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34 Pages

### Notes-Chapter 11 - Optimization

Course: ECON 101, Spring 2011
School: Cambrian College
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Word Count: 1363

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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 &lt; 0 Min: d 2 z &gt; 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...

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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 "
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