1 Objective. 2 Constrained optimization. 2.1 Utility maximization. Dieter Balkenborg Department of Economics

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1 BEE020 { Basic Mathematical Economics Week 2, Lecture Thursday Constrained optimization Dieter Balkenborg Department of Economics University of Exeter Objective We give the \ rst order conditions" for constrained optimization problems like utility maximization and costs minimization. The Lagrangean method to obtain these conditions is introduced and its economic interpretation is discussed. The relevant reading for this handout is Chapter 7, Sections 4. See also the book by Begg on the consumer optimum and on cost minimization. 2 Constrained optimization 2. Utility maximization The preferences of a consumer are described by a family of indi erence curves. A mathematically convenient way to describe a family of indi erence curves is to describe them as the level curves of a utility function u (x; y). The utility u (x; y) can be interpreted as a numerical measure of the satisfaction the consumer has when consuming x units of the rst commodity (say, apples) and y units of the second commodity (say oranges). Along a level curve utility is constant, so all commodity bundles on a level curve bring equal satisfaction, they are hence the indi erence curves. If one bundle has more utility than another, the consumer will prefer it. As an example consider the utility function u (x; y) = xy. The indi erence curves for u =,u =2andu =havetheform x where consumption on higher indi erence curves yields higher utility. Suppose the price of apples is p, the price of oranges is q and the consumer has a budget b. Then the budget constraint px + qy b

2 must be satis ed for the consumption bundle (x; y) which the consumer buys, i.e., his total expenditure on the two commodities cannot exceed his budget. The aim of the consumer is to maximize his utility subject to the budget constraint. This is a constrained optimization problem because the consumer would consume in nite amounts of both commodities if he were not constrained by his budget. u (x; y) isthe objective function for this problem because the objective is to maximize this function. We call total expenditure g (x; y) = px + qy the constraining function because it appears in the constraint. We expect the budget constraint to be binding in optimum, i.e., we expect the budget inquality to hold with equality in the optimum. This is so because by assumption the consumer has no possibility to save and because no other commodities are available. From the Principles lecture you know the essential condition which must hold in an optimum: The budget line must be tangential to the indi erence curve. Now the budget line is given by the equation px + qy = b or y = b q p q x so it has the slope p, which, apart form the sign, is the relative price of apples in terms q of oranges. (To buy an apple more the consumer must give up p units of oranges.) q The slope of the indi erence curve is { apart from the sign { the marginal rate of substitution. In the previous handout we tried to explain that the marginal rate of substitution Thus in optimum the marginal rate of substitution must be to the = p q () This is the rst equation which characterizes the constrained optimum. For the example u (x; y) =xy = y and = x. The y x = p (2) q or qy = px; () i.e., in optimum the consumer spends equal amounts on each commodity. We need a second equation for the optimum. This second equation is the budget equation px + qy = b (4) To repeat: the constrained optimum (x ;y ) is the solution of a simultaneous system of two equations in two unknowns. The rst equation expresses that in the optimum the marginal rate of substitution is equal to the relative price. The second one states that the consumer spends all his money in the optimum. 2

3 gives In our example condition () gives y = p x. Substituting this into the budget equation q µ p px + q q x = px + px =2px = b x = b 2p y = p q x = p q b 2p = b 2q We have found the optimum. For instance, when the budget is b = 00 and the prices are p =2andq = 5 the optimal consumption is x = =25 y = =0: 2.2 Cost minimization The cost minimization problem is in many ways dual to the utility maximization problem and leads to very similar conditions. Consider a price-taking rm with production function Q (K; L) =KL. Suppose the rm wants to nd the least costly way to produce Q 0 units of output given the prices r and w for the inputs. Then she wants to minimize total costs rk + wl subject to the constraint that she produces at least Q 0 units Q 0 Q (K; L) : Here expenditure on inputs is the objective function whereas the production function is the constraining function. Again from the principles lecture we know that in the cost minimum the iso-cost line must be tangential to the isoquant. Since the slope of an iso-cost line rk + wl =constant is the negative of the relative price of inputs r it follows that in the optimum the marginal w rate of must equal the relative In @Q=@L = r w : = = K, so L K = r w wl = rk

4 so in optimum equal amounts are spend on both inputs. This is the rst equation. As the second equation we have that in the optimum the rm will produce exactly Q 0 units and no more, the cost-minimizing optimum (K ;L ) must lie on the isoquant for Q 0 Q 0 = Q (K; L) : In our example suppose that r =2,thatw = 5 and that the rm wants to produce 250 units of output at the lowest costs. Then the optimum (K ;L ) must satisfy the conditions and L K = 2 5 or 250 = KL: Substituting in the last equation L = 2 5 K yields L = 2 5 K 250 = 2 25 = 2 5 = 2 5 K2 5 4 = K 2 K =5 2 =25 L = 2 5 K =0: The optimal input combination is (K ;L )=(5; 0). 2. The general problem In general we want to maximize or to minimize an objective function susbject to a constraint z = f (x; y) g (x; y) c where c is a constant. We are interested in the case where the constraint is binding in the optimum, i.e., where g (x; y) = c holds at the optimum. (Otherwise the constraint is always satis ed near the optimum and the optimum must be a critical point of f.) The optimum must then solve the two g (x; y) = c 4

5 The Lagrangian approach The Lagrangian method is an alternative way to derive these two conditions. The method transform the constrained optimization problem into an unconstrained optimization problem in conjunction with a pricing problem. We discuss here only the maximization problem. First we form out of the two given functions f and g a new function which depends on an additional parameter, the so-called Lagrange multiplier. This new function { called the Lagrangian { is L (x; y) =f (x; y) (g (x; y) c) Then we lookfor an unconstrained maximum or a minimum of this function and determine the Lagrange multiplier such that the constraint holds with equality in the optimum. For a maximization problem the interpretation is as follows: Instead of strictly forbidding that the constraint g (x; y) c ever gets violated, we allow to violate it, but at a price. This price is : It is a ctituous price in a ctitious problem and hence called a \shadow price". Suppose the constraint is violated. g (x; y) c is then the number of units by which the constraint is violated. When each unit of violation costs, the total amount (g (x; y) c) must then be paid for violation. This amount is subtracted from the value of the objective function which we want to maximize. (If the constraint is not violated, there is a reward.) Now we look for an unconstrained maximum (x ;y ) of the Lagrangian. Suppose we have found an absolute maximum (x ;y ) of this function and suppose and we were able to choose a positive such that (x ;y ) satis es the constraint g (x ;y )=c. Then(x ;y ) is automatically an optimum of the constrained optimization problem: We have L (x ;y )=f (x ;y ) (g (x ;y ) c) =f (x ;y ). Now consider any pair (x; y) that does not violate the constraint, i.e., that satis es g (x; y) c. Then L (x; y) L(x ;y )because(x ;y ) is an absolute maximum. Moreover (g (x; y) c) is positive since (x; y) satis es the constraint. Hence f (x; y) L (x; y) and overall we obtain that f (x; y) f (x ;y ) holds for all (x; y) satisfying the constraint. This means that (x ;y ) is a constrained maximum. To nd the maximum of the Lagrangian we solve the rst @x @y @y Notice that if we divide the two equations on the right @g=@y : (7) In addition we want the constraint to hold with equality in the optimum, i.e., we impose the equation (6) g (x; y) =c: (8) 5

6 We then solve the two equations (7) and (8) for x and y. The Lagrangian was just a tool to derive these equations. If the need arises, we can calculate the value of the Lagrange multiplier from equation (5) or (6).. Utility maximization again In this case the Lagrangian is L (x; y) =u (x; y) (px + qy b) =u (x; y)+ (b px + py) It can be interpreted as follows: In the original problem saving is not possible. suppose it is possible to save and also to go into debt. The savings are Now s = b px + py Now suppose there is some future utility from saving because the money can be spend in the future. For simplicity we assume that this future utility is linear in savings and given by s Then is the marginal utility of saving a penny. One also speaks of the marginal utility of money. The Lagrangian can then be interpreted as the total utility from consumption today and in the future. When we solve the constrained optimization problem with the Lagrangian approach we e ectively ask: What would the marginal utility of saving have to be such that it is optimal not to save and not to go into debt..2 Cost minimization again The rm wants to minimize the costs rk + wl subject to the constraint Q 0 Q (K; L). First we transorm the problem such that it ts with our general framework above. Minimizing rk + wl is the same as maximizing rk wl. Now we have a maximization problem. Next we had the constraint in the form g (x; y) c with the constant on the right. Therefore we rewrite Q 0 Q (K; L) as Q (K; L) Q 0. The Lagrangian becomes: L (K; L) = rk wl ( Q (K; L)+Q 0 ) = ( Q (K; L) rk wl) Q 0 We see that { up to a constant which will drop out in the di erentiation { the Lagrangian is just the pro t function of the rm when the output price is. Whenwesolvethecost minimization problem with the Lagrangian method we therefore ask: What would the output price have to be such that it is optimal to produce exactly Q 0 units of output. 6

7 4 Objective Two types of optimization problems frequently occur in economics: unconstrained optimization problems like pro t maximization problems and constrained optimization problems like the utility maximization problem of consumer constrained by a budget. In this handout we discuss the main step to solve an unconstrained optimization problem - namely to nd the two \ rst order conditions" and how to solve them. The solutions for this system of two equations in two unknowns can be local maxima, minima or saddle points. How to distinguish between these types will be discussed in a later handout, after we have discussed constrained optimization problems. The relevant reading for the previous handout is Chapter 7, Sections and 2 in the textbook of Ho mann and Bradley. The relevant reading for this handout is Chapter 7, Sections. 5 Optimization problems 5. Unconstrained optimization Suppose we want to nd the absolute maximum of a function z = f (x; y) in two variables, i.e., we want to nd the pair of numbers (x ;y ) such that f (x ;y ) f (x; y) holds for all (x; y). If (x ;y ) is such a maximum the following must hold: If we freeze the variable y at the optimal value y and vary only the variable x then the function f (x; y ) { which is now just a function in just one variable { has a maximum at x. Typically the maximum of this function in one variable will be a critical point. Hence we expect the partial to be zero at the optimum. Similarly we expect the partial to be zero at the optimum. We are thus led to the rst ;y=y jx=x ;y=y = 0 which must typically be satis ed. We speak of rst order conditions because only the rst derivatives are involved. The conditions do not tell us whether we have a maximum, a minimum or a saddle point. To nd out the latter we will have to look at the second derivatives and the so-called \second order conditions", as will be discussed in a later lecture. 7

8 Example Consider a price-taking rm with the production function Q (K; L). Let r be the interest rate (the price of capital K), let w be the wage rate (the price of labour L) and let P be the price of the output the rm is producing. When the rm uses K units of capital and L units labour she can produce Q (K; L) units of the output and hence make a revenue of P Q (K; L) by selling each unit of output at the market price P. Her production costs will then be her total expenditure on the inputs capital and labour Her pro t will be the di erence rk + wl: (K; L) =PQ(K; L) rk rl: The rm tries to nd the input combination (K ;L ) which maximizes her pro t. To nd it we want to solve the rst = r @L w = 0 (0) These conditions are intuitive: Suppose for r>0: is product of capital, i.e., it tells us how much more can be produced by using one more unit of capital. is the marginal increase in revenue when one more unit of capital used and the additional output is sold on the market. r is the marginal cost of using one more unit of capital. r>0 means that the marginal pro t from using one unit of capital is positive, i.t., it increases pro ts to produce more by using more capital and therefore we cannot be in a pro t optimum. Correspondingly, mean that pro t can be increased by producing less and using less capital. So (9) should hold in the optimum. Similarly (0) should hold. It is useful to rewrite the two rst order = r = w (2) P Thus, for both inputs it must be the case that the marginal product is equal to the price ratio of input- to output price. When we divide here the two left-hand sides and equate them with the quotient of the right hand sides we obtain as a = r. w P P = r () w so the marginal rate of substitution must be equal to the price ratio of the input price (which is the relative price of capital in terms of labour). 8

9 Example 2 Let us be more speci c and assume that the production function is Q (K; L) =K 6 L 2 and that the output- and input prices are P = 2, r =andw = 6 = 6 K 5 6 L = 2 K 6 L Á 2 K 6 L 2 = The rst order conditions () and (2) become 2 K 5 6 L 2 K 6 L 2 = 5 6 K 6 L 2 = (4) 2 2 K 6 L 2 = 2 : (5) This is a simultaneous system of two equations in two unknowns. The implied condition () becomes L K which simpli es to L K = L = K Example A monopolist with total cost function TC(Q) =Q 2 sells his product in two di erent countries. When he sells Q A units of the good in country A he will obtain the price P A =22 Q A for each unit. When he sells Q B units of the good in country B he obtains the price P B =4 4Q B : How much should the monopolist sell in the two countries in order to maximize pro ts? To solve this problem we calculate rst the pro t function as the sum of the revenues in each country minus the production costs. Total revenue in country A is Total revenue in country B is Total production costs are TR A = P A Q A =(22 Q A ) Q A TR B = P B Q B =(4 4Q B ) Q B TC =(Q A + Q B ) 2 9

10 Therefore the pro t is (Q A ;Q B )=(22 Q A ) Q A +(4 4Q B ) Q B (Q A + Q B ) 2 ; a quadratic function in Q A and Q B. In order to nd the pro t maximum we must nd the critical points, i.e., we must solve the rst order A = Q A +(22 Q A ) 2(Q A + Q B )=22 8Q A 2Q B = 4Q B +(4 4Q B ) 2(Q A + Q B )=4 2Q A 0Q B =0 8Q A +2Q B = 22 (6) 2Q A +0Q B = 4: Thus we have to solve a linear simultaneous system of two equations in two unknowns. 6 Simultaneous systems of equations 6. Linear systems We illustrate four methods to solve a linear system of equations using the example 6.. Using the slope-intercept form We rewrite both linear equations in slope-intercept form 5x +7y = 50 (7) 4x 6y = 8 7y = 50 5x y = x 4x 6y = 8 4x +8 = 6y y = 2 x x 0

11 The solutions to each equation form a line, which we have now described as the graph of linear functions. At the intersection point of the two lines the two linear functions must have the same y-value. Hence x = y = 2 x = 2 x + 5 j = 4x +5x 87 = 29x x = = We have found the value of x in a solution. To calculate the value of y we use one of the linear functions in slope-intercept form y = 2 =2 The solution to the system of equations is x =;y =2 It is strongly recommended to check the result for the original system of equations The method of substitution = = = 2 0 = 8 First we solve one of the equations in (7), say the second, for one of the variables, say y: 4x 6y = 8 4x +8 = 6y y = 4 6 x += 2 x + (8) Then we replace y in the other equation by the result. (Do not forget to place brackets aroundtheexpression.) Weobtainanequationinonlyonevariable,x, whichwesolve for x. 5x +7y = 50 µ 2 5x +7 x + = 50 5x + 4 x +2 = 50 5 x + 4 x +2 = x = 50 2 = 29 x = =

12 We have found x and can now use (8) to nd y: Hence the solution is x =;y= The method of elimination y = 2 x += 2 +=5 To eliminate x we proceed in two stages: First we multiply the rst equation by the coef- cient of x in the second equation and we multiply the second equation by the coe±cient of x in the rst equation 5x +7y = 50 j 4 4x 6y = 8 j 5 20x +28y = x 0y = 90 Then we subtract the second equation from the rst equation 20x +28y = x 0y = 90 j 0+28y ( 0y) =200 ( 90) 58y =290 y = 290 =5 58 Having found y we use one of the original equations to nd x 5x +7y = 50 5x +5 = 50 5x = 5 x = Instead of rst eliminating x we could have rst eliminated y: 6..4 Cramer's rule 5x +7y = 50 j 6 4x 6y = 8 j 7 0x +42y = 00 28x 42y = 26 j + 58x = 74 x = This is a little preview on linear algebra. In linear algebra the system of equations is written as 5 7 x 50 = 4 6 y 8 2

13 x 50 where and are so-called columns vectors, simply two numbers written x beloweachotherandsurroundedbysquarebrackets. is the 2 2-matrix 4 6 of coe±cients. A 2 2-matrix is a system of four numbers arranged in a square and surrounded by square brackets. With each 2 2-matrix a b A = c d we associate a new number called the determinant det A = a b c d = ad cb Notice that the determinant is indicated by vertical lines in contrast to square brackets. For instance, =5 ( 6) 4 7= 0 28 = 58 Cramer's rule uses determinants to give an explicit formula for the solution of a linear simultaneous system of equations of the form ax + by = e cx + dy = f or a b b d x y = e f : Namely, x = e f a c b d b d = ed bf ad bc y = a c a c e f b d = ae ce ad bc Thus x and y are calculated as the quotients of two determinants. In both cases we divide by the determinant of the matric of the coe±cients. For x the determinant in the numerator is the determinant of the matrix obtained by replacing in the matrix of coe±cients the coe±cients a and c of x by the constant terms e and f, i.e., the rst row is replaced by the row vector with the constant terms. Similarly, for y the determinant in the numerator is the determinant of the matrix obtained by replacing in the matrix of coe±cients the coe±cients b and d of y by the constant terms e and f, i.e., the second row is replaced by the row vector with the constant terms. The method works only if the determinant in the denominator is not zero.

14 In our example, 50 7 x 8 6 = 5 7 = y 4 8 = 5 7 = ( 6) ( 8) 7 5 ( 6) ( 8) ( 6) 4 7 = = = = = =5 Exercise Use the above methods to nd the optimum in Example Existence and uniqueness. A linear simultaneous system of equations ax + by = e cx + dy = f can have zero, exactly one or in nitely many solution. To see that only these cases can arise, bring both equations into slope-intercept form y = e f a b x y = f d c d x If the slopes of these two linear functions are the same, they describe identical or parallel lines. The slopes are identical when a = c, i.e., when the determinant of the matrix of b d coe±cients ad cb is zero. If in addition the intercepts are equal, both equations describe the same line. In this case all points (x; y) on the line are solutions. If the intercepts are di erent the two equations describe parallel lines. These do not intersect and hence there is no solution. Example 4 has no solution: The two lines x +2y = 2x +4y = 4 y = 2 2 x y = 2 x We assume here for simplicity that the coe±cients b and d are not zero. The arguments can be easily extended to these cases, except when all four coe±cients are zero. The latter case is trivial - either there is no solution or all pairs of numbers (x; y) are solutions. 4

15 are parallel x There is no common solution. Trying to nd one yields a contradiction 2 2 x = y = 2 x j + 2 x 2 = 6.2 One equation non-linear, one linear Consider, for instance, y 2 + x = 0 y + 2 x = In this case we solve the linear equation for one of the variables and substitute the result into the non-linear equation. As a result we obtain one non-linear equation in a single unknown. This makes the problem simpler, but admittedly some luck is needed to solve the non-linear equation in one variable. Solving the linear equation for x can sometimes give a simpler non-linear equation than solving for y and vice versa. One may have to try both possibilities. In our example it is convenient to solve for x: 2 x = y x = 2 2y y 2 +(2 2y) = y 2 2y +=(y ) 2 =0 So the unique solution is y =andx =2 2 =0. 6. Two non-linear equations There is no general method. One needs to memorize some tricks which work in special cases. 5

16 In the example of pro t-maximization with a Cobb-Douglas production function we were led to the system 5 6 K 6 L 2 = 2 2 K 6 L 2 = 2 : This is a simultaneous system of two equations in two unknowns which looks hard to solve. However, as we have seen, division of the two equations shows that L = K must hold in a critical point. We can substitute this into, say, the rst equation and obtain 2 = 6 K K = 2 p = K 2 p K = 2 K = 2 =8 5 6 L 2 = 6 K 5 6 K 2 = 6 K = 6 K 2 6 = 6 K Thus the solution to the rst order conditions (and, in fact, the optimum) is K = L =8. Remark This method works with any Cobb-Douglas production function Q (K; L) = K a L b where the indices a and b are positive and have a sum less than. Remark 2 The rst order conditions (4) and (5) form a \hidden" linear system of equations. Namely, if you take the logarithms (introduced later!) you get the system ln ln K + 2 ln L = ln 2 ln ln K 2 ln L = ln 2 which is linear in ln K and ln L. One can solve this system for ln K and ln L, whichthen givesusthevaluesforl and K. 6