1
Mathematical Economics - revision problems
1. Answer questions (a)and (b) regarding a long-run pro…t maximizing, perfectly competitive
…rm with the following production function
1
2
y = f (K; L) = K 3 + L 3
(a) Solve the …rst-order conditions, and …nd expressions for the input demand functions
L = L (w; v; p) and K = K (w; v; p); where w = price of capital, v = price of labour
and p = price of production good.
(b) Find comparative statics expressions:
@L =@w; @L =@v; @L =@p:
Solution: Pro…t function
(K; L) = py
2
wK
1
vL = p K 3 + L 3
wK
vL
(a) First-order conditions for this problem:
@ (K; L)
@K
@ (K; L)
@L
=
=
1
2
pK 3
3
2
1
pL 3
3
w=0
v=0
Solving …rst-order conditions we get regular demand functions:
K (w; v; p)
L (w; v; p)
=
3w
2p
=
3v
p
3
=
3
2
=
8p3
(from the …rst equation)
27w3
p
3v
3
2
(from the second equation)
To be sure that the regular demand functions are optimal it is necessary to check secondorder conditions. Hessian matrix for this is given by:
00
(K; L) =
"
2
9 pK
4
3
0
0
2
9 pL
5
3
#
Evaluating Hessian matrix at the stationary point of the objective function we get:
00
(K ; L ) =
"
2
9 p(K
0
1
)
4
3
0
2
p(L
)
9
5
3
#
00
Hessian Matrix
(K ; L ) is a symmetric negative de…nite matrix because leading principal minors alternate in sign starting with negative:
l1
=
l2
=
4
2
p(K ) 3 < 0
9
4
4 2
p (K ) 3 (L )
81
5
3
> 0; (p > 0; K > 0; L > 0)
Hence second-order conditions are satis…ed and regular demand functions L = L (w; v; p)
and K = K (w; v; p) solve pro…t maximization problem.
@L (w; v; p)
@w
@L (w; v; p)
@v
@L (w; v; p)
@p
=
=
p
3v
3
2
are:
0
3
2
=
=
3
2
3v
p
(b) Comparative statics expressions for L (w; v; p) =
3
2
1
2
p
3v
p
3v
1
2
p
3v 2
1
3v
2. Determine whether the following functions are homogeneous. If so, of what degree?
(a) f (x; y) =
p
3
xy 2
(b) g(x; y; z) = x3 z + 2y 2 xz
Solution: A function f : Rn ! R is homogeneous of the degree r if and only if for all x 2 Rn
and for all positive real constants > 0
f ( x) =
r
f (x):
p
p
p
p
3
3
xy 2 = 3 xy 2 = f (x; y). Function f (x; y) = 3 xy 2
(a) f ( x; y) = 3 x( y)2 =
is homogeneous of the degree one.
(b) g( x; y; z) = ( x)3 ( z) + 2( y)2 ( x)( z) = 4 x3 z + 2 4 y 2 xz = 4 (x3 z + 2y 2 xz) =
4
g(x; y; z): Function g(x; y; z) = x3 z + 2y 2 xz is homogeneous of the degree four.
3. Let Q = Q(K; L; E) be the output function of three inputs
Q(K; L; E) = AK L E
(a) Is this function homogeneous?
(b) Under what conditions would there be constant returns to scale?
Solution: A function f : Rn ! R is homogeneous of the degree r if and only if for all x 2 Rn
2
and for all positive real constants
>0
f ( x) =
r
f (x):
(a) Q( K; L; E) = A( K) ( L) ( E) = A K
L
E = + + AK L E =
+ +
Q(K; L; E): Hence given output function is homogeneous of the degree + + :
(b) There would be constant returns to scale if the output function will be homogeneous
of the degree one, which means that + + = 1:
4. Answer the following questions about an individual with utility function
U (x1 ; x2 ; x3 ) =
1
1
5
ln x1 + ln x2 +
ln x3 ;
4
3
12
who maximizes utility subject to the standard budget constraint:
p1 x1 + p2 x2 + p3 x3 = M:
(a) Set up the Lagrangian and solve the …rst-order conditions for the regular demand functions x1 (p1 ; p2 ; p3 ; M ), x2 (p1 ; p2 ; p3 ; M ) and x3 (p1 ; p2 ; p3 ; M )
(b) Are these demand functions homogeneous of any degree? (Show.)
(c) Find an expression for the optimal value of the Lagrange multiplier
(p1 ; p2 ; M ):
(d) Find all the comparative statics terms
@xi
for i = 1; 2; 3:
@M
Solution: The Lagrangian for this problem is
L(x1 ; x2 ; x3 ; ) =
1
1
5
ln x1 + ln x2 +
ln x3 + (M
4
3
12
p1 x1
p2 x2
First-order condition:
@L(x1 ; x2 ; x3 ;
@x1
@L(x1 ; x2 ; x3 ;
@x2
@L(x1 ; x2 ; x3 ;
@x3
@L(x1 ; x2 ; x3 ;
@
)
)
)
)
=
=
=
1
4x1
1
3x2
5
12x3
= M
3
p1 = 0
p2 = 0
p3 = 0
p1 x1
p2 x2
p3 x3 = 0
p3 x3 )
From …rst, second, and third equation we get:
x1
=
x2
=
x3
=
1
4 p1
1
3 p2
5
12 p3
Substituting formulas for x1 ; x2 ; x3 into the budget constraint we get:
M
1
5
3
12
3+4+5
12
1
4
M
=
0
=
0
=
1
M
Now we substitute the expression for the Lagrange multiplier into formulas for x1 ; x2 ;
x3 and get demand functions:
x1 (p1 ; p2 ; p3 ; M )
=
x2 (p1 ; p2 ; p3 ; M )
=
x3 (p1 ; p2 ; p3 ; M )
=
M
4p1
M
3p2
5M
12p3
Stationary point of the Lagrange function is optimal because the Lagrange function is
concave as a sum of concave functions.
(b) All demand functions are homogeneous of the degree zero because:
x1 ( p1 ; p2 ; p3 ; M )
=
x2 ( p1 ; p2 ; p3 ; M )
=
x3 ( p1 ; p2 ; p3 ; M )
=
(c) Optimal value of
M
M
=
= x1 (p1 ; p2 ; p3 ; M )
4 p1
4p1
M
M
=
= x2 (p1 ; p2 ; p3 ; M )
3 p2
3p2
5 M
5M
=
= x3 (p1 ; p2 ; p3 ; M )
12 p3
12p3
was found in (a) and
(p1 ; p2 ; M ) =
4
1
M
(d) Comparative statics terms:
@x1
@M
@x2
@M
@x3
@M
=
=
=
1
4p1
1
3p2
5
12p3
5. Find the general solution of the following di¤erence equations:
xt+2
6xt+1 + 8xt = 0
Solution: Given di¤erence equation is homogeneous and of the second degree. To get general
solution of such equation it is necessary to solve the characteristic equation
2
6 +8=0
Characteristic equation has two real roots: 1 = 2 and
of the given equation is
xt = A1 2t + A2 4t
2
= 4 and the general solution
where A1 and A2 are undermined real constants.
6. Let us consider the second order non-homogeneous di¤erence equation which is a mathematical
representation of a version of the Samuelson model
Yt
c(1 + v)Yt
1
+ cvYt
2
= 1 0 < c < 1; 0 < v:
Determine the particular solution. Use the characteristic equation to determine the stability
frontier. The latter is the curve which separates parameter con…gurations leading to roots
with modulus less than one from those with modulus greater than one. Determine also the
‡utter boundary, that is, the curve separating real and complex roots.
Solution: There is a constant function on the right hand side of the given equation, hence as
a …rst approximation of particular solution we will take a constant function too. We are
looking for a function Yt = k which satis…es non-homogeneous equation. Substituting
this function into the di¤erence equation we get
k
c(1 + v)k + cvk
k(1
5
c)
=
1
=
1
and …nally remembering that 0 < c < 1 we obtain particular solution
Yt =
1
1
c
To determine the stability frontier we will consider the characteristic equation
2
c(1 + v) + cv = 0
with discriminant
= c2 (1 + v)2
4cv
Coe¢ cients of the characteristic equation are
a1 =
c(1 + v); a0 = cv
Roots of the characteristic equation are less than one in modulus if and only if the
following inequalities are simultaneously satis…ed
1 + a1 + a0
> 0
1
> 0
a1 + a0
1
a0
> 0
In our case we have
1 + a1 + a0
=
1
1
=
1 + 2cv + c > 0
=
1
a1 + a0
1
a0
c>0
cv > 0
First and second inequality is satis…ed, so the stability frontier is determined by the third
inequality and given by
1
c= ; v>0
v
The ‡utter boundary is the curve separating real and complex roots. Characteristic
equation has real roots if
0 and complex roots if < 0: Hence the ‡utter boundary
is de…ned by = 0
= c2 (1 + v)2
4cv = 0 () c =
where 0 < c < 1 and v > 0:
6
4v
(1 + v)2
7. A model by J.R. Hicks uses the following di¤erence equation:
Yt+2
t
(b + k) Yt+1 + kYt = a (1 + g) ;
t = 0; 1; :::
where a; b; g; k are positive constants.
(a) Find a particular solution Yt of the equation assuming that (1+g)2 (b+k)(1+g)+k 6= 0.
(b) Give conditions for the characteristic equation to have two complex roots.
(c) Find the growth factor r of the oscillations when the conditions obtained in part (b) are
satis…ed, and determine when the oscillations are damped.
Solution: (a) There is an expotential function on the right hand side of the given di¤erence
equation, so we will be looking for a particular solution as an expotential function
Yt = D(1 + g)t
Substituting this function into the non-homogeneous equation we get
D(1 + g)t+2
(b + k)D(1 + g)t+1 + kD(1 + g)t
= a(1 + g)t
D (1 + g)2
= a
(b + k)(1 + g) + k
D
Hence
Yt =
(1 +
=
(1 + g)2
a
(b + k)(1 + g) + k
a
(1 + g)t
(b + k)(1 + g) + k
g)2
(b) Characteristic equation
2
(b + k) + k = 0
has complex conjugate roots if and only if discriminant
= (b + k)2
4k
p
is negative, that is, if jb + kj < 2 k; but b and k are positive, so …nally
< 0 if and
p
only if b < k + 2 k:
(c) The growth factor r is equal to the absolute value of the complex conjugate roots of
characteristic equation, that is r = j 1 j and r = j 2 j: Using the Viette’a formula we get
that
1 2 =k
7
Hence
j
1 2j
=j
1 jj 2 j
and
r=
p
= r2 = k
k
Oscillations are damped if r < 1; which means that k < 1:
8. Consider the model
Ct
=
0:75Yt
It
=
4 (Yt
Et
= Ct + It
Yt
= Et
Yt
1)
1
where C = consumption, Y = income, E = total expenditure, I = investment.
(a) Show that this results in a second order di¤erence equation for income.
(b) Solve this equation.
Solution:
Yt
(a)
4:75Yt
1
Et
= Ct + It = 0:75Yt + 4 (Yt
Yt
= Et
+ 4Yt
=
2
1
= 4:75Yt
4Yt
1
Yt
1)
= 4:75Yt
4Yt
1
2
0
Derived di¤erence equation for income is homogeneous and of the second degree. To get
general solution of such equation it is necessary to solve the characteristic equation
2
4:75 + 4 = 0
Characteristic equation has two real roots:
1
=
p
4:75
105
4
2
=
19
p
105
8
and
2
=
19 +
p
8
and the general solution of the derived equation is
xt = A1
19
p
105
8
!t
+ A2
where A1 and A2 are undermined real constants.
8
19 +
p
8
105
!t
105
9. For the following equation
xt+1 =
x2t ;
4 + xt
2R
(a) Determine the intervals of
over which (real) …xed points exist.
(b) Determine the intervals of
for which they are stable or unstable.
Solution: (a) To …nd …xed points of the given di¤erence equation we must …nd all real roots
of the following equation (the so called equilibrium condition)
xt+1
2
4 +x
(x )
(x )2 + 4
= xt = x
= x
=
0
p
The last equation has one real root x = 0 for = 0 and two real roots x1 = 2
p
and x2 = 2
for < 0: Hence the given equation has two …xed points if and only if
< 1:
(b) Let F (x; ) = 4 +x x2 denote the right hand side of the given di¤erence equation
and let x be an equilibrium point of this equation. Then
0
(i) if jFx (x ; )j < 1 then x is locally asymptotically stable (attracting …xed point)
0
(ii) if jFx (x ; )j > 1 then x is unstable (repelling …xed point)
To check stability of the …xed points we must calculate partial derivative of the function
F (x; ) with respect to x
0
Fx (x; ) = 1 2x
and evaluate this derivative at each equilibrium point
0
Fx (x1 ; )
0
Fx (x2 ; )
=
=
p
1+4
p
1 4
p
0
First …xed point x1 is unstable for each < 0 because jFx (x )j = 1 + 4
> 1 for
< 0:
p
0
If jFx (x2 ; )j = j1 4
j < 1 then second …xed point x2 is stable. Solving this double
inequality we get that equilibrium point x2 is locally asypmtotically stable if 41 < < 0
and unstable if < 14 :
9
10. Consider the price-adjustment demand and supply model:
Qd
=
Qs
dp
dt
=
2
3p
3+p
2(Qd
=
Qs )
(a) Derive the …rst-order linear di¤erential equation for price p = p(t):
(b) Solve the di¤erential equation from (a).
Solution: (a)
dp
= p = 2(Qd
dt
Qs ) = 2 [2
3p
( 3 + p)] = 2(5
4p)
Hence we must solve linear …rst-order di¤erential equation of the form
p + 4p = 10
(b) We will use the integrating factor method. Integrating factor for this equation
Z
4dt
=
4t + C
= e4t
Now we multiply both sides of the di¤erential equation by the integrating factor
pe4t + 4pe4t = 10e4t
After this operation we can rewrite the di¤erential equation as
d
pe4t = 10e4t
dt
Integrating above equation with respect to time we obtain
pe4t
=
p(t)
=
Z
10e4t dt =
5
+ Ce
2
4t
5 4t
e +C
2
; C2R
11. The following di¤erential equation is
1
k = 0:2k 3
0:1k
is a version of the Solow growth model. This is a Bernoulli di¤erential equation. Solve it.
10
is:
Solution: Given equation is a Bernoulli di¤erential equation and its standard form is
1
k + 0:1k = 0:2k 3
1
Now we divide both sides of this equation by k 3 and get
kk
1
3
2
+ 0:1k 3 = 0:2
Following Bernoulli’s way of solving we introduce auxiliary variable
2
w = k3
Di¤erentiating this equation with respect to time we get
w
=
3
w
2
2
k
3
= k
1
3
1
3
k
k
Now we substitute this auxiliary variable into the Bernoulli di¤erential equation
3
w + 0:1w
2
1
w+ w
15
=
0:2
=
2
15
and get linear …rst-order di¤erential equation which can be solved using integrating factor
procedure.
Integrating factor for this equation is:
Z
1
dt
15
=
1
t+C
15
1
= e 15 t
Now we multiply both sides of the di¤erential equation by the integrating factor
1
we 15 t +
1
2 1t
1
we 15 t =
e 15
15
15
After this operation we can rewrite the di¤erential equation as
1
d
2 1t
we 15 t =
e 15
dt
15
11
Integrating above equation with respect to time we obtain
we
1
15 t
w(t)
=
=
Z
1
2 1t
e 15 dt = 2e 15 t + C
15
2 + Ce
1
15 t
; C2R
We know that
2
w
= k3
k
= w2
3
and …nally
k(t) = 2 + Ce
3
2
1
15 t
; C2R
12. For the following di¤erential equation
x = x3 + x2
4(x + 1)
(a) Find all steady-state(s) (equilibria, …xed points).
(b) Using linear approximation describe local asymptotic stability of the steady-state(s).
Solution: (a) To …nd equilibria of the given di¤erential equation we must …nd all real roots
of the following equation (the so called equilibrium condition)
x =
x3 + x2
4(x + 1)
2
(x + 1)(x
4)
0
= 0
= 0
The last equation has three real roots x1 = 0; x2 = 2 and x3 = 2:
(b) Let F (x) = x3 + x2 4(x + 1) denote the right hand side of the given di¤erential
equation and let x be an equilibrium point of this equation. Then
0
(i) if F (x ) < 0 then x is locally asymptotically stable (attracting …xed point)
0
(ii) if F (x ) > 0 then x is unstable (repelling …xed point)
To check stability of the …xed points we must calculate …rst-order derivative of the
function F (x)
0
F (x) = 3x2 + 2x 4
12
and evaluate this derivative at each equilibrium point
0
F (x1 )
=
0
F (x2 )
0
F (x3 )
4
=
4
=
12
0
First equilibrium point x1 is stable becauseF (x1 ) < 0:Second and third equilibrium
0
0
points are unstable because F (x2 ) > 0 and F (x3 ) > 0:
13. In a macroeconomic model C(t); I(t); and Y (t) denote respectively the consumption, investment, and national product in a country at time t: Assume that, for all t :
Y (t)
= C(t) + I(t)
I(t)
=
C(t)
2C(t)
1
Y (t) + b
=
2
where b is a positive constant.
(a) Derive the following di¤erential equation for Y (t) : Y (t) = 12 Y (t)
b:
(b) Solve this equation when Y (0) = Y0; > 0 and then …nd the corresponding I(t):
Solution:
(a)
Y (t)
= C(t) + I(t)
1
Y (t) =
Y (t) + b + 2C(t)
2
1
Y (t) =
Y (t) + b + Y (t)
2
1
Y (t) =
Y (t) b
2
(b) Firstly we must get the general solution of the following di¤erential equation
Y
1
Y =
2
b
We will use the integrating factor method. Integrating factor for this equation
Z
1
dt
2
=
= e
13
1
t+C
2
1
2 t
is:
Now we multiply both sides of the di¤erential equation by the integrating factor
Ye
1
Ye
2
1
2 t
1
2 t
= be
1
2 t
After this operation we can rewrite the di¤erential equation as
d
Ye
dt
1
2 t
1
2 t
= be
Integrating above equation with respect to time we obtain
1
2 t
Ye
=
Y (t)
=
Z
be
1
2 t
dt =
2be
1
2 t
+C
1
2b + Ce 2 t ; C 2 R
Now we can use initial condition to get particular solution
Y (0)
C
2b + Ce0 =
=
2b + C = Y0
= Y0 + 2b
Hence particular solution is
1
2b + (Y0 + 2b) e 2 t
Y (t) =
Finally we can derive the time path for I(t)
I(t)
=
I(t)
=
2C(t) = Y (t) =
2b +
1
Y (t)
2
b
1
1
(Y0 + 2b) e 2 t
2
14. For the following di¤erential equation
p
x=2 x
0:25x; x > 0
(a) Find all steady-state(s) (equilibria, …xed points).
(b) Using linear approximation describe local asymptotic stability of the steady-state(s).
Solution:
(a) To …nd equilibria of the given di¤erential equation we must …nd all real roots
14
of the following equation (the so called equilibrium condition)
x =
p
0
2 x
p
x(2
0:25x = 0
p
0:25 x) = 0
The last equation has only one positive root x = 64:
p
(b) Let F (x) = 2 x 0:25x denote the right hand side of the given di¤erential equation
and let x be an equilibrium point of this equation. Then
0
(i) if F (x ) < 0 then x is locally asymptotically stable (attracting …xed point)
0
(ii) if F (x ) > 0 then x is unstable (repelling …xed point)
To check stability of the …xed points we must calculate …rst-order derivative of the
function F (x)
0
1
1
F (x) = p
x 4
and evaluate this derivative at the equilibrium point
0
1
F (x ) = p
64
0
1
1
=
4
8
Equilibrium point x1 is stable becauseF (x ) =
15
1
=
4
1
8
< 0:
1
8