
(i)

9
1
2
2 32
52
x dx  x 
3 1 3
9
1
1
x5
1
7  x dx  7 x 
7 .
5 0
5
0
(ii)

(iii)

(iv)

2

1
1
4
2
3
0
5
2
 x  1 dx   x  1 2
5
1
3
x 5x2  2 
1
0
2
 .
5
1
5x2  2
4 2
 5856.4  60.025  5796.375
40
1
(v)
 x  1
2
dx    x  1
1 2
0
2

0
1
2
 .
x 1 0 3
8.4
1.
(i)
(ii)
(iii)
(iv)
(v)
(vi)

I  1  x  dx  2 1  x   C.

3
I  3xe dx  e  C.

2
x
2
I  x  2 x dx  
 x C

2 5
2
I  x  x dx  x  2 x  C

3
1
I   x  2 x  7   x  1 dx   x  2 x  7 

5
9
7
I  9 x 5  7 x 2 dx   x 4  x 3  C.
4
3

1
2
x2
x2
3
1
2
(vii) I 

(viii) I 

2
6

2
1
2
1
2
3
2
3
2
5
1
2
2
5
1


2
2
 10 x  6 x 


C
5
3 2
5
3 1
 2



2
2
2
 x  x  dx  4  x  x 




1
4
x
2
dx   1  x5  2  C
1
5
1  x5  2
96
5
2
 C.
2.
(i)
Let u  x  1 so x  u 2  1 , dx/du  2u and dx  2udu. Then
x
x  1dx 
2
 u
2
 1 2u 2 du  2 u 6  2u 4  u 2 du

2
7
3


5
2
2
x

1
x

1
 u7 2 5 1 3 





  C.
 2  u  u   C  2
 2  x  1 2 5 


7
5
3
7
3




2
 2
2udu
then
3
3
2
2
2
2
x 2 / 3 x  2 dx 
u

2
u 1udu 
u 4  4u 2  4du


27
27
2 1 5 4 3


 u  u  4u   C
27  5
3

(ii) Let u  3x  2

u
so x 
and dx =


Substituting u   3x  2  2 into this expression gives the result.
1
u
x
(iii) Let u  4 x  14 so
2
 14 
and dx  udu . As x ranges from 0 to 2 u ranges from
4
2
22. Moreover dx does not change sign. Thus
du
22
2

x
1 22  u  14  udu 1  u 3
dx 
   14u 
8 14
u
8 3
4 x  14
 14
3
3
11
1

   22  2  14 22  14  2  14 14 
83
3

1

23 14  20 22  0.4566
24
14 to


2
0

3.
4.

Thus substitution clearly will not work as when x ranges from 1 to 1 u remains at the value 2 and we
no longer have a definite integral.
In the formula for integrations by parts,
(i) Let u  cos x and v  x. Then
 x cos xdx   uvdx  x sin x   sin xdx  x sin x  cos x  C
(ii) Let u  x, v  log 3x. Then

x2
x2
x2
x2
x log 3xdx  log 3x 
dx  C  log 3x   C
2
3x
2
6

(iii) Let u  e x , v  x. Then
 xe dx   uvdx  xe   e dx  xe
x
x
x
x
 e x  C.
1
(iv) Let u  1  x  2 , v  x. Then

1
x 1  x  2 dx 
3
2
2
1

x
 2 x 
3
3

3
1  x  2 dx 
97
3
5
2
4
2
1

x
x

1

x
 
 2  C
3
15
(v) Let v  log x and u  1. Then
 log xdx  x log x   ldx  x log x  x  C
(vi) Let v  x3 , u  e3 x . Then
1
x3e3x dx  e3 x x 3  x 2 e3 x dx.
3


For the integral on the right hand side let u  e3 x and v  x 2 . Then
1
2
xe3 x dx  e3 x x 2 
xe3 x dx.
3
3


Again for the integral on the right hand side let u  e3 x , v  x. Then
1
1 3x
1
1
xe x dx  e3 x x 
e dx  e3 x x  e3 x  C
3
3
3
9


Substituting (3) into (2) the result thus obtained with (1) gives
1 
2
2
x3e3dx  e3 x  x3  x 2  x    C.
3 
3
9

(vii) Let u  eax and v  sin bx. Then
eax
b ax
I  eax sin bx 
sin bx 
e cos bxdx
a
a


(4)
For the integral on the right hand side let u  eax and v  cos bx. Then

eax
b ax
e cos bxdx 
cos bx 
e sin bx
a
a
1
  eax cos bx  bI  .
a

ax
Substituting back into (4) gives
ax
I  e sin bx  b2  eax cos bx  bI  .
a
a
Solving for I and adding a constant gives
ax
I  2e 2  a sin bx  b cos bx   C
a b
8.5
1.
(i)
(ii)



e x dx  lim
b 
0

e x dx

b
e x dx  lim eb  1  , so our integral is divergent.
b 
0
 lim
b 
0

b
0
b
e  x dx  lim  e  x   lim  eb  1  1
b 
0
b 
so the integral is convergent.
(iii)
x
1
0

1
2
dx  lim
 0

1

1
1


 1
x dx  lim  2 x 2   lim  2  2 2   2
 0


  0 

1
2
98
(2)
(3)
so the integral is convergent.
(iv)

1
x 2 dx  lim
 0
0

1
 1
x 2 dx   lim x 1  lim 1   ,


0
 0

 
1

so the integral is divergent.


Now
 x dx  lim  x
(v)
0
2
1
x 2 dx 
0
1

2
 0
1
x 2 dx 
2
1

2
x 2 dx.
0

1 
dx  lim x 1   lim   1  ,
 0
 1  0   
so our integral is divergent.
2.
(i)

2
1
(ii)
0

2

x
1

0

0
3 2


0

2
y
3
x2 y 
x
1
2
 3x  2 y  dx dy  
2
1
1
0

y2
2
1 y
1
1 
1
 dy  y  y 2   1
3 2
3
4 1
12
1
7
x3 7 x
8
x 2  dx  
 .
3
3 3 0 3
0
y3
dx 
x  y dydx  x y 
3 0
0
1
2
1

y2
1

1
x3 x 2 y

dy 
3
2 0
dx 
x 2  y 2 dydx 
2
0

2
1
1
0
(iv)
x  xydxdy 
2
1
0
(iii)
1
1
1
0
1
4 3
x4
1
x dx 
 .
3
3 0 3
9 x  6 xy  4 y dx dy 
2
2
0
 3x  3x y  4xy
3
2
0
1
1
3
1
4
121
 3y  3 y  4 y dy  y 7  y 6  y 5 
.
7
2
5 0 70
0
8.6
1. (i)
6
5
4
The Lagrangian function for this problem is
Z  x1 x2   Y  p1x1  p2 x2  .
The first order conditions are
Z1  x2   p1  0
Z 2  x1   p2  0
Z   Y  p1 x1  p2 x2  0.
The first two conditions imply that
px
x2  1 1
p2
so substituting into the third equation gives the Marshallian demand functions
x1*  Y 2 p1 , x2*  Y 2 p2 .
(ii) The indirect utility function is the maximum value function of this problem so it is
V  p, Y   x1* x2*  Y 2 p1 p2 .
99
y2
1
2
dy
0
(iii) CS 

1.2
1
1.2
Y
Y
 log p1  Y log1.2  0.09116Y .
2
2 p1dp1 2
1
C is defined by
V  p1 , Y  C   V  p 0 , Y 
Y  C 

4 1.2  p2
2
C Y

2
 Y  Y  C  Y 1.2
4 p2

1.2  1  0.09544Y .
E is defined by
V  p1 , Y   V  p 0 , Y  E 
Y  E 
Y2


4 p2
4 1.2  p2

2

 E  Y 1  1 1.2  0.08713Y.
(iv) As income Y  0 we have
E  CS  C.
2.
Now
(i)
By Roy’s identity the Marshallian demand functions are given by
V
pj
*
xj 
.
V
Y
V
1
Y
,
n
βi
 pi
i=1
and
n
γ
βj
V
 n j   y   γ i pi 
βi
p j
 p n p βi
 pi  i=1
j i
i 1
i 1
so
n


x*j   j   j  y    i pi   j 1   j    j  y    i pi 
i j
i 1



.
pj
pj
100
(ii) The loss in consumers’ welfare using consumers’ surplus is given by
CS 
where

1.1 p10
p10
x dp1 
*
1

1.1 p10
p10
1 1  1   p11dp1
 1  y    i pi  . Thus

i 1

CS   1 1  1  p1  log p1 1
1.1
  1 1  1 1.1 p10  log 1.1 p10   1 1  1  p10  log p10
  0.1  1 p10 1  1   log1.1.
(iii) The loss in consumer welfare measured by compensation variation C is defined by
V  p1 , y  C   V  p 0 , y  ,
where the new prices p1 are the same as the old prices p except p11  1.1 p10 . So
y  C    i pi0  0.1 1 p10
V  p1 , y  C  
i
1.1  pi0 βi
i
and C is given by
y  C    i pi0  0.1 1 p10
i
1.1  p
i
1
0 i
i
1

y    i pi0
i
0 i
 pi
,
i
that is,

y  C    i pi0  0.1 1 p10  1.1 1  y    i pi0  ,
i


that is,



C  0.1 1 p10  1.1 1  1  y    i pi0  .
i


The loss in consumer welfare measured by equivalence variation is defined by
V  p1 , y   V  p 0 , y  E 
where
V  p1 , y  
y    i pi0  0.1 1 p10
i
1.1  pi0 i
i
1
and
V  p0 , y  E  
y  E    i pi0
i
0i
i
p
i
.
101
Thus E is given by
y    i pi0  0.1 1 p10
y  E    i pi0 
i
i
that is,

1.1
,
1


0.1 1 p10  1.1 1  1  y    i pi0 
i

 C .
E
1

1.1
1.1 1
Exercises for Chapter 9
9.4
1. All the equations are first order linear differential equations with constant coefficients.
(i)
The general solution to the homogeneous equation is y  Ce4x with C an arbitrary constant.
A particular solution to the nonhomogeneous equation is
x
4 x
4 x
x
y  e 4 x e 4 s e 2 e ds  e
e6 s  0  e
e6 x  1  1 e 2 x  e 4 x .
6
6
6
0
 





The general solution to our equation then is
y  x  ce4 x  1 e2 x  e4 x .
6


Using the initial condition we get c  2 and the required particular solution of
2e4 x  e2 x  e 4 x
y  x 
.
6


(ii) The general solution to the homogeneous equation is y  ce6c with c an arbitrary constant. A
particular solution to the nonhomogeneous equation is
x
e7 x 6 x
1
y  e7 x e7 s e s ds 
e  1  e x  e7 x .
6
6
0





The general solution to our equation is then
y  x  ce7 x  1 e x  e7 x .
6


Using the initial condition we get c  1 so the required particular solution is
7 x
y  x   5e  1 e  x
6
6
102
(iii) The general solution to the homogenous equation is y  ce x , c an arbitrary constant.
particular solution to the nonhomogeneous equation is
y  e x

x
s 2e  s ds.
0
Now using integration by parts we have
s e
2 s

dx   s 2e s  2 se s ds
and
 se
s

ds   se s  e s ds   se  s  e  s
so
s e
2 s
ds   s 2e  s  2e  s  s  1  c.
It follows that

x
0
s 2e  s ds   s 2e  s  2e  s  s  1 0   x 2e  x  2e x  x  1  2
x
and that a particular solution to the nonhomogeneous equation is
y  x2  2  x  1  2e x .
The general solution to our equation then is
y  x   x2  2  x  1  2ex  cex  x2  2  x  1  cex
where c is an arbitrary constant.
y 1  1  1  5  ce  c  4e1
so required particular solution is
y  x   x2  2  x  1  4e x 1.
2.
(i)
We have
Y  Y  v
dY
I
dt
so
v
dY
 Y 1     I
dt
or
dY
I
 1
.
 aY   where a 
v
dt
v
103
A
(ii) Potential equilibrium where Y  Y and dY  0.
dt
That is at
Y 
aI
I

.
v 1
This is a particular solution to the nonhomogeneous equation. The general solution to the homogeneous
equation is Y  ce  at with c an arbitrary constant.
Adding we get the general solution to the nonhomogeneous equation given by
Y  Y  ce  at .
and
Y  Y if and only if e at  0. That is, if and only if a  0.
(iii) Want a  0 where a    1 .
v
As  is the marginal property to consume, 0    1 and   1  0. The accelerator v is v  0, so this
condition will not hold and Y will be subject to explosive growth. This may be a good representation in a
growth model.
3.
(i)
Consumers when faced with rising prices may buy now to avoid higher prices in the future. If this
rationale is correct we would expect c  0. Alternatively they may reason that rising prices today
must be matched by prices falling in the future so they will hold off buying now, in which case we
would expect c  0.
(ii) Equating Qs to Qd we have
dP
a  bP  c
  P
dt
or
dP
 a
b
 dP 
,
with d 
dt
c
c
which is a first order linear differential equation with constant coefficients. A particular solution to the
dP
 0 in the equation to give
nonhomogeneous equation is found by letting P  P and
dt
 a
.
P
b
As P is at rest at this value P is a potential equilibrium.
The general solution to the homogeneous equation is
P  t   edt  P  0 edt , an arbitrary constant.
Adding gives the general solution to the nonhomogenous equation
P  t   P  P  0 e dt .
104
(iii) Now
P  P iff d  0,
that is,
iff
b
 0.
c
We are given b  0 and   0 so the condition we require is c  0.
(  a )
If this is the case the resultant equilibrium for P is P 
.
(b   )
4.
(i)
Consider


1
Y   K ,  L     K    L    K L1
so Y  K , L  is homogeneous of degree one.
 Y   K  1L1  0  2Y     1 K  1L1  0.
K
 K2
2
Similarly  Y  0 and  Y2  0.
K
K
Now
Y  0, L   0 L1  0
Y  K ,0   K  0  0,
so the function satisfies all the requirements of a neoclassical production function.
(ii)
   
Y  Y K ,1  K
K
L
L

so  k   k

 k
(iii) The differential equation is
k   sk    n   k
or
k    n   k  sk 
which is clearly a Bernoulli Equation. Dividing through by k  gives
k k    n   k 1 α  s
and letting z  k 1 so
z  1   k k 
we write our equation in terms of z as
z   n   1    z  s 1   .
105
The general solution to this equation is
z  k 1  c1e
 1  n   t

s ,
n 
where c1 is an arbitrary constant.
(iv) As 0    1, 1    n    is positive so
e  1  n  t  0 as t  .
Thus
k 1 
s
n 
and

s
k 
n 
*

1
1
is the steady state level of k.
(v) The steady state level o consumption is
c*    k *    n    k *

s

n 


1
 n   

s
n 

1
1
.
The golden rule level of k occurs when
 k *   n   .
That is when
k * 1  n  
so

k  n 
*
g


1
 1

n  

1
.
comparing k * with k g* we see that the golden rule level of s is
s*g  .
9.6
All these equations are second order linear differential equations with constant coefficients.
1.
(i)
Homogeneous equation with auxiliary equation
m2  4m  4  0,
which has repeated roots of m1  m2  2, so the general solution is y  c1e2 x  c2 xe2 x , c1 and c2
arbitrary constants.
106
Now
y  0  c1  2
and
y  2ce2 x  2c2 xe2 x  c2e 2 x  4e 2 x  2c2 xe 2 x  c2e 2 x
so
y  0  4  c2  5  c2  1.
The required particular solution is then
y  2e2 x  xe2 x .
(ii) This homogeneous equation has an auxiliary equation given by
m2  m  2  0
which has roots m1  2, m2  1 so the general solution to the differential equation is y  c1e2 x  c2e  x ,
c1 and c2 arbitrary constants.
Now
y  0  c1  c2  1, y  2c1e2 x  c2e x , y  0  2c1  c2  5.
Solving the two equations in c1 and c2 gives c1   4 and c2  7 so the required particular solution is
3
3
y   4 e2 x  7 e x .
3
3
(iii) Nonhomogeneous equation with auxiliary equation
m 2  2m  3  0
which has roots m1  3,
m1  1 so the general solution to the homogeneous equation is
y  c1e3 x  c2e  x , with c1 and c2 arbitrary constants. For a particular solution of the nonhomogeneous
equation try
y  a0  a1 x  a2 x 2
so
y  a1  2a2 x and y  2a2 .
Substituting into our differential equation gives
2a2  2  a1  2a2 x   3  a0  a1 x  a2 x 2   9 x 2 .
Equating coefficients gives
2a2  2a1  3a0  0,  4a2  3a1  0,  3a2  9
107
so
a2  3, a1  4 and a0   14 .
3
Therefore a particular solution to the nonhomogeneous equation is
y   14  4 x  3x2
3
and the general solution is
y   14  4 x  3x2  c1e3 x  c2e x .
3
(iv) The auxiliary equation is
m 2  2m  3  0
which has roots m1  3 and m2  1 so the general solution to the homogeneous equation is
y  c1e3 x  c2e  x , c1 and c2 arbitrary constants.
For a particular solution to the nonhomogeneous equation we try
y  cxe x
so
y  ce x  cxe x , y  2ce x  cxe x .
Substituting into our differential equation gives
2ce  x  cxe  x  2  ce  x  cxe  x   3cxe  x  8e  x ,
that is,
4ce  x  8e  x  c  2.
Thus a particular solution to the nonhomogeneous equation is
y  2 xe x
and the general solution is
y  2 xe  x  c1e3 x  c2e  x .
(v) This equation has auxiliary equation given by
m2  m  2  0
which has roots m1  2, m2  1, so the general solution to the homogeneous equation is
y  c1e x , c1 and c2 arbitrary constants.
For a particular solution of the nonhomogeneous equation by
y  cxe2 x
108
with
y  ce2 x  2cxe2 x , y  4ce2 x  4cxe2 x .
Substituting in our equation gives
e2 x  4c  4cx  2cx  2cx   4e2 x  c  4 ,
3
so the general solution to our equation is
y  4 xe2 x  c1e2 x  c2e x .
3
Now
y  0   c1  c2  5
y  4 e 2 x  8 xe 2xc  2c1e 2 x  c2e  x
3
3
4
y  0    2c1  c2  1,
3
so
c1  14 , c2  31
9
9
and our required particular solution is
(12 xe2 x  14e2 x  31e x )
y
.
9
(vi) The auxiliary equation is
2m 2  2m  1  0
which has conjugate complex roots
m1  1  i , m2  1  i
2 2
2 2
so the general solution to our equation is
 c cos x c2 sin x 
y  2  ex 2  1

2 
 2
with
y  e
Hence
 c1 cos x c2 sin x  x 2  c1 sin x c2 cos x 

e 

.
2  2
2 
2 2 
 2 2
x 2
y  0   2c1  4  c1  2
y  0  
(c1  c2 )
 2  c2  2
2
so the required particular solution is
y  e x 2 2cos x  2sin x .
2
2


109
(vii) The auxiliary equation is
m2  m  0
which has roots m1  0, m2  1 so the general solution to the homogeneous equation is y  c1  c2e  x ,
where c1 and c2 are arbitrary constants. For a particular solution to the nonhomogeneous equation try
y  a0  a1 x.
But a constant appears in the general solution of the homogeneous equation so instead try
y  a0 x  a1 x 2 .
Now
y  a0  2a1x, y  2a1
so substituting into our differential equation gives
2a1  a0  2a1x  x  a1  1 , a0  1
2
and the general solution to the nonhomogeneous equation is
2
y   x  x  c1  c2e  x .
2
Differentiating we have
y  1  2 x  c2e  x .
Using the initial conditions we have
y  0   c1  c2  1
y  0   1  c2  0
 c2  1 and c1  2,
so our required particular solution is
2
y   x  x  2  e x .
2
(viii) The auxiliary equation is
m 2  2m  1  0
which has repeated roots m1  m2  1, so the general solution to the homogeneous equation is
y  c1e  x  c2 xe  x , c1 and c2 arbitrary constants.
For a particular solution tempted to try y  ce x but e  x appears in the general solution of the
homogeneous equation as does xe  x so instead we try
y  cx 2e x .
110
Now
y  2cxe  x  cx 2e  x  ce x  2 x  x 2 
y  ce x  2 x  x 2   ce  x  2  2 x   ce  x  2  4 x  x 2  .
so substituting into our differential equation gives
ce x  2  4 x  x2  4 x  2 x 2  x 2   e x  c  1 .
2
The general solution to our equation is then
2 x
y  x e  c1e  x  c2 xe  x .
2
2.
Consider y  y1  y2 and
ay  by  cy  a  y1  y2   b  y1  y2   c  y1  y2 
  ay1  by1  cy1    ay2  by2  cy2 
 f1  x   f 2  x  .
(i)
which
The auxiliary equation is
m2  3m  0
has
roots
m1  0, m2  3
so
the
solution
to
the
homogeneous
y  c1  c2e , where c1 , c2 are arbitrary constants.
3x
For f1  x   6 try y  a0 x. Then y  a0 , y  0 and we have 3a0  6  a0  2 .
For f 2  x   3e3 x try y  cxe3 x . Then
y  ce3 x  3cxe3 x  ce3 x 1  3x 
y  3ce3 x 1  3x   3ce3 x  3ce3 x  2  3x 
so
3ce3x  2  3x   3ce3x 1  3x   3e3x  c  1.
Combining we have that a particular solution to our differential equation is
2 x  xe3 x
and the general solution is
y  2 x  xe3 x  c1  c2e3 x .
111
equation
is
(ii) From (vii) of question 1 the general solution to the homogeneous equation is
2 x
y  c1e  x  c2 xe  x , c1 and c2 arbitrary constants, and a particular solution is y  x e . For
2
f2  x   3x we try y  a0  a1x as a particular solution so y  a1 , y  0 and
2a1  a0  a1x  3x  a1  3 and a0  6.
Combining we have a particular solution to the nonhomogeneous equation as
2 x
y  6  3x  x e  c1e  x  c2 xe  x .
2
(iii) From (v) of question 1 the general solution to the homogeneous equation is
y  c1e 2 x  c2e  x , with c1 and c2 as arbitrary constants.
For f1  x   6 x, we try y  a0  a1x so y  a1 , y  0 and then a2  2a0  2a1 x  6 x  a  3,
a0  3 .
2
For f 2  x   4e x try y  cxe x
so
y  ce x 1  x  , y  ce x  x  2
and we have
ce x  x  2  1  x  2 x   e x  c   4 .
3
Combining we have the particular solution to the nonhomogeneous equation as
y  3  3x  4 xe x
2
3
and the general solution is
y  3  3x  4 xe x  c1e2 x  c2e x .
2
3
9.7
Routh’s theorem states real roots and the real parts of imaginary roots of the nth degree polynomial
equation
a0 mn  a1m n 1   an 1m  an  0
are negative if and only if the first n of the determinants
a1 a3 a5 a7
a1 a3 a5
a a3
a a2 a4 a6
a1 , 1
, a0 a2 a4 , 0
a0 a2
0 a1 a3 a5
0 a1 a3
0 a0 a2 a4
112
are all positive, where it is understood that in these determinants we set ar  0 for all r  n needed to
obtain these determinants.
We know y  x  will be convergent if and only if the real roots or the real parts of complex roots are all
negative. Applying Routh’s theorem to our equation, this will be the case if and only if
a 0
a,
are  0  a  0, ab  0  a  0 and b  0.
1 b
2.
(i)
In this equation a1  21 so by Routh’s theorem the time path is not convergent.
(ii) From this equation
a1  21  0
a1
a3
a0
a2
a1
a3
0
a0
a2
0  1 36
0
a1
a3

21 20
1 36
0
21 20
0
0
0  20  1
3 3
21 20
1 36
21 20
The time path for y is convergent.
(iii) From this equation
a1  8
a1
a3
a0
a2

8
4
1 7
 60,
so the time path for y is divergent.
9.8
1. (i)
In this example
 ve    ve
dy2
f
 1 
dy1 y   0
f2
 ve
1
so y1  0 has a positive slope.
113
 0.
Similarly y2  0 has a positive slope, so we have two possible phase diagrams:
y2
y2
y1  0
(1)
y2  0
(4)
y1  0
y2  0
(2)
(3)
(a)
y1
(b)
y1
Also,
 y1
 f1  0
 y 2
so y1 and y1 move in opposite directions so as y1 increases y1 must decrease from positive to zero, to
negative which give the directional arrows for y1 in the two diagrams.
 y2
 g2  0
 y2
so as y 2 increases, y 2 must decrease from positive, to zero to negative, which gives the directional
arrows for y 2 as shown in the diagrams.
Phase diagram (a) gives convergent time paths for our variables without cycles.
Phase diagram (b) gives the same in sections (1) or (2) but divergent time paths in sections (3) or (4).
(ii) In this example
 ve   ve
 y2
f
 1 
 y2 y 0
f2
  ve 
1
and
 y2
 y1

y2  0
 ve    ve
g1

g2
  ve 
114
so the slope of y1  0 is negative and the slope of y2  0 is positive as shown in the diagram:
y2
y2  0
y1  0
y1
Also
 y1
 f1  0
 y1
so as y1 increases y1 must decrease from positive to zero to negative which gives the directional arrows
for y1 as shown in the diagram.
Similarly
 y2
 g2  0
 y2
so as y2 increases, y2 must also increase from negative to zero to positive which gives the directional
arrows for y2 as shown in the diagram.
Thus the time paths for our variables is divergent without cycles.
(iii) In this example
  ve   ve
 y2
f
 1 
 y1 y   0
f2
  ve 
1
115
so the slope of y1  0 is negative, so we have two possible phase diagrams:
y2
y2
y1  0
y2  0
(2)
(3)
y2  0
y1  0
(1)
(a)
y1
(4)
(b)
y1
The directional arrows are obtained by considering
 y1
 f1  0;
 y1
so as y1 increases y1 must increase going from negative to zero to positive.
Thus the directional arrows for y1 are as given in the diagrams.
By the same reasoning, as y2 increases y2 must increase going from negative to zero to positive thus
giving the directional arrows for y2 as shown in the diagrams.
Phase diagram (a) gives rise to divergent time paths for our variables without cycles.
Phase diagrams (b) gives convergent time paths for our variables, without cycles in sections (3) and (4) but
divergent time paths without cycles in sections (1) and (2).
(iv) In this example
  ve   ve,
 y2
f
 1 
 y1 y   0
f2
  ve 
1
116
so the slope of y1  0 is negative
and
  ve    ve,
 y2
g
 1 
 y1 y   0
g2
 ve 
2
so the slope of y2  0 is positive. Thus we get the following phase diagram:
y2
y2  0
y1  0
y1
As
 y1
 f1  0 ,
 y1
as we increase y1, y1 must also increase going from negative to zero to positive which gives the
directional arrows for y1 as shown in the diagram.
 y2
 f2  0
 y2
as we increase y2 , y2 must decrease giving the directional arrow as shown in the diagram.
Thus the time paths of our variables diverge without cycles.
117
Exercises for Chapter 10
10.2
1. (i)
y t 1
f
yt
Clearly we require f   0 and f   1.
(ii)
yt 1
f
yt
We require f   0 but f   1.
118
(iii)
f
y t 1
yt
We require f   1.
(iv)
y t 1
yt
We require f   1.
119
f  10  5 log yt so f   5  1 for 0  yt  1.
yt
The time path of y is divergent without oscillations.
(a)
(b)
f  3  8 cos yt so f   8 sin yt  0 for 0  yt  , so the time path has no oscillations.
For 0.122  yt  .878, 8 sin yt  1 and the time path is divergent, otherwise convergent.
(c)
f  10  8 yt5 so f   40 yt4  1 for yt  1 so the time path has no oscillations and is divergent.
10.3
1. All the equations are second order linear difference equations with constant coefficients.
(i)
We have
y  x   y  x  1  y  x  , 2 y  x   y  x  2  2 y  x  1  y  x 
so writing our difference equation in standard format gives
y  x  2  4 yy  x  1  3 y  x   0
which is a homogeneous equation whose auxiliary equation is
m2  4m  3  0,
with roots m1  1 and m2  3. The general solution is
y  c1  c2 3x , c1 and c2 arbitrary constants.
(ii) We have
y  x  1  y  x  1  y  x  2 
 2 y  x   y  x   2 y  x  1  y  x  2 
so our difference equation is
y  x   4 y  x 1  4 y  x  2  x2
with auxiliary equation
m 2  4m  4  0.
This equation has repeated roots
m1  m2  2
so the general solution to the homogeneous equation is
y   c1  c2 x  2x.
For a particular solution try
y  a0  a1 x  a2 x 2 .
120
Substituting into our equation gives
2
a0  a1 x  a2 x 2  4  a0  a1  x  1  a2  x  1 


2
2
4  a0  a1  x  2   a2  x  2    x .


Equating coefficients gives
a0  4a1  12a2  0
a1  8a2  0
a2  1
 a1  8, a0  20,
so a particular solution to the nonhomogeneous equation is
y  20  8 x  x 2
and the general solution is
y  20  8x  x2   c1  c2 x  2x.
(iii) Writing our equation in standard format gives
y  x  2  2 y  x  1  2 y  x   0
which has an auxiliary equation
m 2  2m  2  0.
The roots of this equation are conjugate complex numbers
m1  1  i and m2  1  i
so the general solution to the homogeneous equation is
x

 

y  2 2  c1 cos x  c2 sin x  .
4
4 

For a particular solution of the nonhomogeneous equation try
y  a0  a1 x  a2 x 2 .
Substituting into our differential equation gives
2
2
a0  a1  x  2   a2  x  2   2 a0  a1  x  1  a2  x  1 


2  a0  a1 x  a2 x 2   6 x  x 2 .
Equating coefficients gives
a0  2a2  0, a1  6, a2  1
so a particular solution to the nonhomogeneous equation is
y  2  6 x  x 2
121
and the general solution is


y  2  6 x  x 2  2 2 c1 cos  x  c2 sin  x .
4
4
x
(iv) The auxiliary equation is
m 2  4m  4  0
which has repeated roots m1  m2  2 so the general solution to the homogeneous equation is
y  c1 2 x  c2 x 2 x , c1 and c2 arbitrary constants.
For f1  x   2x try y  cx 2 2x which gives
cx 2 2 x  4c  x  1 2 x 1  4  x  2  2 x  2  2 x.
2
2
Equating the coefficient of 2 x gives
2c  1  c  1 .
2
For f2  x   8x try y  a0  a1 x which gives
a0  a1 x  4  a0  a1  x  1   4 a0  a1  x  2    8 x.
Equating coefficients gives
a0  4a1 , a1  8, a0  32 .
Combining we have that a particular solution for the nonhomogeneous equation is
y  cx 2 2 x 1  32  8 x
and the general solution is
y  cx 2 2 x 1  32  8 x  c1 2 x  c2 x 2 x.
(v) The auxiliary equation is
m 2  3m  2  0
which has roots m1  2, m2  1 so the general solution to the homogeneous equation is
y  c1 2 x  c2 , c1 and c2 arbitrary constants.
For f1  x   c2 try y  a0 x which gives
2a0  x  2  6a0  x  1  4a0 x  5  a0   5 .
2
For f 2  x   2x try y  cx 2 x which gives
2c  x  2 2x2  6cx2x1  4cx2x  2x.
122
Equating the coefficients of 2 x gives
16c  11  c  1 .
16
Combining we have that a particular solution to the nonhomogeneous equation is
y   5 x  x2 x 4
2
and the general solution is
y  c1 2x  c2  5 x  x2x4.
2
Now
y  0   c1  c2  0
y 1  2c1  c2  5 2  1 8  5 8
 c2  3, c1  3.
Hence the required solution is
y  3.2x  3  5 x  x2x4.
2
2.
In this model we have the following second order linear difference equation with constant coefficients
y  t     2    y  t 1   1    y t  2  v
which has an auxiliary equation
m2    2    m   1     0.
(i)
The roots of the auxiliary equation are real and distinct if
4 1   
2
as   0.
 2     2  4 1       
2
2  
Now
4 1   
2  
2

22  
as   0
2  
 2 2  ,
2
so our condition on  implies that
  2  2   .
Let m1  m2 be the roots of the auxiliary equation. Then
m1  m2   2      2
by our implied condition on  and
m1m2   1     0 ,
123
so both roots are positive and at least one root is  1 . Hence the time path is divergent and nonoscillatory.
(ii) The roots of the auxiliary equation are real and equal if
4(1+ α )

,
(2 +  ) 2
in which case the common root is
m  2    2  1
from our implied condition on  . Again the time path for output is divergent and nonoscillatory.
(iii) We get cycles in this model only if the roots of the auxiliary equation are conjugate complex and
the condition that ensures this is
4 1+ α 

(2 + α)2 .
The absolute value of these roots will be
2
2
 2     2   1      2     2
4
4
  1   
For convergent cycles we require then
 1     1   1     1 as  1     0.
For divergent cycles we require
 1     1.
(iv.)

Explosive Growth
without Cycles
Explosive
Cycles
  2     4 1   
2
Convergent
Cycles
 1     1

124
3.
When time is treated as a discrete variable we regard our economic variables as changing only after
the passage of discrete intervals of time. If y(x) is our economic variables then y(x) is defined for x 
0, 1, 2, . . .
For continuous time the economic variables are regarded as changing continuously.
In a second order linear differential equation with constant coefficients cycles can only occur in the case
where the roots of the auxiliary equation are conjugate complex numbers. In a second order linear
difference equation with constant coefficients, cycles can occur in all three cases: where the roots of the
auxiliary equation are real and distinct, real and equal, and conjugate complex. For example, in the first
case if m1 and m2 are the distinct roots and m1  m2 with m1 negative then eventually m1 dominates
and we have cycles. This means that for difference equations, unlike differential equations we must look at
the conditions on the parameters of our model that ensure
(i)
m1 and m2 are both positive
(ii)
m1 and m2 are both negative
(iii) m1 negative, m2 positive but m1  m2 .
10.5
(i)
Substituting into the definitional equation we have
Yt  a  bYt  v Yt 1  Yt 2  ,
that is,
Yt  cYt 1  cYt 2 
a where c  v
(1  b)
(1  b).
The auxiliary equation is
m2  cm  c  0.
The potential equilibrium is found by setting Yt  Yt 1  Yt  2  Y in our definitional equation to obtain
Y  a .
1  b 
(ii) The auxiliary equation has real and distinct roots if c 2  4c  0 ,
that is,
c  c  4  0 .
As c is positive, this requires c  4.
125
If m1 and m2 are the roots of the auxiliary equation then
m1  m2  c  0
m1m2  c 2  c 2  4c  4c  0.
so both roots are positive and the time path for Y is not subject to oscillations.
The larger root is
c  c  c  4
 1 as c  4, so we have explosive growth for Y.
2
(iii) Real and equal roots of the auxiliary equation if
c  c  4  0  c  4.
The common root then is
m  c  2  1,
2
so we have explosive growth with no oscillations.
(iv) We have cycles if the roots of the auxiliary equation are conjugate complex numbers. The
condition needed to ensure this case is
c  c  4  0  c  4,
and the roots will then be given by
m1 , m2 
c  c 4  c
i.
2
The absolute value of these roots is
c 2  4c  c 2  c.
4
4
If
c  1  c  1 then we have convergent cycles. If c  1 divergent cycles.
v)
c  v , where v is the accelerator and s  1  b is the marginal propensity to save. Then we
s
have the following cases:
(a)
v  4s
Explosive growth for Y with no oscillations
(b)
v  4s
Oscillations for Y
v  s exp losive oscillations
v  s damped oscillations.
126
10.6
1. (i)
Substituting 2 y  x   y  x  2  2 y  x  1  y  x  and y  x   y  x  1  y  x  into the
equation gives the difference equation in standard format:
y  x  2  3 y  x  1  2 y  x   6.
The auxiliary equation is
β 2  3  2  0.
Consider
1 2
2 1
 3,
so by Schur’s theorem the time path for y will be divergent.
(ii) Substituting 2 y  x   y  x   2 y  x  1  y  x  2 and y  x   y  x   y  x 1 gives the
equation in standard format:
 y  x   y  x 1  y  x  2  7.
Consider
1 1
1

 0,
so by Schur’s theorem the time path for y is divergent.
(iii) The auxiliary equation is
3m 2  2m  1  0.
Consider
3 1
1 3
8
3 0 1 2
2 3 0
1
1 0 3 2
2 1 0
3 0 1 2

3
2 3 0
1
8 0 0
4
2 1 0
3
2 3 1
 1 1
1 3
 1 1
3 2
4 0 8
8 0 4  8 0
4
2 1 3
3
2 1
4 8
8
4
 48,
127
so by Schur’s theorem the time path for y is convergent.
3. We must first write our difference equation in its alternative form. Now
 3 y  x   y  x  3  3 y  x  2   3 y  x  1  y  x 
 2 y  x   y  x  1  2 y  x  1  y  x 
y  x   y  x  1  y  x  ,
so substituting back into our difference equation gives
y  x  3  3 y  x  2   3 y  x  1  y  x 
 a1  y  x  2   2 y  x  1  y  x  
 a2  y  x  1  y  x  
 a3 y  x   c.
Collecting terms gives
y  x  3  y  x  2  a1  3  y  x  1 3  2a1  a2 
 y  x  a1  a3  a2  1  c.
The auxiliary equation is then
m3   a1  3 m 2   3  2a1  a2  m
  a1  a3  a2  1  0.
Schur’s theorem states that the roots of this polynomial will all have absolute values less than 1 if and only
if
1 
2 
3 
1
a1  a3  a2  1
a1  a3  a2  1
1
1
0
a1  a3  a2  1
3  2a1  a2
a1  3
1
0
a1  a3  a2  1
a1  a3  a2  1
0
1
a1  3
3  2a1  a2
a1  a3  a2  1
0
1
1
0
0
a1  a3  a2  1
3  2a1  a2
a1  3
a1  3
1
0
0
a1  a3  a2  1
3  2a1  a2
3  2a1  a2
a1  3
1
0
0
a1  a3  a2  1
a1  a3  a2  1
0
0
1
a1  3
3  2a1  a2
3  2a1  a2
a1  a3  a2  1
0
0
1
a1  3
a1  3
3  2a1  a2
a1  a3  a2  1
0
0
1
are all positive.
128
Exercises for Chapter 11
11.3
1. (i)
The Hamiltonian is
H  u  u2  u,
which is concave in u. Moreover
 H  1  2u   ,
u
 2 H  2,
 u2
so the value of u that minimizes H is
   1 .
u
2
Now
     H  0,
x
so   c, an arbitrary constant, and
 c  1 .
x  u  1
2
c2   c1  1 t
, where c2 is another arbitrary constant. But
2
x(0)  4  c2
Hence x 
x(1)  2 
4   c1  1
 c1  5,
2
so the optimal time paths are
x * ( t )  4  2t
u * ( t )  2
 * (t )  5.
(ii) The Hamiltonian is
H  4u  u2    x  2u 
which is concave in u and x. Now
 H  4  2 u  2  ,
u
 2 H  2,
 u2
so the value of u that maximizes H is
u  2  .
129
The costate equation is
     H   ,
x
so   c1e  t ,
where c1 is an arbitrary constant. The state equation is
x  x  2u  x  4  2c1e  t
so we have the first order linear differential equation
x  x  4  2c1e  t .
A particular solution to this equation is
x  et

t
0
e s  4  2c1e s   et

t
0
4e  s  2c1e 2 s ds  et 4e  s  c1e 2 s 
t
0
 et  4et  c1e2t  4  c1   4  c1et   4  c1  et ,
so the general solution is
x  c2et  4  c1et   4  c1  et  4  c1et   4  c1  c2  et ,
where c2 is an arbitrary constant. Using the end point conditions we have
x  0   2  4  c1  4  c1  c2  2
 c2  2,
and
x 1  10  4  c1e 1   6  c1  e  10  c1  e  e 1   14  6e  c1  0.9827.
So our optimal time paths are
x*  t   4  0.9827e  t  5.0173et
 *  t   0.9827e  t
u*  t   2  0.9827e  t .
(iii) The Hamiltonian is
H  x  u2    u  3x  ,
which is concave in x and u. As
 H  2u  
u
and
 2 H  2,
 u2
the value of u that maximizes H is
u .
2
130
The costate equation is
    H  1  3
x
yielding the differential equation
  3  1
whose general solution is
  1  c1e3t ,
3
with c1 being an arbitrary constant. The state equation can then be written as
c
x  3x  1  1 eet .
6 2
The general solution to the homogeneous form of this equation is x  c2 e 3t with c2 an arbitrary constant.
A particular solution to the nonhomogeneous equation can be obtained as in 1 (ii). Alternatively we can
use the method of undetermined coefficients and use as our trial particular solution
x  k 0  k1e3t .
Then x  3k1e3t and substituting into our equation we get
c
6k1e3t  3k 0  1  1 e3t
6 2
c
 k 0  1 , k1  1 .
18
12
Thus the general solution to the state equation is
c1e3t 1
3t
x  c2e 
 .
12 18
Using the end point conditions we have
c
x  0   1  c2  1  1  1
12 18
(1)
c1e3 1
x 1  4  c2e 
  4.
12 18
3
(2)
c
From equation (1) we have c2  17  1 . Substituting into equation (2) we have
18 12
3
 17 c1  3 c1e

e

 71
 18 12 
12
18


c
 1  e3  e 3   1  71  17e 3 
12
18
2  71  17e 3  140.3071
 c1 

 2.3343
60.1071
3  e3  e 3 
c2  0.9444  0.1945  0.7499.
131
Thus our optimal time paths are
x *  t   0.7499e 3t  0.1945e3t  0.0556
 *  t   0.3333  2.3343e3t
u*  t   0.1667  1.1672e3t .
2.
(i)
The consumer’s problem is
max
subject to

40
log C  t  e 0.02 t dt
0
W  t   0.05W  t   C  t 
W  0   1, W  40   0.5,
W and C expressed in millions of dollars. The control variable is consumption C  t  , the state variable is
wealth W  t  .
(ii) The Hamiltonian is
H  log Ce0.02t    0.05W  C  .
The necessary conditions are
 H  e 0.02 t    0
C
C
2
as  H2  0,
C
     H  0.05
w
W   0.05 W  C
W  0   1, W  40   0.5.
As the Hameltonian is concave in C and W, (the sum of concave functions) these conditions are also
sufficient.
(iii) The value of C that maximizes the Hameltonian is
0.02 t
Ce
.

From the costate equation we have the first order linear differential equation
  0.05  0
which has a general solution given by
  t   c1e0.05t
132
where c1 is an arbitrary constant, so
C  1 e0.03t  c2e0.03t ,
c1
say.
The state equation gives the differential equation
W   0.05W  c2e0.03t
which has a general solution given by
W  c3e0.05t  50c2  e0.03t  e0.05t  .
W 0  1
 c3  1
W  40   0.5  0.5  e 2  50c2  e1.2  e 2   c2  0.03386.
The consumer’s optimal time path for consumption is then
C*  t   0.03386e0.03t
and the accompanying time path for the consumer’s wealth is
W *  t   1.693e0.03t  0.693e0.05t .
3.
(i)
Substituting u  t  for x  t  in the objective function we have
Maximize

b
a
f  u  t  , x  t  , t  dt
subject to x  t   u  t 
x  a   x a , x  b   xb .
(ii) The Hamiltonian for this problem is
H  f  u  t  , x  t  , t   u  t  ,
and the necessary conditions for an optimal solution are
 H   f  u, x , t     0
u
u
   H  
x
 f  u, x , t 
x
(1)
(2)
x  u
(3)
x  a   xa , x  b   xb .
133
Differentiating both sides of equation (1) with respect to t yields
d   f  u, x, t     .

dt 
u

Using equations (2) and (3) renders Euler’s equation
11.4
1. (i)
The Hamiltonian is
2
H  2x  u  u    x  u
2
which is concave in x and u and as
 H  1  u    0
u
 2 H  1,
 u2
the value of u that maximizes H is
u   1    .
The costate equation is
     H  2  
x
which has a solution given by
  2  c1e  t
with c1 an arbitrary constant. As x 10  is free the transversality condition is
 10  0  0  2  c1e10  c1  2e10 ,
so
  2  2e10t .
The state equation is
x  x  u  x  1  2e10t
which is a nonhomogeneous linear differential equation of degree 1 whose general solution is
x  c2et  e10t  1,
with c2 an arbitrary constant. But
x  0  5  c2  4  e10 .
134
Hence our optimal time paths are
x*  t   4et  et 10  e10t  1
 *  t   2  2e10t
u*  t   1  2e10t .
(ii) The Hamiltonian is
H  4 x  3u  2u2    x  u 
which is concave in x and u and has derivatives
 H  3  4u   ,  2 H  4,
u
 u2
so the value of u that maximizes H is
 x  3 .
u
4
The costate equation is
    H    4    ,
x
which has as its general solution
  4  c1e  t .
As x(1) is free
 1  0  c1  4e
so we have
  4  4e1t .
The state equation is
x  x  u  x  7  e1t
4
which has as its solution
x  c2et  7  1 e1t .
4 2
As x  0  2 we have c2  1  1 e so our optimal solutions are
4 2
*
t
t 1
1
1
x  t   e  e  1 e1t  7
4
2
2
4
*
1 t
  t   4  4 e
u*  t    7  e1t .
4
(iii) Proceeding as we did for exercise 1.(iii) of Section 11.3 we have
  1  c1e3t
3
135
where c1 is an arbitrary constant. Try  1  0. Then
3
c1   e
3
and using the end point condition x  0  1 as we did in the previous exercise we have
3
c2  17  e .
18 36
These values for the constants in our solutions would yield
3 t 1
3
x*  t   17  e e 3t  e
 1.
18 36
36
18
But then




3
x* 1  17  e e 3  1
18 36
36
which clearly is not greater or equal to 3.
Rerunning the problem as we did for exercise 1. (iii) of Section 11.3 with x  0  1 and x 1  3, we get
after a little arithmetic that
2  53  17e 3  104.3071
c1 

 1.7354
60.1071
3  e3  e 3 
c2  0.9444  0.1446  0.7999.
Thus our optimal time paths are
x *  t   0.7998e 3t  0.1446e3t  0.0556
 *  t   0.3333  1.7354e3t
u*  t   0.1667  0.8677e3t .
(iv) The Hamiltonian is
H  tu  x  u2    u  x 
which is concave in u and x. Differentiating with respect to u we have
 H  t  2u   ,  2 H  2,
u
 u2
so the value of u that maximizes H is
t    .
u
2
The costate equation is
     H   1   
x
which has as its solution
  1  c1e  t ,
136
where c1 is an arbitrary constant. The state equation is
u  x   t  1  c1e  t 
x 
,
2 x
which has as its general solution
c
x  c2et  1 e-t  t
4
2
where c2 is another arbitrary constant.
Try  1  0  c1  e which would yield
1t
x  c2et  e  t .
4 2
From the end point condition x  0  1 we have that
1  c2  e  c2  1  e  1.6796
4
4
rendering
1t
x  1.6796et  e  t .
4 2
Now
x 1  1.6796e  3  3.8156  2
4
so our option yields the optimal time paths which are
1 t
x *  t   1.6796et  e  t
4 2
*
1t
  t   1  e
t  1  e  .
t  
1 t
u
*
2
(v) The Hamiltonian is
H  10 x  20u    x  u   10    x  u    20
which being linear in x and u is concave in x and u. Moreover it is an increasing function in u for   20
but a decreasing function in u for   20. Thus
u*  3   20
u*  0   20.
The costate equation is
     H   10   
x
which has as its general solution
  10  c1e  t ,
137
where c1 is an arbitrary constant. But as x 1 is free  1  0 so
c1  10e  27.1828
and
 *  10  27.1828e  t .
This is a monotonically decreasing function in t and its maximum value on our time horizon is
*  0  10  27.1828  17.1828.
Thus  * is always less than 20 and
u *  0.
The state equation is
x  x  u  x
so the general solution for x is
x  c2et
where c2 is an arbitrary constant. But x  0  2 so c2  2. The optimal time paths then are
u*  0
x *  t   2e t
 *  t   10  27.1828e  t .
(vi) The Hamiltonian is
H  10 x  50u    x  u   10    x  u    50
which is concave in x and u and is a monotonically increasing function in u for   50 but a monotonically
decreasing function in u for   50. So
u*  4   50
u*  2   50.
The costate equation is
     H   10   
x
which has as its solution
  10  c1e  t .
As x  2  is free   2   0 so
c1  10e2  73.8904.
It follows that
*  t   10  73.8904et ,
138
a monotonically decreasing function in t starting from the value  *  0  63.8904, so  * becomes 50 at
t  t where
*  t   50  10  73.8904e t
rendering t  0.2082. Thus
u*  4 0  t  0.2082
u*  2 0.2082  t  2.
The state equation is
x  x  u* .
For the time interval [0, 0.2082]
x  4  c1e t
and as x  0  1, c1  3. So for this interval
x*  4  3et .
For the interval (0.2082, 2]
x  2  c2et .
But
x*  0.2082   4  3e0.2082  0.3056
so
0.3056  2  c2e0.2082  c2  1.3758.
For this interval
x*  2  1.3758et .
2.
(i)
The consumer’s problem is
  logC t  e
T
max imize
 t
dt
0
subject to W  rW  C
W  0   W0 , W  T   0.
(ii) The Hamiltonian is
H  log C  t  e t    rW  C  .
The function log C is concave in C, and the linear function rW  C is concave in W and C. Hence H is
concave in W and C.
139
(iii) The necessary conditions are
 H  e  t    0
C
C

    H   r
W
W   rW  C
W  0  W0 , W T   0,  T   0  T W T   0.
2
 t
Now  H2   e 2  0
C
C
and H is concave in W and C so these conditions are also sufficient.
(iv) The value of C that maximizes H is
 t
Ce .

From the costate equation
  c1e  rt
where c1 is an arbitrary constant of integration and c1    0  0 say. It follows that
t r 
C* t   1 e   .
0
(v) As  T   0e rt and the transversality condition requires  T   0 we must have 0  0.
Moreover the optimal solution for consumption rules out 0  0. Hence 0  0 and  T   0.
The transversality condition then requires that W T   0.
(vi) Optimal consumption will rise over time if r   . That is if the rate of interest is greater than the
consumer’s personal rate of time preference. It will fall if r   .
3.
(i)
The problem facing the community is
max imize

100
log au dt
0
subject to x  u
x  0   1000, x 100   0.
(ii) The Hamiltonian is
H  log a   log u  u
140
which is clearly concave in u. A set of necessary and sufficient conditions for an optimal solution is
H    0
u u
    H  0
x
x  u
x  0   1000, x 100   0,  100   0, x 100   100   0.
(iii) The value of u that maximizes H is u   . From the costate equation

  t   c1
an arbitrary constant. The transversality condition requires that c1  0 and u   prohibits c1  0 so
c1
 100  c  0. The state equation renders
x    ,
c1
which has as its solution
x    t  c2 ,
c1
where c2 is another arbitrary constant. The end point condition x  0  1000 yields c2  1000 and as
 100  0 the transversality condition requires that
x 100    100  1000  0
c1
so c1   .
100
The optimal rate of extraction is then
u*  100
with the accompanying time path for the resource given by
x*  t   1000  100t.
11.5
(i)
The elasticity of marginal utility is
U   c  c
 c  
.
U c
Now U   c   1 and U   c    12 so   c   1.
c
c
The instantaneous elasticity of substitution is
  c   1  1.
 c
141
(ii) We have
  
Y  K
L
L
3
4
L
L
1
4
so
3
y  k 4.
The problem facing the economy is

Maximize

e 0.05t log c dt
30
subject to k   k 4  c  0.11k
k  0   k 0 , k     0, 0  c  t    k  .
(iii) The Hamiltonian is
3
H  e0.05t log c    k 4  c  0.11k 


and the current value Hameltonian is
3
H c  He0.05t  log c    k 4  c  0.11k 


(iv) The necessary conditions for an optimal solution are
 HC 1
  0
C C
  
1
1
 HC
 0.05     3 k 4  0.11  0.05     3 k 4  0.16 
k
4

4

3
k   k 4  c  0.11k
k  0   k 0 , lim   T   0, lim k T   0, lim   T  k T   0.
T 
T 
T 
From equation (1)
  1 and     12 c
c
c
so equation (2) yields
1
c  t   3 k  t  4  0.16 c  t  .
4


This equation with the state equation gives the required system of differential equations.
1
c  t   0  3 k  t  4  0.16.
4
142
(1)
(2)
Solving for k gives the steady state level if k
k *  482.7976.
The steady state level of c is found by substituting this value for k  t  in the equation k   t   0, and
solving for c. That is
c**   482.7976  4  0.11  482.7976   49.8893.
3
(v) At the steady state
  1**  0.02,   0.02e 0.05t
c
so
  T   0.02e 0.05T  0
as T  
k  T   k *  482.7976
k  T   T   9.6774e 0.05T  0 as T  .
It follows that the transversality condition holds at the steady state.
143