11 Chapter - Optimization:
More Than One Choice Variable
Alpha Chiang, Fundamentals of
Mathematical Economics, 3rd Edition
One variable case
z = f ( x)
dz = 0
dz = f '( x )dx
Max: d 2 z < 0
Min: d 2 z > 0
necessary but not sufficient
for either maximum or minimum
Function of 2 variables
z = f ( x, y )
dz = f x dx + f y dy
dz = 0 f x = f y = 0 for dx and dy not both zero
fx
f xx
z
,
x
f xy
z
y
first partials
or
2 z
z
x 2 x x
( f y ) or
x
z
z
y 2 y y
( fx )
x
f yy
fy =
z
or
x y
2
2 z
xy
z
2 z
f yx or
y x
yx
Young's theorem: f xy = f yx
second order partials
Examples:
1.
3
z = x + 5 xy y
2
f x = 3x + 5 y
f y = 5x 2 y
f xx = 6 x
f yy = 2
f xy = 5
2
Examples:
2.
2 y
z=x e
f x = 2 xe
y
f xx = 2 xe
y
f xy = 2 xe y
2 y
f y = 1x e
2 y
f yy = x e
2nd Order Total Differential
dz = f x dx + f y dy
d z = d (dz ) = (dz )dx + (dz )dy
x
y
= ( f x dx + f y dy )dx + ( f x dx + f y dy )dy
x
y
= ( f xx dx + f yx dy )dx + ( f xy dx + f yy dy )dy
2
= f xx dx 2 + f yx dydx + f xy dxdy + f yy dy 2
= f xx dx 2 + 2 f xy dxdy + f yy dy 2
Examples:
3.
3
z = x + 5 xy y
2
2
Find dz, d z
dz = f x dx + f y dy
(
)
= 3 x + 5 y dx + ( 5 x 2 y ) dy
2
2
2
d z = 6 xdx + 2(5)dxdy 2dy
2
Second Order Condition:
z = f ( x, y )
d 2z < 0
for maximum, arbitrary values of dx and dy
d 2z > 0
for minimum
d 2z < 0
iff
f xx
f yx
d 2z > 0
iff
f xx < 0, f yy < 0, f xx f yy > f xy 2
f xy
>0
f yy
f xx > 0, f yy > 0, f xx f yy > f xy 2
Conditions for Relative Extremum
CONDITIONS FOR RELATIVE EXTREMUM
Maximum
FOC - first order
conditions
SOC - 2nd order
conditions
Minimum
fx = f y = 0
fx = f y = 0
f xx < 0, f yy < 0
f xx > 0, f yy > 0
f xx f yy > f xy 2
f xx f yy > f xy 2
Objective Functions With
More Than Two Variables
Objective Functions With
More Than Two Variables
3 variables:
dz = f1dx1 + f 2 dx2 + f 3dx3
(dz )
(dz )
( dz )
dx1 +
dx2 +
dx3
x1
x2
x3
( f1dx1 + f 2 dx2 + f 3dx3 )dx1
x1
+
( f1dx1 + f 2 dx2 + f3dx3 )dx2
x2
+
d 2z
=
=
d 2z
( f1dx1 + f 2 dx2 + f3dx3 )dx3
x3
= f11dx12 + f12 dx1dx2 + f13dx1dx3
+ f 21dx2 dx1 + f 22 dx2 2 + f 23dx2 dx3
+ f31dx3dx1 + f 32 dx3dx2 + f 33dx32
3 variables:
dz = f1dx1 + f 2 dx2 + f 3dx3
(dz )
(dz )
( dz )
dx1 +
dx2 +
dx3
x1
x2
x3
( f1dx1 + f 2 dx2 + f 3dx3 )dx1
x1
+
( f1dx1 + f 2 dx2 + f3dx3 )dx2
x2
+
d 2z
=
=
d 2z
( f1dx1 + f 2 dx2 + f3dx3 )dx3
x3
= f11dx12 + f12 dx1dx2 + f13dx1dx3
+ f 21dx2 dx1 + f 22 dx2 2 + f 23dx2 dx3
+ f31dx3dx1 + f 32 dx3dx2 + f 33dx32
This is a quadratic form
3 variables:
Technique: regard dxi as variables that can take on any values, fij as coefficients
We have a symmetric Hessian Determinant
f11
f12
f13
H = f 21
f31
f 22
f 32
f 23
f33
Successive principal minors
H1 = f11 ,
f11
H2 =
f 21
f12
,
f 22
H3 = H
3 variables:
Rules:
H1 < 0, H 2 > 0, H 3 < 0
2
maximum
d z is negative definite
z is a
if
or 2
minimum
d z is positive definite
H1 > 0, H 2 > 0, H 3 > 0
3 variables:
Example: Find the extreme values of
z=
2
2 x1
2
2
+ x1 x2 + 4 x2 + x1 x3 + x3 + 2
f1 =
4 x1 + x2 +
f2 =
x1 + 8 x2
f3 =
x1
x3 = 0
=0
+ 2 x3 = 0
3 variables:
Solution:
x1
x1
x2 =
,
x3 =
8
2
x1 x1
4 x1
=0
8
2
32 x1 x1 4 x1 = 0
27 x1 = 0
x1 = 0,
x2 = 0,
x3 = 0,
z =2
3 variables:
Solution: contd
f11
H = f 21
f31
f12
f 22
f32
f13
411
f 23 = 1 8 0
f33
102
H1 = f11 = 4
f11
H2 =
f 21
f12 4 1
=
= 31
f 22 1 8
H 3 = H = 4(16 0) 1(2 0) + 1(0 8) = 64 2 8 = 54
H1 > 0, H 2 > 0, H 3 > 0
positive definite, z*=2 is a minimum
n-variable case:
z = f ( x1 , x2 ,K , xn )
dz = f1dx1 + f 2 dx2 + K + f n dxn
f11
f12 K
f1n
f 21
H=
M
f 22 K
M
f2n
M
f n1
fn2 K
f nn
Principal minors: H1 , H 2 ,K , H n
F.O.C.:
f1 = f 2 = K = f n = 0
S.O.C.:
n principal minors must be positive (minimum)
n principal minors must alternate in sign (maximum)
Economic Applications
Multiproduct firm:
Assume a two-product firm under pure competition.
Prices are exogenous P10 P20
Firms and Revenue function:
R1 = P10Q1+ P20Q2
where Qi is output level of the ith product per unit time.
Firms cost function:
C = 2Q12 + Q1Q2 +2 Q22
Note: C/Q1 = 4Q1+Q2 and C/Q2 = Q1+2Q2. . The
marginal costs of one product also depends on the output of
the other product.
Economic Applications
Multiproduct firm:
Revenue:
R1 = P Q1 + P20Q2
10
Cost:
C = 2Q12 + Q1Q2 + 2Q22
= TR TC = P Q1 + P20Q2 2Q12 Q1Q2 2Q22
10
Problem: find the levels of Q1 and Q2 which in combination,
will maximize .
Take first partials and equate to zero:
1
= P 4Q1 Q2 = 0
10
Q1
2
= P20 Q1 4Q2 = 0
Q2
Profit:
Economic Applications
Multiproduct firm:
2 simultaneous equations:
4Q1 + Q2 = P
10
Q1 + 4Q2 = P20
Solution:
Q1* =
4 P P20
10
15
and
Q2 * =
4 P20 P
10
15
Second order condition:
11 12 4 1
H=
=
21 22 1 4
H1 = 4 < 0
and
H 2 = 15 > 0
Conclusion: the Hessian matrix or d 2 z is negative definite,
and the solution does maximize profit.
Economic Applications
Multiproduct firm: - monopolist
Demands facing the monopolist:
Q1 = 40 2 P + P2
1
Q2 = 15 + P P2
1
Prices are related to output levels.
Why are they substitute goods? P1 , Q 2 , P2 , Q1
Economic Applications
Multiproduct firm: - monopolist
Express prices in terms of sales volumes:
2 P + P2 = Q1 40
1
P P2 = Q2 15
1
Solving P1 and P2 by Cramer's rule or by substitution,
P = 55 Q1 Q2
1
P2 = 70 Q1 2Q2
Economic Applications
Multiproduct firm: - monopolist
Since P AR1 and P2 AR2 ,
1
Total Revenue:
R = PQ1 + P2Q2
1
= (55 Q1 Q2 )Q1 + (70 Q1 2Q2 )Q2
= 55Q1 + 70Q2 2Q1Q2 Q12 2Q22
Total Cost:
2
C = Q12 + Q1Q2 + Q2
Profit Function:
= R C = (55Q1 + 70Q2 2Q1Q2 Q12 2Q22 ) (Q12 + Q1Q2 + Q22 )
2
= 55Q1 + 70Q2 3Q1Q2 2Q12 3Q2
Economic Applications
Multiproduct firm: - monopolist
Objective Function:
= 55Q1 + 70Q2 3Q1Q2 2Q12 3Q22
First partial derivatives, equated to zero (FOC)
1 = 55 3Q2 4Q1 = 0
2 = 70 3Q1 6Q2 = 0
4Q1 + 3Q2 = 55
3Q1 + 6Q2 = 70
Solution:
Q1* = 8, Q2 * = 7 2
3
2
P * = 39 1 , P2 * = 46 3 , * = 488 1
1
3
3
Economic Applications
Multiproduct firm: - monopolist
SOC:
Second partial derivatives: 11 = 4, 12 = 21 = 3, 22 = 6
4 3
H=
,
3 6
H1 = 4 < 0, H 2 = 15 > 0,
Therefore * is a maximum
Economic Applications
Price Discriminating Monopolist 3 markets
Total Revenue:
R = R1 (Q1 ) + R2 (Q2 ) + R3 (Q3 )
Total Cost:
C = C (Q)
where Q = Q1 + Q2 + Q3
Profit Function:
= R1 (Q1 ) + R2 (Q2 ) + R3 (Q3 ) C (Q)
Economic Applications
Price Discriminating Monopolist 3 markets
= R1 (Q1 ) + R2 (Q2 ) + R3 (Q3 ) C (Q)
Q
'
1 = R1 (Q1 ) C '(Q)
= R1' (Q1 ) C '(Q) = 0
Q1
Q
'
'
2 = R2 (Q2 ) C '(Q )
= R2 (Q2 ) C '(Q) = 0
Q2
Q
'
3 = R3 (Q3 ) C '(Q)
= R3' (Q3 ) C '(Q) = 0
Q3
'
C '(Q ) = R1' (Q1 ) = R2 (Q2 ) = R3' (Q3 )
or MC = MR 1 = MR 2 = MR 3
Economic Applications
Price Discriminating Monopolist 3 markets
MC = MR 1 = MR 2 = MR 3
The levels of Q1, Q2,Q3 should be chosen such that the marginal
revenue in each market is equal to the marginal cost of total
output Q.
Economic Applications
Relationship of MR and P (digression)
R = P
Q
MR =
Q dP
dR
dQ
dP
=P
+Q
= P 1 +
dQ
dQ
dQ
P dQ
1
1
= P 1 + = P 1
1
MRi = Pi 1
i
Economic Applications
Relationship of MR and P (digression)
MR 1 = MR 2 = MR 3 can be written as
1
1
1
P 1
= P2 1
= P3 1
1
1
2
3
Implication:
The smaller the value of in a particular market,
the higher the price charged in that market.
Economic Applications
Price Discriminating Monopolist 3 markets
Second order condition:
Q
11 = R1 "(Q1 ) C " (Q)
= R1 "(Q1 ) C "(Q)
Q1
22 = R2 "(Q2 ) C " (Q)
Q
= R2 "(Q2 ) C "(Q)
Q2
33 = R3 "(Q2 ) C " (Q)
Q
= R3 "(Q3 ) C "(Q)
Q3
and 12 = 21 = 13 = 31 = 23 = 32 = C "(Q)
Q
= 1
since
Qi
Economic Applications
Price Discriminating Monopolist 3 markets
Second order condition:
R1 " C "
C "
C "
H = C "
R3 " C "
C "
C "
C "
R3 " C "