Tutorial 3: Optimisation

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2 Tutorial : Optimisation ECO411F Find and classify the extrema of the cubic cost function C = C (Q) = Q 5Q +.. Find and classify the extreme values of the following functions (a) y = x 1 + x x 1x + 4x x + 6x (b) y = x 1 x + x 1 x + x x + x + x. State whether each of the following functions are concave, strictly concave, convex, strictly convex or neither: (a) y = x (b) y = x xz + z (c) y = x (d) y = xz 4. Find the extreme values of the function z = 8x + xy x + y + 1 and show that only one of the two extrema can be characterised as a local minimum and neither of the extrema can be characterised as a local maximum. If x = 0, how would you characterise the curvature of z at x? 5. Identify the stationary point of the function f (x 1, x, x ) = x 1 1x 1 x 1 x + x x x + x and show that this stationary point is a unique global minimum. 6. Identify the stationary point of the function f (x 1, x, x ) = x 1 x 1 x 1 x + x x 1.5x + 10x and determine if it represents a maximum or minimum. 7. Find the extremum of the following functions (a) z = xy subject to x + y = (b) z = x y xy subject to x + y = 6 (c) y = x + xz + 4z subject to x + z = 8 (d) y = x + z xz subject to x + z = 1 (e) y = ln x + ln z subject to z = 16 4x 1

3 8. Find the optimal values of each of the choice variables in the functions below, and solve for the Lagrangian multiplier too. (a) U(x, z) = x + x + z 6z + x subject to x + z =. (b) U(a, b, c) = 1 a + 4b 4a + 8c subject to 1 a + b + c. 9. Let A, α, β be positive constants. Show that the Cobb-Douglas production function is concave if and only if α + β 1. Q = AK α L β 10. A monopolist can produce quantities x and y of two products X and Y at cost 4x + xy + y. The inverse demand functions are p X = 150 5x + y p Y = 0 + x y where p X and p Y are the prices charged for X and Y. (a) Find the values of x, y, p X and p Y which maximise profit, and the maximal profit. (b) Confirm that your answer to (a) represents a maximum. 11. Find the extrema of z = x 1x + 8xy + y y + 6y and classify the extrema as minima or maxima, and as global (unique or not unique) or local. 1. Billy has a Cobb-Douglas utility function U(x 1, x ) = x /4 1 x 1/4 where x 1, x denote the consumption of cars and jets respectively. goods are p 1, p and Billy s income is m. The prices of the (a) Find the demand functions for cars and jets that maximise Billy s utility. (b) Confirm that your answer to (a) represents a constrained maximum. (c) What is the economic interpretation of the Lagrangian multiplier in this case?

4 1. The production of Jimmy Choo shoes (Q) requires two inputs, capital (K) and labour (L): Q = f(k, L) = 9K 1/ L / A unit of capital costs R4 and a unit of labour costs R16. (a) Does the production function exhibit increasing, decreasing or constant returns to scale? Explain your answer. (b) Show that the production function exhibits diminishing returns to each factor of production. (c) What is the cost minimising bundle of capital and labour to produce any fixed Q = Q? (d) Confirm that your answer to (a) represents a constrained minimum. (e) What is the economic interpretation of the Lagrangian multiplier in this case? 14. Solve the following maximisation problem by applying the Kuhn-Tucker theorem: max x,y 4x xy 4y subject to x + y x y Solve the following maximisation problem by applying the Kuhn-Tucker theorem: max x,y.6x 0.4x + 1.6y 0.y subject to x + y Consider the issue of optimal time allocation: x 0 y 0 Suppose that you have a utility function defined over two goods, x 1 and x, such that your utility function is u (x 1, x ). Suppose it takes t 1 units of time to consume x 1 units of Swiss chocolate and t units of time to consume x units of champagne. The total time available for consumption is T. Your challenge is to maximize your utility subject to your income constraint p 1 x 1 + p x m and time constraint t 1 x 1 + t x T. Assume that both x 1 and x are strictly positive at the optimum and u 1 and u are positive, reflecting non-satiation.

5 (a) Set up a Lagrangian function in which utility is maximized subject to both the time and income constraint and specify the first order conditions for the optimum. (b) Consider the four solution possibilities to this problem based on the four possible ways in which the complementary slackness conditions might be met. Which is most likely to yield the optimal solution? λ > 0 and µ > 0 λ > 0 and µ = 0 λ = 0 and µ > 0 λ = 0 and µ = 0 4

6 Tutorial : Optimisation SELECTED SOLUTIONS ECO411F Q = 0 is a maximum Q = 10 is a minimum 6.. (a) y = 0 : minimum (b) y = : minimum (a) strictly convex (b) strictly convex (c) strictly concave (d) neither 4. FOC z x = 4x + y 6x = 0 z y = x + y = 0 Solve simultaneously to get the extrema: x = y = 0 and x = y = 1 : SOC: zxx H = z yx z xy z yy = 48x 6 jh 1 j = j48x 6j = 48x 6 < 0 when x < 6 48 > 0 when x > 6 48 Thus, the Hessian cannot be everywhere positive or negative de nite and so neither of our extrema can be global. 1

7 We evaluate the Hessian at each critical point. Consider x = y = 0 : 6 H = Thus, H is inde nite at x = y = 0. jh 1 j = j 6j = 6 < 0 jh j = jhj = 16 < 0 Consider x = y = 1 : H = 10 jh 1 j = j10j = 10 > 0 jh j = jhj = 16 > 0 Thus, H is positive de nite at x = y = 1 and so this point is a local minimum. 1 = 4x 1 1 x = x 1 + 6x x = x + x = 0 Solve these simultaneously to get the stationary point. SOC: f 11 f 1 f H = 4f 1 f f 5 = f 1 f f 0 jh 1 j = j4j = 4 > 0 jh j = 4 6 = 15 > 0 jh j = jhj = 46 > 0 The Hessian matrix is everywhere positive de nite, and so the stationary point represents a unique global minimum.

8 6. x 1 = 4; x = ; x = 4. This is a unique global maximum (a) z = 1 when = 1 ; x = 1; y = 1. H = 4 > 0 and so this is a constrained maximum. (b) z = 19 when = 4; x = 1; y = 5. H = < 0 and so this is a constrained minimum. (c) (x ; z ) = (8; 0) Constrained minimum 1 (d) (x ; z ) = ; Constrained minimum 4 (e) (x ; z ) = ; Constrained maximum (a) (x ; z ; ) = (1; 4; 15) (b) (a ; b ; c ; ) = (1; 6; 1; 16) 9. First order derivatives: Hessian: H = = @K = AK 1 = AK Q # ( 1) AK L AK 1 L 1 AK 1 L 1 ( 1) AK L For production function to be concave, the Hessian must be negative de nite, i.e. the signs of the principal minors must alternate in sign, starting with a non-positive. Since A, K and L are positive jh 1 j = ( 1) AK L ( 1) 0 ) 0 1

9 jh j = ( 1) AK L AK 1 L 1 AK 1 L 1 ( 1) AK L = h ( 1) AK L i h ( 1) AK L i h AK 1 L 1i h AK 1 L 1i = ( 1) ( 1) A K L A K L = A K L ( 1) ( 1) Since A, K and L are positive, ( 1) ( 1) ( + 1) 0 ) + 1 Therefore the function is concave i + 1: 10. (a) = xp X + yp Y 4x + xy + y = x (150 5x + y) + y (0 + x y) 4x xy y = 150x 5x + xy + 0y + xy y 4x xy y = 9x + xy + 150x 4y + 0y FOC: Solving simultaneously: D (x; y) 18x + y ) x 8y + 0 x = 9 y = 6 = 0 = 0 p X = (9) + 6 = 111 p Y = 0 + (9) (6) = 6 4

10 And: max = xp X + yp Y 4x + xy + y = 9 (111) + 6 (6) 4 (9) + 9 (6) + (6) = 765 (b) SOC: H = D (x; y) = =@y 18 8 jh 1 j = j 18j < 0 jh j = 18 8 = 140 > 0 Thus, H is negative de nite and hence we have a maximum. 11. Dz 4x 1 + 8y 8x + y 4y + 6 H = D z = z=@y y 4 FOCs 4x 1 + 8y = 0 (1) 8x + y 4y + 6 = 0 () 5

11 Solve simultaneously: From (1) Substitute () into () 8 y x = y y 4y + 6 = 0 4 ) y + 1y 6 = 0 ) y + 4y 1 = 0 ) (y + 6) (y ) = 0 ) y = 6; () Therefore, candidates for extrema are x = 69 4 and x = 5 11 ; y = ; z = 4 8 : SOCs: ; y = 6; z = The de niteness of the Hessian matrix depends on the value of y, so the extrema cannot be absolute (global) but only relative (local) extrema. We must determine the de niteness of the Hessian matrix at the candidate extrema: At x = 69 4 ; y = 6 : H = jh 1 j = j 4j = 4 < 0 jh j = = 96 > 0 At this point, H is negative de nite. So this point is a local maximum. At x = 5 4 ; y = : H = jh 1 j = j 4j = 4 < 0 jh j = = 96 < 0 6

12 At this point, H is neither positive nor negative de nite. So this point is neither a local minimum nor a local maximum. 1. (a) FOCs: max L = x =4 x 1 ;x 1 x 1=4 + (m p 1 x 1 p x ) 1 4 x 1=4 1 x 1=4 p 1 = 0 = 4 x=4 1 x =4 p = = m p 1x 1 p x = 0 Solving simultaneously: (b) SOC: 6 BH = 4 x 1 = m 4p 1 x = m 4p 0 p 1 p p 1 16 x 5=4 1 x 1=4 16 x 1=4 p 16 x 1=4 1 x =4 1 x =4 16 x=4 1 x 7=4 7 5 jbh j = 0 p 1 p p 1 16 x 5=4 1 x 1=4 16 x 1=4 p 16 x 1=4 1 x =4 1 x =4 1 x 7=4 16 x=4 = 1 16 x 5=4 1 x 7=4 p 1x 1 + 6p 1 p x 1 x + p x > 0 Therefore bordered Hessian is negative de nite, and we have a constrained maximum. (c) It is the marginal utility of money. 7

13 1. (a) Constant returns to scale. 9 (K) 1= (L) = = 9K 1= L = ) constant returns to scale. OR Sum the exponents 1 + = 1 ) constant returns to scale. (b) f K = K = L = f KK = K 5= L = < 0 ) diminishing returns to capital f L = 6K 1= L 1= f LL = K 1= L 4= < 0 ) diminishing returns to labour (c) Problem: min 4K + 16L subject to Q = K;L 9K1= L = Lagrangian: L = 4K + 16L + Q 9K 1= L = FOCs: (1) = 4 K = L = = 0 (1) = 16 6K1= L 1= = 0 = Q 9K1= L = = 0 () 1 4 = 1L K ) L = 1 K (4) Sub (4) into () Q 9K 1= 1 K = = = K = Q K = = 9 Q = 0:176Q

14 L = 1 =! = 9 Q 1 9 1=Q = 0:088Q 4 K = L = = 0 ) = 4 K= L = = 4 = 9 Q! =! = 1 9 1=Q = 4 4=9 9 =! 9 = =9 7=9 = 4 = 5=9 = :86 (d) SOC: 0 K = L = 6K 1= L 1= H = 4K L = K 5= L = K = L 1= 5 6K 1= L 1= K = L 1= K 1= L 4= H = H = 0 K = L = 6K 1= L = + 1K 1= L = +6K 1= L 1= 6K 4= L 1= 1K 4= L 1= = K = L = 18K 1= L = + 6K 1= L 1= 18K 4= L 1= = 54K 1 108K 1 = 16K 1 16 = K 9

15 At K ; H = 16 K < 0 Therefore, the point is a constrained minimum. (e) It is the marginal cost. 14. Rewrite the constraints so they are in the form g i (x) 0 : (a) Set up the Lagrangian x + y 0 x y L = 4x xy 4y (x + y ) (x y + 1) The Kuhn-Tucker conditions are: Consider the possible cases in turn: = 8x y = 0 = x 8y + = 0 0; x + y and (x + y ) = 0 0; x y 1 and (x y + 1) = 0 (a) Case 1: > 0; > 0 If > 0, then complementary slackness implies x + y = 0: If > 0, then complementary slackness implies x y + 1 = 0. Solving these equations simultaneously gives x = 0; y = 1: Substituting x = 0; y = 1 into equations (1) and () gives = = 0 Solving simultaneously gives = 18 5 ; = 4 5 which violates > 0: Case : > 0; = 0 If > 0, then complementary slackness implies x + y = 0: Substituting = 0 into equations (1) and () gives 8x y = 0 x 8y = 0 10

16 Solving these three equations simultaneously gives x = 1 4 ; y = 7 8 ; = 15 4 which violates > 0: Case : = 0; > 0 If > 0, then complementary slackness implies x y + 1 = 0. Substituting = 0 into equations (1) and () gives 8x y = 0 x 8y + = 0 Solving these three equations simultaneously gives x = 8 ; y = 1 4 ; = 5 4 which satis es all conditions. Case 4: = 0; = 0 Substituting = 0; = 0 into equations (1) and () gives 8x y = 0 x 8y = 0 which is satis ed if and only if x = 0; y = 0 which violates x no solution candidates here. y + 1 0: So Therefore: The Kuhn Tucker necessary conditions for a maximum are satis ed under Case where x = 8 ; y = 1 4 ; = 0; = 5 4 : (This point also satis es the su cient conditions for a maximum and so is the optimum point.) 15. Two possible Lagrangians and associated FOCs: (a) L = :6x 0:4x + 1:6y 0:y + (10 x y) = :6 0:8x 0; x 0; (x) (:6 0:8x ) = = 1:6 0:4y 0; y 0; (y) (1:6 0:4y ) = 0 0; x + y 10; () (x + y 10) = 0 OR 11

17 (b) L = :6x 0:4x + 1:6y 0:y + 1 (10 x y) + (x) + (y) = :6 0:8x 1 + = = 1:6 0:4y 1 + = 0 1 0; x + y 10; ( 1 ) (x + y 10) = 0 0; x 0; ( ) (x) = 0 0; y 0; ( ) (y) = 0 Consider the various cases (Working now with the Lagrangian and FOCs from (a)): Case (I): x + y = 10; x = 0; y = 0 (I) : x + y = 10; x = 0; y = 0 (II) : x + y < 10; x = 0; y = 0 (III) : x + y = 10; x > 0; y = 0 (IV) : x + y = 10; x = 0; y > 0 (V) : x + y < 10; x > 0; y = 0 (VI) : x + y < 10; x = 0; y > 0 (VII) : x + y = 10; x > 0; y > 0 (VIII) : x + y < 10; x > 0; y > 0 Inconsistent, since x = 0; y = 0 violates x + y = 10 Case (II): x + y < 10; x = 0; y = 0 x + y < 10 ) = 0 (from complementary slackness condition) Substituting = 0; x = 0; y = 0 = :6 0 = 1:6 0 So no solution here. Case (III): x + y = 10; x > 0; y = 0 x + y = 10; y = 0 ) x = 5 1

18 x > 0 ) :6 0:8x = 0 (from complementary slackness condition) Sub in x = 5 : :6 0:8 (5) = 0 ) = 0: violates 0 Case (IV): x + y = 10; x = 0; y > 0 x + y = 10; x = 0 ) y = 10 y > 0 ) 1:6 0:4y = 0 (from complementary slackness condition) Sub in y = 10 : 1:6 0:4 (10) = 0 ) = :4 violates 0 Case (V): x + y < 10; x > 0; y = 0 x + y < 10 ) = 0 (from complementary slackness condition) Case (VI): x + y < 10; x = 0; y > 0 Sub in = 0; y = 0 into = 1:6 0:4y = 1:6 So no solution here x + y < 10 ) = 0 (from complementary slackness condition) Case (VII): x + y = 10; x > 0; y > 0 Sub in = 0; x = 0 into = :6 0:8x = :6 So no solution here x > 0 ) :6 0:8x = 0 (from complementary slackness condition) y > 0 ) 1:6 0:4y = 0 (from complementary slackness condition) x + y = 10 1

19 This gives the augmented matrix: 0: : :6 1: Solving yields: x :5 4y5 = 4 5 0:4 This satis es all conditions and is a possible solution. Case (VIII): x + y < 10; x > 0; y > 0 x > 0 ) :6 0:8x = 0 (from complementary slackness condition) y > 0 ) 1:6 0:4y = 0 (from complementary slackness condition) x + y < 10 ) = 0 (from complementary slackness condition) Solving: But then: :6 0:8x (0) = 0 ) x = 4:5 1:6 0:4y 0 = 0 ) y = x Therefore, the only solution is 4y5 x + y = (4:5) + 4 = 1 10 So no solution here. (a) L = u (x 1 ; x ) (p 1 x 1 + p x m) (t 1 x 1 + t x T ) The solution that provides the maximum value must 1 = u 1 p 1 t 1 = = u p t = 0 0; p 1 x 1 + p x m and (p 1 x 1 + p x m) = 0 0; t 1 x 1 + t x T and (t 1 x 1 + t x T ) = 0 14

20 (b) It will not be the case that = 0 and = 0 because complementary slackness implies that p 1 x 1 + p x < m and t 1 x 1 + t x < T and then more of one or both of the goods can be consumed, which would raise utility. It is also unlikely that both > 0 and > 0 since this would mean that both constraints would be binding, i.e. p 1 x 1 + p x = m and t 1 x 1 + t x = T: It is more likely that either > 0 and = 0, in which case the budget constraint is binding, or = 0 and > 0 in which case the time constraint is binding. 15