BEE1020 Basic Mathematical Economics Dieter Balkenborg Week 14, Lecture Thursday Department of Economics

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1 BEE1020 Basic Mathematical Economics Dieter Balkenborg Week 14, Lecture Thursday Department of Economics Optimization University of Exeter Sincethefabricoftheuniverseismostperfect,and is the work of a most perfect creator, nothing whatsoever takes place in the universe in which some form of maximum or minimum does not appear. Leonhard Euler, 1744 optimization problems: unconstrained(profit maximization) constrained(utility maximization with budget constraint) first order conditions for optimum yield simultaneous system of equations( Review) Lagrangian approach for constrained problems

2 Unconstrained optimization Find absolute maximum of function i.e.,findpair(x,y )suchthat z=f(x,y) f(x,y ) f(x,y) forall(x,y). The following must hold: freeze the variable y at optimal value y,varyonlyxthen f(x,y ) Musthavemaximumatx. Obtainfirstorderconditions z x x=x,y=y = 0 z y x=x,y=y = 0 2

3 x y

4 Example: Production function Q(K, L). r interest rate wwagerate P priceofoutput profit Π(K,L)=PQ(K,L) rk rl. FOC for profit maximum: Π K = P Q K Π L = P Q L Intuition: SupposeP Q K r>0... r=0 (1) w=0 (2) 4

5 Rewrite as Q K = r P Q L = w P Division yields: / Q Q K L P/ = r w P = r w Marginal rate of substitution must equal ratio input prices. (3) (4) (5) 5

6 Example: P =12,r=1andw=3. Then Q(K,L)=K 1 6L 1 2 Q K = 1 6 K 5 6L 1 Q 2 L =1 2 K1 6L 1 2 / Q Q K L = 1 / 6 K 5 6L K1 6L 1 2= 1 3 K 5 6L 1 2K 1 6L 1 2= 1 L 3K FOC s: 1 6 K 5 6L 1 2 = K1 6L 1 2 = 3 12 K 5 6L 1 2= 1 2 K 1 6L 1 2= 1 2 (6) (7) 6

7 FOC s: Division yields Hence 1 6 K 5 6L 1 2 = K1 6L 1 2 = L 3K =1 3 K 5 6L 1 2=K 5 6K 3 6=K 1 3= 1 2 K 5 6L 1 2= 1 2 K 1 6L 1 2= 1 2 L=K (8) (9) or 3 1 K =1 2 1 K =1 8 K=L=8 7

8 z K L

9 Example: AmonopolistwithtotalcostfunctionTC(Q)=Q 2 sellshisproductintwodifferentcountries. WhenhesellsQ A units ofthegoodincountryahewillobtaintheprice P A =22 3Q A foreachunit. WhenhesellsQ B unitsofthegoodincountryb he obtains the price P B =34 4Q B. Howmuchshouldthemonopolistsellinthetwocountriesinorder to maximize profits? Solution: Total revenue in country A: Total revenue in country B: TR A =P A Q A =(22 3Q A )Q A TR B =P B Q B =(34 4Q B )Q B 9

10 Total production costs are: Profit: TC=(Q A +Q B ) 2 Π(Q A,Q B )=(22 3Q A )Q A +(34 4Q B )Q B (Q A +Q B ) 2, FOC: Π Q A = 3Q A +(22 3Q A ) 2(Q A +Q B )=22 8Q A 2Q B =0 Π Q B = 4Q B +(34 4Q B ) 2(Q A +Q B )=34 2Q A 10Q B =0 or linear simultaneous system 8Q A +2Q B = 22 (10) 2Q A +10Q B =

11 Simultaneous systems of equations Linear systems example Using the slope-intercept form 5x+7y = 50 (11) 4x 6y = 18 7y = 50 5x y = x 4x 6y = 18 4x+18 = 6y y = 2 3 x+3 11

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13 At intersection point two linear functions must have same y-value. Hence x = y=2 3 x = 2 3 x+5 7 x = 14x+15x 87 = 29x x = =3 foundvalueofx. calculatethevalueofy: solution: x =3,y =5 y= =5 13

14 strongly recommended to check: = 15+35= = 12 30= 18 14

15 The method of substitution solveoneoftheequationsforoneofthevariables: 4x 6y = 18 4x+18 = 6y y = 4 6 x+3=2 x+3 (12) 3 replace y in the other equation. (brackets!), obtain equation in one variable. 15

16 use(12)tofindy: solution: x=3,y=5. ( 5x+7y ) = x+7 3 x+3 = 50 5x+ 14 x+21 = x+14 x+21 = x = 50 21=29 x = =3 y= 2 3 x+3= =5 16

17 The method of elimination multiply first equation by coefficient of x in second equation multiply second equation by the coefficient of x in first equation 5x+7y = x 6y = x+28y = x 30y = 90 subtract second equation from first equation 20x+28y=200 20x 30y= y ( 30y)=200 ( 90) 58y=290 y= =5 17

18 useoneoftheoriginalequationstofindx 5x+7y = 50 5x+35 = 50 5x = 15 x = 3 We could have eliminated y first: 5x+7y = x 6y = x+42y = x 42y = x = 174 x = 3 18

19 Cramer s rule linear algebra. Write [ system of ] equations [ ] as 5 7 x 50 = 4 6][ y 18 [ ] [ ] x1 50 where and columns vectors x 2 18 [ ] isthe2 2- matrixofcoefficients. Witheach2 2-matrix [ ] a b A= c d associate a new number called the determinant deta= a b c d =ad cb 19

20 vertical lines, not square brackets!!! For instance, =5 ( 6) 4 7= 30 28= 58 or Cramer s rule for system x = e b f d a b c d ax+by = e cx+dy = f [ a b b d][ x y = ed bf ad bc 20 ] [ ] e =. f y a e c f = a b c d = ae ce ad bc

21 In our example, 50 7 x 18 6 = y = Exercise: = 50 ( 6) ( 18) 7 5 ( 6) 4 7 = 5 ( 18) ( 6) 4 7 = Q A +2Q B = 22 2Q A +10Q B = 34. = = =3 = =5 21

22 Existence and uniqueness. linear simultaneous system of equations Rewrite as ax+by = e cx+dy = f y = e b a b x y = f d c d x slopes identical when a b = d c, i.e., when determinant ad cb is zero. If in addition intercepts are equal, both equations describe the same line. Example: x+2y = 3 2x+4y = 4 22

23 hasnosolution: Thetwolines y = x y = x are parallel x Thereisnocommonsolution. Tryingtofindoneyieldsacontra- 23

24 diction x = y=1 1 2 x +1 2 x 3 2 = 1 24

25 One equation non-linear, one linear Consider, for instance, y 2 +x 1 = 0 y+ 1 2 x = 1 Inourexampleitisconvenienttosolvesecondequationforx: 1 2 x = 1 y x = 2 2y y 2 +(2 2y) 1 = y 2 2y+1=(y 1) 2 =0 Sotheuniquesolutionisy =1andx =2 2 1=0. Two non-linear equations no general method. See example above 25

26 Constrained optimization Example: Utility maximization preferences of a consumer described by family of indifference curves. mathematically convenient as level curves u(x, y) =constant of utility function. example: u(x,y)=xy. indifferencecurvesforu=1,u=2and u=3: x priceofapplesisp,priceoforangesisqbudgetb. 26

27 budget constraint px+qy b 27

28 maximize utility subject to the budget constraint! constrained optimization problem u(x, y) objective function total expenditure g(x,y) = px+qy is called the constraining function. Principles lecture budget line must be tangential to the indifference curve marginal rate of substitution must equal relative price: budget line: px+qy=b or y= b q p q x p q =negativeofrelativepriceofapplesintermsoforanges. slope of the indifference curve is negative of marginal rate of substitution u/ x u/ y In optimum: 28

29 u/ x u/ y =p (13) q Fortheexampleu(x,y)=xy: u x =yand u y =x y x =p (14) q or qy=px, (15) budget equation px+qy=b (16) To repeat: the constrained optimum (x,y ) is the solution to a simultaneous system of two equations in two unknowns. The first equation expresses that in the optimum the marginal rate of substitution is equal to the relative price. The second one states thattheconsumerspendsallhismoneyintheoptimum. 29

30 y= p q x. Substitutingintobudgetequationgives ( ) p px+q q x = px+px=2px=b x = b 2p y = p q x = p b q2p = b 2q Whenthebudgetisb=100andp=2andq=5then x = =25 y= =10. Cost minimization dual to the utility maximization problem productionfunctionq(k,l)=kl. leastcostlywaytoproduceq 0 unitsgivenpricesrandw 30

31 minimize total costs subject to rk+wl Q 0 Q(K,L). at cost minimum iso-cost line must be tangential to the isoquant. iso-cost line rk+wl=constant hasslop w r slopeofisoquantis Q/ K Q/ L In optimum: Q/ K Q/ L = r w. 31

32 Inourexample Q K or =Land Q L =K,so L K = r w wl=rk soinoptimumequalamountsarespendonbothinputs. Thisis the first equation. Asthesecondequationwehavethatthefirmwillproduceexactly Q 0 units: Q 0 =Q(K,L). supposer=2,w=5,q 0 =250 optimum(k,l ): L K =2 5 or L= 2 5 K 32

33 and 250=KL. SubstitutinginthelastequationL= 2 5 Kyields 250 = 2 125=2 5 3 = 2 5 K2 5 4 =K 2 K =5 2 =25 L = 2 5 K =10. Theoptimalinputcombinationis(K,L )=(25,10). The general problem maximize or minimize an objective function subject to a constraint wherecisaconstant. z=f(x,y) g(x,y) c 33

34 interested in case where constraint is binding in optimum, i.e., g(x,y)=c The optimum must then solve the two conditions f/ x f/ y = g/ x g/ y g(x,y) = c The Lagrangian approach alternative way to derive these two conditions. The method transforms the constrained optimization problem into an unconstrained optimization problem in conjunction with a pricing problem. λ 0 is the so-called Lagrange multiplier. Lagrangian L(x,y)=f(x,y) λ(g(x,y) c) lookforanunconstrained maximumoraminimumofthisfunction 34

35 determine the Lagrange multiplier such that constraint holds with equality Supposefoundabsolutemaximum(x,y )oflagrangian. Supposeg(x,y )=c. L(x,y )=f(x,y ) λ(g(x,y ) c)=f(x,y ). consider(x,y)withg(x,y) c. ThenL(x,y) L(x,y )because(x,y )isabsolutemaximum. λ(g(x,y) c)isnotpositive Hencef(x,y) L(x,y) L(x,y ) f(x,y ) FOC for maximum of Lagrangian: L x = f x λ g x =0 or f x =λ g x L y = f y λ g y =0 or f y =λ g y 35 (17) (18)

36 divide: f/ x f/ y = g/ x g/ y. (19) In addition constraint must hold in optimum Utility maximization again Lagrangian g(x,y)=c. (20) L(x,y)=u(x,y) λ(px+qy b)=u(x,y)+λ(b px+py) The savings: future utility from saving: s=b px+py λs λisthemarginalutilityofsavingapenny. Cost minimization again 36

37 The Lagrangian L(K,L) = rk wl λ( Q(K,L)+Q 0 ) = (λq(k,l) rk wl) λq 0 UptoaconstanttheLagrangianisprofitfunctionforoutputprice λ. 37