BEEM103 UNIVERSITY OF EXETER. BUSINESS School. January 2009 Mock Exam, Part A. OPTIMIZATION TECHNIQUES FOR ECONOMISTS solutions

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1 BEEM03 UNIVERSITY OF EXETER BUSINESS School January 009 Mock Exam, Part A OPTIMIZATION TECHNIQUES FOR ECONOMISTS solutions Duration : TWO HOURS The paper has 3 parts. Your marks on the rst part will be rounded down to 55 marks: Your marks on the second part will be rounded down to 5 marks. Obviously, you cannot get over 00 marks overall. This examination is closed book and no materials are allowed. Full work must be shown on your script. Please write legibly. Sumit your answers in the rst lecture in January th, Room 004, in the Harrison Building.

2 Part A (You can gain no more than 55 marks on this part.) Problem (5 marks) Simplify Solution x p y x p y y p x x p y + y p x 3p a (4a) 6 4 p a 3 (4a 5 ) p a x p y y p x x p y + y p x x p p p p y y x x y y x x p y + y p x x p y y p x x p y (y p x) + (y p x) p p x p y (y p x y xy x y + xy x) x y xy y p p p xy + x x y x y x y 4 6 3p a (4a) 6 4 p a 3 (4a 5 ) p a a Problem (5 marks) Solve Solution 5 x y 5 x 5y (y + 4) 5y 4y 8y 6 0 3y y 0 y + a 3 (4 6 a 6 ) a 3 4 (4a 5 ) a 4 6 a 3 a 6 a a 5 a 5 a p a 5 3p a ln x ln y ln 5 x y 4 q ( ) p Thus y 6 6 or y 46 3 and correspondingly x 4 + y 5 or x 3 3. Problem 3 (0 marks) Consider the function y (x) e x x 4 i) Calculate and draw a sign diagram for the rst derivative. Where is the function increasing or decreasing. Are there any peaks or troughs? Does the function have a (global) maximum. Is the function quasi concave?

3 ii) Show that the second derivative contains the factor x 4x 4 + 6x + Draw a sign diagram for the second derivative. Where is the function convex or concave. Are there any in ection points? Solution 3 y (x) e x x 4 y 0 (x) x + x e x (+x ) y 00 (x) x 4x 4 + 6x + e x (+x ) y x a) critical point zero, y 0 positive to the left, negative to the right of zero. Function therefore increasing to the left, decreasing to the right. critical point hence unique maximum. y (x) g (h (x)) where g (u) e u and h (x) x x 4. As a monotone transformation of a concave function the function is quasi concave. b) Second derivative is zero at x p ; in between it is negative, outside positive. x p are hence in ection points with the function concave in between the roots and convex outside. Problem 4 (5 marks) For the function y (x ) nd the (global) maxima and minima a) on the interval [ [ ; 8] : ; 0] and b) on the interval Solution 4 y0 (x ), y 00. The function is decreasing for x and increasing thereafter. On the interval [ ; 0] the function has therefore its maximum at and its minimum at x 0. On the interval [ ; 8] it has its minimum at x and its maximum at x 8 because y (8) (8 ) 6 > 6 ( ) y ( ) : 3

4 Problem 5 (5 marks) Find the equation of the tangent plane of at the point (x ; y ; z ) (; 3; z (; 3)). Solution 5 z 3 @z (x + y) (x y) (x + y) z (x; y) x y x + y (x + y) (x y) (x + y) The formula for the total di erential at x ; y 3 y (x + 3 jx;y3 5 5 dx jx;y3 jx;y3 x (x + jx;y3 5 5 yields the formula for the tangent z (x ) + 4 (y 3) 5 z 6 5 x 4 5 y 5 Problem 6 (0 marks) Show that the function is concave. u (x; y) ln (x + 5y) + ln (5x + y) Solution 6 x + 5y + 5 5x + 5 x + 5y + 5x + u (x + 5y) (5x + y) < u 5 (x + 5y) (5x + y) < u 5 5 (x + 5y) (5x + y) < 0 Set a (x + 5y) and b (5x + y). The Hessian of ^u is a H 5b 5a 5b 5a 5b 5a b 4

5 and hence det H a + 5b 5a + b 5 a + b 5a a b + a b + 5b 4 5a 4 50a b 5b a b 576a b > 0 We see that the leading principle minors of the Hessian have the right sign and so u is concave, as was to be proved. Problem 7 (0 marks) A consumer has the utility function u (x; y) 4 (3x ) (y 0) a) Determine the marginal utility for the two commodities. Is more always better for the consumer? b) The consumer has a budget of 40. A unit of the rst commodity costs 0 and a unit of the second 5. Write down the budget equation. c) The consumer wants to maximize his utility subject to his budget constraint. Write down the Lagrangian for this problem. d) Calculate the rst order conditions for a critical point of the Lagrangian. e) Assume only the budget constraint binds. Derive a linear equation to be satis ed by a critical point that does not involve the Lagrange multiplier for the budget constraint. f) Use the equation from e) and the budget equation to nd the constrained optimum. Solution @u No, more is not always better. b) c etc) 4 (3x ) > 0 only for x < 4 4 (y 0) > 0 only for y < 5 L 4 (3x ) x (y 0) + (40 0x 4 (3x ) (y 0) (3x ) 0 8 (y 0) 9 (x 4) (y 5) 9x y (3x ) 8 (y 0) 40 0x 5y 0 Solution is: x 4;

6 Problem 8 (5 marks) A monopolist with inverse demand function p f (Q) and with total cost function T C (Q) has to pay an excise tax t to be subtracted from the price p paid by consumers. Assuming an interior equilibrium show that the pro ts of the pro t maximizing rm decrease as the excise tax increases. Solution 8 The pro t function is (Q; t) Q (f (Q) t) T C (Q) Q > 0 By the envelope theorem a marginal increase in tax changes pro t Q < 0, pro ts decrease. Problem 9 (0 marks) Solve the problem max u t[0;] TX p ut x t, x t+ ( u t ) x t for t 0; ; T, x 0 > 0 given. t0 Solution 9 The value function takes the form V t (x) p c t x t with c T and c t + c t+ for 0 t < T. (At hindsight I realize that this means that c t T + t, so the whole question is simpler than I thought.) This is clear for t T since obviously p u T x T is maximized over all u T [0; ] at u T. Suppose it holds for t + with 0 t < T. Then we must nd h put V t (x t ) max W t (x t ; u t ) max x t + p i c t+ ( u t ) x t u t[0;] u t[0;] The FOC t r r xt ct+ x t 0, p u t p c t+ u t, u t u t u t and so c t+ + c t+ u t + c t+ r s V t (x t ) W t (x t ; u xt c t ) + t+x t + c t+ p p xt + c t+ + c t+ + ct+ p + c t+ p xt p c t x t since c t was de ned as + c t+. We obtain u t + c t+ + (T + (t + )) T + x t+ ( u t ) x t c t+ x t T t + c t+ T + from which we can inductively infer all x t starting with x 0. t t 6

7 Problem 0 (5 marks) Find a solution to the di erential equation with x (0). dx dt x ( x) + x x Solution 0 Z x + Z dx dt x ln x ln ( x) t + C x ln x t + C x x et+c x e t+c e t+c x + e t+c x e t+c x e t+c + e t+c Since x (0) is x (t) et. +e t ec we obtain +e C ec + e C, hence e C, hence C 0. the solution Problem (0 marks) Solve the problem Z max u (t) dt, _x (t) x (t) + u (t), x (0), x () free: Solution The Hamiltonian is H u + p (x + u) The maximum of the Hamiltonian with respect to u is given by the FOC u + p 0, so p u. The solution to _p H x is _p dp p. Thus dt dp dt p Z Z p dp dt t + C ln p t + C p e t+c C e t Because x () is free we must have p () 0 and hence C 0. Thus p (t) 0 and therefore p (t) 0 for all t. Since p u we get u (t) 0. So _x (t) x (t) and x (0), which has the unique solution x (t) e t. Part B (You can gain no more than 5 marks on this part.) 7

8 Problem (5 marks) For a consumer with the utility function u (x; y) (0 x) (0 y) derive his demand function. What is the marginal increase in utility if the budget of the consumer is slightly increased? Solution The Lagrangian is the FOC are L (0 x) (0 y) + (b p x x p y y) + x + 3 y (0 x) p x + 0 (0 y) p y It is clear that the consumer never wants to consume zero of one or both commodities unless the budget constraint is binding since one of the marginal utilities is strictly positive along the axes. We must distinguish four more remaining (assuming p x ; p y ; b > 0):. No constraint binds. Then all Lagrange multipliers are zero and x 0, y 0. This is the case if (0; 0) is a ordable, i.e. 0p x + 8p y b. The consumer is satiated. an increase in his budget does not make him better o.. The non-negativity constraint x 0 binds with the budget constraint. So 3 0 by complementarity. y bp y, (0 bp y ) p y 0, so y bp y 0; the consumer is not able to buy his ideal amount of the y commodity. Moreover, (0 bp y ) p x p y 0 0, 0p y 0p x p y + bp x 0 The relative price of the rst commodity is higher than the marginal rate of substitution, so that it is not worth to buy of it. 3. The non-negativity constraint y 0 binds with the budget constraint. So 0 by complementarity. x bp x, (0 x ) p x 0, so x bp x 0; the consumer is not able to buy his ideal amount of the y commodity. Moreover, 3 (0 x ) p y p x 0 0, 0p x 0p x p y + bp y 0 The relative price of the second commodity is so high that it is not worth to buy of it. 8

9 4. Only the budget constraint binds. So 3 0. We must have x 0 and y 0 to have the Lagrange multiplier non negative. We obtain (0 x) p x (0 y) p y p x x + p y y b The solution is p x (0 y) p y (0 x) p y x p x y 0 (p y p x ) p y p y x p y p x y 0p y (p y p x ) p x p x x + p x p y y p x b p x + p y x 0p y 0p x p y + p x b x 0p y 0p x p y + p x b p x + p y y 0p x 0p x p y + p y b p x + p y These are non-negative if we are not in cases ) or 3) Concerning the last question in the problem, let u (p x ; p y ; b) be the optimal utility achievable for the given budget. Then, by the @b so a marginal increase in the budget raises utility by the value of the Lagrange multiplier for the budget constraint. We obtain in the four cases from above: ) 0 ) (0 bp y ) p y 3) (0 bp x ) p x 3) (0 x ) p x p x 0 0p y! 0p x p y + p x b p x + p y Problem 3 (5 marks) Solve the problem (with 0 < < ) max X t ln (x t u t ) subject to x t+ u t, x 0 > 0, u t > 0 t0 Solution 3 H t ln (x u)+pu. t 0 and p x t u t. x t u t Also the transversality condition lim t! p t (x t x t ) 0 must hold for any admissible 9 t

10 sequence (x t ) and we must have x t+ u t, x 0 x 0. We obtain from the rst two conditions p t p t p, so, since x t+ u t x t u t x t x t+ p t x t+ x t x 0 p t : : : x 0 t+ p p t One can choose an admissible sequence x t > 0 with lim x t arbitrarily small. Transversality implies hence (since x t > 0 must hold for all t) that lim t! p t x t plim t! x t 0. Now x t x t 0! x 0 p p 0, p x 0 ( ) and then Part C x t x 0 x 0 ( ) t t x 0 Problem 4 (0 marks) Sketch the graph of the area A carved out by the two inequalities (x ) + y 5 x + (y ) 5 For any point (a; b) in the plane with non-negative coordinates use the Lagrangian approach to determine the point closest to (a; b) within or on the boundary of A. How many cases do we have to consider? Solution 4 The Lagrangian is L (x a) (y b) + 5 (x ) y + 5 x (y ) the FOC (x a) (x ) x 0 (y b) y (y ) 0 We have four cases: None of the constraints are binding, only one is binding, both are binding. To have both constraints binding 0

11 A) If no constraint is binding the FOC yield x a, y b and consequently (a; b) must be in the region A, indicated by I in the following graph. 8 6 y x II 6 V III I IV. B) If both constraints are binding then x + (y ) (x ) + y 5 y + x + y x (x ) + x 5 0 x x 4 (x 3) (x + 4) Thus the two possible solutions are (x ; y ) ( 3; 3) and (x ; y ) (4; 4). In the rst case we have the two FOC which have the ( 3 a) ( 3 b) b 4 7 a a 4 7 b Since we must have ; 0 it follows that (x ; y ) (3; 3) is the solution if 3b 3 + 4a 3a 3 + 4b

12 This gives area V in the above graph. In the second case we have the two FOC which have the (4 a) 6 (x ) 8 x 0 (4 b) y (y ) a b 3 7 b a Since we must have ; 0 it follows that (x ; y ) (3; 3) is the solution if 4b 4 + 3a 4a 4 + 3b This gives area IV in the above graph. B) Suppose only the constraint 5 (x ) + y is binding. Hence 0 by complementarity the FOC are Division (x a) (x ) 0 (y b) y 0 x a y b x y, xy ay xy y bx + b, bx (a ) y + b and therefore 5 (a ) y + b b 5b y q (a ) + b x a b 5b q (a ) + b! + y (a ) + b b 5 (a q ) (a ) + b y The rst order conditions can be rewritten as a ( + ) (x ) b ( + ) y

13 which, since 0 implies that (x ) and y are positive multiples of a and b, so the optimum is x 5 (a ) q(a ) + b, y To be admissible the solution must satisfy 5 (x ) + (y ) 5 (x ) + (y ) Subtracting from the inequality the equation above yields 0 x y +, y x 5b q (a ) + b Hence (x ; y ) lies aboe the diagonal y x. Since the vector (a; b) is obtained from (x ; y ) byt multiplying with ( + ) 0, it follows that (a; b) must be in the area II in the graph. C) Only the constraint 5 x +(y ) is binding. Hence 3 0 by complementarity the FOC are Division yields and (x a) x 0 (y b) (y ) 0 x a y b x, xy ay x + a xy bx, (b ) x a (y ) y 5! a (y ) + (y ) a + (b ) b (b ) (y ) 5 (b ) y q a + (b ) 0 a 5 (b ) A 5a q b a + (b ) a + (b ) It follows as above that x 5a qa y + (b ) and that (a; b) must be in the area III in the graph. 3 5 (b ) q a + (b )

14 Problem 5 (0 marks) Suppose that for a twice continuously di erentiable univariate function f (x) de ned on an interval there is a unique point x which solves the equation f 0 (x ) 0. Suppose f 00 (x ) < 0. Sketch an argument why x must be an absolute maximum of the function. Is f (x) necessarily concave? Solution 5 Because f 00 (x) is continuous there is a neighborhood around x where f 00 (x) > 0 holds:f 0 (x) is decreasing near x. Hence f 0 (x) > 0 for all x < x near to x. But then we must have f 0 (x) > 0 for ALL x < x because otherwise there would be by the intermediate value theorem a x < x with f 0 (x) 0; contradicting the assumptions. Hence f (x) is increasing to the left of x and hence f (x) < f (x ). It follows by a symmetric argument that f (x) is decreasing for x > x and hence f (x ) > f (x) for all x > x. Thus x is an absolute maximum. Problem 3 above shows that f does not necessarily have to be concave. 4

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