ECON0702: Mathematical Methods in Economics Yulei Luo SEF of HKU February 15, 2009 Luo, Y. (SEF of HKU) MME February15,2009 1/81 Constrained Static Optimization So far we have focused on …nding the maximum or minimum value of a function without restricting the choice variables. In many economic problems, the choice variables must be constrained by economic considerations. E.g., consumers maximizes their utility functions with the budget constraints. A …rm minimizes the cost of production with the constraint of production technology. Such constraints may lower the maximum (or increase the minimum) for the objective function being maximized (or minimized). Because we are not able to choose freely among all choice variables, the objective function may not be as large as it can be. The constraints would be said to be nonbinding if we could obtain the same level of the objective function with or without imposing the constraint. Luo, Y. (SEF of HKU) MME February15,2009 2/81 Finding the Stationary Values For illustration, consider a consumer’schoice problem: maximize his utility as follows u (x1, x2) = x1x2 + 2x1, (1) subject to the budget constraint 4x1 + 2x2 = 60. (2) For this simple constrained problem, we can solve it by substituting the budget constraint into the objective utility function without using any new technique: The budget constraint can be rewritten as 60 4x1 x2 = 2 = 30 2x1,which combining with the utility function gives u (x1) = x1 (30 2x1) + 2x1. (3) By setting ∂u = 32 4x = 0, we get x = 8 and x = 14. Since ∂x1 1 1 2 d 2u dx 2 = 4 < 0, the stationary value constitutes a constrained 1 maximum. Luo, Y. (SEF of HKU) MME February15,2009 3/81 The Lagrange multiplier method However, when the constraint is itself a complicated function or when the constraint cannot be solved to express one variable as an explicit function of the other variables, the technique of substitution and elimination is not enough to solve the constrained optimization problem. We will introduce a new method: the Lagrange multiplier method to solve the constrained optimization problem. The essence of the LM method is to convert a constrained extremum problem into a form such that the FOC of the free extremum problem can still be applied. In general, given an objective function z = f (x, y) (4) subject to the constraint g (x, y) = c where c is a constant, (5) we can write the Lagrange function as follows Z = L (x, y, λ) = f (x, y) + λ [c g (x, y)] . (6) Luo, Y. (SEF of HKU) MME February15,2009 4/81 (conti.) The symbol, λ, is called a Lagrange multiplier and will be determined and discussed later. If c g (x, y) = 0 always holds, the last term of Z will vanish regardless of the value of λ. Hence, …nding the constrained maximum value of z is equivalent to …nding a critical value of Z. The question now is: How can we make the parenthetical expression in (6) vanish? Let’sproceed to do so, treating λ also as an additional choice variable (in addition to x and y). From the Lagrange function (6), the FOCs are ∂L = fx λgx = 0, (7) ∂x ∂L = fy λgy = 0, (8) ∂y ∂L = c g (x, y) = 0. (9) ∂λ The …nal equation will automatically guarantee the satisfaction of the constraint. Since λ [c g (x, y)] = 0, the stationary values of Z in (6) must be identical with those of (4), subject to (5). Luo, Y. (SEF of HKU) MME February15,2009 5/81 Reconsider the above consumer’schoice problem, …rst, we can write the Lagrange function as follows Z = L (x1, x2, λ) = x1x2 + 2x1 + λ [60 (4x1 + 2x2)] . (10) The FOCs are: ∂L = x2 + 2 4λ = 0, (11) ∂x1 ∂L = x1 2λ = 0, (12) ∂x2 ∂L = 60 (4x1 + 2x2) = 0, (13) ∂λ solving the above equation for the critical values gives x1 = 8, x2 = 14, and λ = 4. (14) Luo, Y. (SEF of HKU) MME February15,2009 6/81 Summary of the Procedure Step 1: Forming the Lagrange function Z = L(x, y, λ) = f (x, y) + λ[c g(x, y)]. (15) Step 2: Find the critical points of the Lagrangian function L(x, y, λ) by computing ∂L/∂x, ∂L/∂y, and ∂L/∂λ and setting each equal to 0 to solve for optimal (x , y , λ) : ∂L ∂L ∂L = 0, = 0, and = 0. ∂x y ∂λ Note that since λ just multiplies the constraint in the de…nition of L, the equation ∂L/∂λ = 0 is equivalent to the constraint c g(x, y) = 0. Note that by introducing the Lagrange multiplier λ into the constrained problem, we have transformed a two-variable constrained problem to the three-variable unconstrained problem of …nding the critical points of a function L(x, y, λ). Luo, Y. (SEF of HKU) MME February15,2009 7/81 Total Di¤erential Approach In the discussion of the free extremum of z = f (x, y), it is learned that the necessary FOC can be stated in terms of the total di¤erential dz : dz = fx dx + fy dy = 0. (16) This statement remains valid after adding the constraint g(x, y) = c. However, with the constraint, we can no longer take both dx and dy as arbitrary change as before because now dx and dy are dependent each other: g(x, y) = c = (dg =) gx dx + gy dy = 0. (17) ) Luo, Y. (SEF of HKU) MME February15,2009 8/81 (conti.) The FOC in terms of total di¤erential becomes dz = 0 subject to g(x, y) = c and gx dx + gy dy = 0. (18) In order to satisfy this necessary FOC, we must have f f x = y , (19) gx gy which together with the constraint g(x, y) = c will provide two equations to solve for the critical values of x and y. Hence, the total di¤erential approach yields the same FOC conditions as the Lagrange multiplier method. Note that the LM method gives the value of λ as a direct by-product. Luo, Y. (SEF of HKU) MME February15,2009 9/81 An Interpretation of the Lagrange Multiplier The Lagrange multiplier λ measures the sensitivity of Z (which is the value of Z in optimum) to the change in the constraint. In other words, it gives us a new measure of value of the scarce resources (The e¤ect of an increase in c would indicate how the optimal solution is a¤ected by a relaxation of the constraint). If we can express the optimal values of (x , y , λ) as implicit functions of c : x = x (c) ; y = y (c) ; λ = λ (c) and all of which have continuous derivatives. Further, we have the following identities: fx (x , y ) λgx (x , y ) = 0, (20) fy (x , y ) λgy (x , y ) = 0, (21) c g (x , y ) = 0. (22) Luo, Y. (SEF of HKU) MME February15,2009 10/81 (conti.) and the Lagrange function in optimum can be written as Z = L(x , y , λ) = f (x , y ) + λ[c g(x , y )], (23) which means that dZ dx dy dλ = fx + fy + [c g(x , y )] dc dc dc dc =0 dx dy +λ 1 gx + gy| {z } dc dc dx dy = fx λgx + fy λgy + λ 0 1 dc 0 1 dc =0 (FOC for x) B C B=0 (FOC for x)C @ A @ A = λ.| {z } | {z } Luo, Y. (SEF of HKU) MME February15,2009 11/81 Generalization: n-Variable and Multi-constraint Case The optimization problem can be formed as follows: max min z = f (x1, , xn), (24) x1, ,xn x1, ,xn f g f g subject to : g(x1, , xn) = c (25) It follows that the Lagrange function is Z = L (x1, , xn, λ) = f (x1, , xn) + λ [c g(x1, , xn)] , (26) for which the FOCs are fi (x1, , xn) = λgi (x1, , xn), i = 1, , n. g(x1, , xn) = c Luo, Y. (SEF of HKU) MME February15,2009 12/81 (conti.) If the objective function has more than one variable, say, two constraints g(x1, , xn) = c and h(x1, , xn) = d (27) The Lagrange function is Z = f (x1, , xn) + λ [c g(x1, , xn)] (28) +µ [d h(x1, , xn)] , for which the FOCs are fi (x1, , xn) = λgi (x1, , xn) + µhi (x1, , xn), i = 1, , n. g(x1, , xn) = c h(x1, , xn) = d. Luo, Y. (SEF of HKU) MME February15,2009 13/81 Second-Order Conditions Note that even though Z is indeed a standard type of extremum w.r.t. the choice variables, it is not so w.r.t. the Lagrange multiplier. (23) shows that unlike (x , y ), if λ is replaced by any other value of λ, no e¤ect will be produced on Z since c g(x , y ) = 0. Thus the role played by λ in the optimal solution is di¤ers basically from that of x and y . While it is safe to treat λ as another choice variable in the discussion of FOCs, we should treat λ di¤erently in the discussion of SOCs. The new SOCs can again be stated in terms of the SO total di¤erential d2z. The presence of the constraint will entail certain signi…cant modi…cations. Luo, Y. (SEF of HKU) MME February15,2009 14/81 Second-Order Total Di¤erential The constraint g(x, y) = c implies that dg = gx dx + gy dy = 0. dx and dy are no longer both arbitrary: we may take dx as an arbitrary change, but then dy is dependent on dx, i.e., dy = (gx /gy ) dx.