z = f (x; y) f (x ; y ) f (x; y) f (x; y )

Transcription

1 BEEM0 Optimization Techiniques for Economists Lecture Week 4 Dieter Balkenborg Departments of Economics University of Exeter Since the fabric of the universe is most perfect, 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, 744 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 ( second order conditions ) will be discussed in the next handout. In order to solve the rst order conditions one has to solve a simultaneous system of equations. We will review methods to solve linear simultaneous systems. Thereafter we develop the Lagrangian approach for constrained optimization problems. Optimization problems. 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

2 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 jxx jxx ;yy 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. 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 prot will be the di erence rk + wl: (K; L) P Q (K; L) rk rl: The rm tries to nd the input combination (K ; L ) which maximizes her prot. To nd it we want to solve the rst r w 0 () These conditions are intuitive: Suppose for instance P r > 0: is product of capital, i.e., it tells us how much more can be produced by using one more unit of capital. P is the marginal increase in revenue when one more unit of capital is and the additional output is sold on the market. r is the marginal cost of using one more unit of capital. P r > 0 means that the marginal prot from using one more of capital is positive, i.e., it increases prots to produce more by using more capital and therefore we cannot be in a prot optimum. Correspondingly, P r < 0 would that prot can be increased by producing less and using less capital. So () should hold in the optimum. Similarly () should hold.

3 It is useful to rewrite the two rst order conditions r w 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 consequence r P. w 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). Example Let us be more specic and assume that the production function is Q (K; L) K 6 L and that the output- and input prices are P, r and 6 K 6 6 K 6 L The rst order conditions () and (4) K 6 L K 6 L K 6 L K 6 L L K 6 K 6 L (6) K 6 L : (7) This is a simultaneous system of two equations in two unknowns. The implied condition () becomes L K which simplies to L K Example A monopolist with total cost function T C (Q) Q 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 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 :

4 How much should the monopolist sell in the two countries in order to maximize prots? To solve this problem we calculate rst the prot 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 Therefore the prot is T R A P A Q A ( T R B P B Q B (4 T C (Q A + Q B ) Q A ) Q A 4Q B ) Q B (Q A ; Q B ) ( Q A ) Q A + (4 4Q B ) Q B (Q A + Q B ) ; a quadratic function in Q A and Q B. In order to nd the prot maximum we must nd the critical points, i.e., we must solve the rst order conditions A Q A + ( Q A ) (Q A + Q B ) 8Q A Q B B 4Q B + (4 4Q B ) (Q A + Q B ) 4 Q A 0Q B 0 8Q A + Q B (8) Q A + 0Q B 4: Thus we have to solve a linear simultaneous system of two equations in two unknowns. Simultaneous systems of equations. Linear systems We illustrate four methods to solve a linear system of equations using the example.. Using the slope-intercept form x + 7y 0 (9) 4x 6y 8 We rewrite both linear equations in slope-intercept form 7y 0 x y x 4x 6y 8 4x + 8 6y y x + 4

5 x 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 x x + 7 x j x + x 87 9x x 87 9 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 + The solution to the system of equations is x ; y It is strongly recommended to check the result for the original system of equations... The method of substitution First we solve one of the equations in (9), say the second, for one of the variables, say y: 4x 6y 8 4x + 8 6y y 4 6 x + x + (0) Then we replace y in the other equation by the result. (Do not forget to place brackets around the expression.) We obtain an equation in only one variable, x, which we solve for x.

6 x + 7y 0 x + 7 x + 0 x + 4 x + 0 x + 4 x x 0 9 We have found x and can now use (0) to nd y: Hence the solution is x ; y... The method of elimination x 9 9 y x + + 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 x + 7y 0 j 4 4x 6y 8 j 0x + 8y 00 0x 0y 90 Then we subtract the second equation from the rst equation 0x + 8y 00 0x 0y 90 j 0 + 8y ( 0y) 00 ( 90) 8y 90 y 90 8 Having found y we use one of the original equations to nd x x + 7y 0 x + 0 x x 6

7 Instead of rst eliminating x we could have rst eliminated y:..4 Cramer s rule x + 7y 0 j 6 4x 6y 8 j 7 0x + 4y 00 8x 4y 6 j + 8x 74 x This is a little preview on linear algebra. In linear algebra the system of equations is written as 7 x y 8 x 0 where and are so-called columns vectors, simply two numbers written x 8 7 below each other and surrounded by square brackets. is the -matrix 4 6 of coe cients. A -matrix is a system of four numbers arranged in a square and surrounded by square brackets. With each -matrix a b A c d we associate a new number called the determinant det A a b c d ad Notice that the determinant is indicated by vertical lines in contrast to square brackets for a matrix. For instance, ( 6) 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 cb or a b c d x y e f : 7

8 Namely, x e f a c b d b d ed bf ad bc y a c a c e f b d af 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 matrix 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 column is replaced by the column 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 column is replaced by the column vector with the constant terms. The method works only if the determinant in the denominator is not zero. In our example, 0 7 x y ( 6) ( 8) 7 ( 6) 4 7 ( 8) 4 0 ( 6) Exercise 4 Use the above methods to nd the optimum in Example 8.. 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 b y f d 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 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. 8 a b x c d x

9 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 x + y x + 4y 4 has no solution: The two lines are parallel y y x x 0 x 4 There is no common solution. Trying to nd one yields a contradiction x y x j + x. One equation non-linear, one linear Consider, for instance, y + x 0 y + 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 9

10 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: x y x y y + ( y) y y + (y ) 0 So the unique solution is y and x 0.. Two non-linear equations There is no general method. One needs to memorize some tricks which work in special cases. In the example of pro t-maximization with a Cobb-Douglas production function we were led to the system 6 K 6 L K 6 L : 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 6 K 6 L 6 K 6 K K p K p K K 8 6 K K 6 6 K Thus the solution to the rst order conditions (and, in fact, the optimum) is K L 8. Remark 6 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 sum to less than. Remark 7 The rst order conditions (6) and (7) form a hiddenlinear system of equations. Namely, if you take the logarithms (introduced later!) you get the system ln 6 6 ln K + ln L ln ln + ln K ln L ln which is linear in ln K and ln L. One can solve this system for ln K and ln L, which then gives us the values for L and K. 0

11 4 Youngs theorem Example: z f (x; y) x + x y + y 4 This is a function in two variables. When we calculate the two partial derivatives we obtain two new functions in x 6x y + 4y Each of these two functions can again be partially di erentiated in the two directions. So there are four second partial x + 6xy 6x + x + 6xy 6x y + 4y 6x y + 4y 6x + y We observe that out of these four new functions two are the same, namely the cross z z. Thus it does not matter whether we rst take with respect to x and then with respect to y or vice versa. Youngs theorem states the two cross z z are always the same, provided some are satis ed.. The four second derivatives are conveniently arranged in the form of a -matrix called the Hessian matrix. In our example: " 6x + 6y xy xy 6x + z Second order conditions The second derivatives must exist and must be continuous functions.

12 A critical point of a function z f (x; y) can be a peak, a trough or a saddle point: 4 y x z 0 z x y peak y 0 x 0 z 00 z x y saddle z 0 4 x y z x + y trough As in the case of a function in one variable, the second derivatives can be used to decide which of the three cases occur. However, the criterion that separates a saddle points from peaks and troughs may come as a surprise. Theorem 8 Suppose that the function z f (x; y) has a critical point at (x ; y ). If the determinant of the is negative at (x ; y ) then (x ; y ) is a saddle point.

13 If this determinant is positive then at (x ; y ) then (x ; y ) is a peak or a trough. In this case the signs z z are the same. If both signs are positive, j(x ;y j(x ;y ) (x ; y ) is a trough. If both signs are negative, then (x ; y ) is a peak. Nothing can be said if the determinant is zero. 8 < < 0: saddle point det z > 0: trough : z < 0: Remark 9 z is positive z is negative or vice versa then the automatically negative and hence one has a saddle point. This is so z is negative and therefore Remark 0 When the determinant is zero more complicated saddle points are possible. For instance, z yx y has a critical point at (0; 0) where all second derivatives are zero. Its graph is called a monkey saddle. Example z f (x; y) x y. The partial x Clearly, (0; 0) is the only critical point. The determinant of the Hessian det 0 0 () ( ) < z Hence the function has a saddle point at (0; 0). Example z 0 x + 0yx 00 0 y has the two partial x + 0x y. It is not di cult to show that (0; 0) is the only critical point. 00 determinant of the Hessian det < z y

14 Hence the function has a saddle point at (0; 0) and not a minimum, although z are negative. The graph of the function and the level curves are as x0 0 yz y 0 0 x 4 Exercise Show that the critical point in the example on page is a peak. Example 4 We have shown in the handouts of the previous weeks that the prot function (K; L) K 6 L K L has a critical point at K L 8. Is it a relative maximum? We have and K For K L 8 we obtain (since 8 8 ( p 8) 64 ) det H j K8 L Thus the determinant is j K8 L8 is 6K 6 L 6 L K 6 L K 6 L K 6 L p > 0 6 Absolute maxima or minima 8 4 > 0 we can conclude that K L 8 The role of intervals for functions in one variable is taken over by convex sets when considering functions in two variables. A region of the (x; y)-plane is called convex if it contains with every two points (x ; y ) and (x ; y ) in the region the line segment 4

15 connecting these two points. (Notice: There are concave and convex functions and there are concave and convex lenses, but there is no such thing as a concave set.) two convex sets two non-convex sets The (x; y)-plane and the the set of all non-negative coordinate pairs (x; y) are also convex sets. An absolute maximum of a function f (x; y) in a region is a point (x ; y ) in the region such that f (x ; y ) f (x; y) holds with respect to all other points in the region. Theorem Suppose the function f (x; y) is dened on a convex set and has there a unique critical point (x ; y ). If the determinant of the Hessian at this point is positive and f is negative (positive) then (x ; y ) is an absolute maximum (minimum) on convex set. The conditions are satis ed in many examples we have discussed, for instance in the pro t-maximization problem above. Remark 6 Suppose the function f (x; y) is dened on a convex set. Suppose the determinant of the Hessian is always positive in the region and f is always (respectively, negative). Then the function is convex (respectively, negative). Hereby the function is convex (concave) if the area above (below) the graph of the function is convex. 7 Quadratic forms nx nx nx z (~x) b ij x i x j b ii x i + i j i in matrix form: B is a symmetric matrix (B B 0 ) nx nx b ij x i x j i i<j z (~x) ~x 0 B~x Quadratic function: nx nx ~z (~x) b ij x i x j + i j ~x 0 B~x + A~x + c nx a i x i + c i translation

16 positive/negative (semi-) de nite ~x 0 B~x > 0 for all ~x R n, ~x 6 0 ~x 0 B~x 0 for all ~x R n, ~x 6 0 ~x 0 B~x < 0 for all ~x R n, ~x 6 0 ~x 0 B~x 0 for all ~x R n, ~x 6 0 inde nite quadratic form ~x 0 B~x < 0 for some ~x R n and ~y 0 B~y > 0 for some ~x R n Theorem 7 (Sylvesters inertia law) There exists a linear change of coordinates x i nx g ij y j j that transforms the quadratic form into Xn ^z (~y) i y i Xn in + y i with n ; n 0 and n + n n: n, n are invariants, i.e. do not depend on the specic coordinate change. Notice: P 0 i y i 0: In matrix notation ~y 0 ~ B~y (G~x) 0 B (G~x) with G invertible and B a diagonal matrix with +; meaning of positive de nite n assume a 6 0 First coordinate change z ax + bx x + cx a x + ba x x + cx a x + b a x x + b a a x + b a x + c! b a! b x a x ; 0 on the diagonal. x + cx u x + b a x u x 6

17 brings quadratic form into au + c! b u a Second coordinate change y p a u and (if necessary) y r c quadratic form into desired form. Notice: det ^B (det G) det B k k minors, k k principle minors, k k not-so principle minors. ( b a) u brings Theorem 8 The quadratic form B is negative denite if and only if the sign of the determinant of the principle k k minors is ( ) k. Theorem 9 z f (~x) twice di erentiable on a convex set. Assume the Hessian matrix is negative denite excepts for a set of measure zero. Then f is strictly concave. Theorem 0 Let h (u) be an increasing function and g (x) concave on a concave domain. Then the composite function h (g (x)) is quasi-concave. 7