ECON 301 Two Variable Optimization (with- and without- constraints) Review of Some Derivative Rules 1. Partial Derivative Rules: U = xy ∂U/∂x = Ux = y ∂U/∂y = Uy = x a b a 1 b a b 1 U = x y ∂U/∂x = Ux = ax − y ∂U/∂y = Uy = bx y − a b xa a 1 b a b 1 U = x y− = b ∂U/∂x = Ux = ax − y− ∂U/∂y = Uy = bx y− − y − U = ax + by ∂U/∂x = Ux = a ∂U/∂y = Uy = b 1/2 1/2 1 1/2 1 1/2 U = ax + by ∂U/∂x = Ux = a 2 x− ∂U/∂y = Uy = b 2 y− 2. Logarithm (Natural log) ln x ¡ ¢ ¡ ¢ (a) Rules of natural log If Then y = AB ln y =ln(AB)=lnA +lnB y = A/B ln y =lnA ln B y = Ab ln y =ln(Ab−)=b ln A NOTE: ln(A + B) =lnA +lnB 6 (b) derivatives IF THEN dy 1 y =lnx dx = x dy 1 y =ln(f(x)) = f 0(x) dx f(x) · (c) Examples If Then 2 1 y =ln(x 2x) dy/dx = (x2 2x) (2x 2) 1/−2 1 1 − 1 −1 y =ln(x )= 2 ln xdy/dx= 2 x = 2x ¡ ¢¡ ¢ 3. The Number e dy if y = ex then = ex dx f(x) dy f(x) if y = e then = e f 0(x) dx · (a) Examples 3x dy 3x y = e dx = e (3) 7x3 dy 7x3 2 y = e dx = e (21x ) rt dy rt y = e dt = re 1 Using Calculus For Maximization Problems OneVariableCase If we have the following function y =10x x2 − we have an example of a dome shaped function. To find the maximum of the dome, we simply need to findthepointwheretheslopeofthedomeiszero,or dy dx =10 2x =0 10 =− 2x x =5 and y =25 Two Variable Case Suppose we want to maximize the following function z = f(x, y)=10x +10y + xy x2 y2 − − Notethattherearetwounknownsthatmustbesolvedfor:x and y. This function is an example of a three-dimensional dome. (i.e. the roof of BC Place) To solve this maximization problem we use partial derivatives. We take a partial derivative for each of the unknown choice variables and set them equal to zero ∂z ∂x = fx =10+y 2x =0 The slope in the ”x” direction = 0 ∂z − = fy =10+x 2y =0 The slope in the ”y” direction = 0 ∂y − This gives us a set of equations, one equation for each of the unknown variables. When you have the same number of independent equations as unknowns, you can solve for each of the unknowns. rewrite each equation as y =2x 10 − x =2y 10 − substitute one into the other x =2(2x 10) 10 − − x =4x 30 − 3x =30 x =10 similarly, y =10 2 REMEMBER: To maximize (minimize) a function of many variables you use the technique of partial differentiation. This produces a set of equations, one equation for each of the unknowns. You then solve the set of equations simulaneously to derive solutions for each of the unknowns. Second order Conditions (second derivative Test) To test for a maximum or minimum we need to check the second partial derivatives. Since we have two first partial derivative equations (fx,fy) and two variable in each equation, we will get four second partials ( fxx,fyy,fxy,fyx) Using our original first order equations and taking the partial derivatives for each of them (a second time) yields: fx =10+y 2x =0 fy =10+x 2y =0 − − fxx = 2 fyy = 2 − − fxy =1 fyx =1 The two partials,fxx,andfyy are the direct effects of of a small change in x and y on the respective slopes in in the x and y direction. The partials, fxy and fyx are the indirect effects, or the cross effects of one variable on the slope in the other variable’s direction. For both Maximums and Minimums, the direct effects must outweigh the cross effects Rules for two variable Maximums and Minimums1 1. Maximum fxx < 0 fyy < 0 fyyfxx fxyfyx > 0 − 2. Minimum fxx > 0 fyy > 0 fyyfxx fxyfyx > 0 − 3. Otherwise, we have a Saddle Point From our second order conditions, above, fxx = 2 < 0 fyy = 2 < 0 − − fxy =1 fyx =1 and fyyfxx fxyfyx =( 2)( 2) (1)(1) = 3 > 0 − − − − therefore we have a maximum. 1 Advanced Topic: This section is optional for ECON 301. For most applications, the structure of the problem will make it clear that this is a max or min problem 3 Example: Profit Max Capital and Labour Suppose we have the following production function q = Output 1 1 q = f(K, L)=L 2 + K 2 L = Labour K = Capital Then the profit function for a competitive firm is π = Pq wL rK P = Market Price or− − w = Wage Rate 1 1 π = PL2 + PK2 wL rK r = Rental Rate − − First order conditions General Form 1 ∂π P −2 1. ∂L = 2 L w =0 PfL w =0 1 ∂π P − − − 2. = K 2 r =0 PfK r =0 ∂k 2 − − Solving (1) and (2), we get 2w 2 2r 2 L∗ =(P )− K∗ =(P )− Example: If P =1000, w =20,andr =10 1. Find the optimal K, L,andπ 2. Check second order conditions Example: Cobb-Douglas production function and a com- petitive firm Consider a competitive firm with the following profitfunction π = TR TC = PQ wL rK (1) − − − where P is price, Q is output, L is labour and K is capital, and w and r are the input prices for L and K respectively. Since the firm operates in a competitive market, the exogenous variables are P,w and r. There are three endogenous variables, K, L and Q. However output, Q, is in turn a function of K and L via the production function Q = f(K, L) which in this case, is the Cobb-Douglas function Q = LaKb (2) 4 where a and b are positive parameters. If we further assume decreasing returns to scale, 1 then a + b < 1. For simplicity, let’s consider the symmetric case where a = b = 4 1 1 Q = L 4 K 4 (3) Substituting Equation 3 into Equation 1 gives us 1 1 π(K, L)=PL4 K 4 wL rK (4) − − The first order conditions are 3 1 ∂π 1 4 4 ∂L = P 4 L− K w =0 ∂π 1 1 3 − (5) = P L 4 K− 4 r =0 ∂K ¡ 4¢ − This system of equations define the optimal¡ ¢ L and K for profit maximization. Rewriting the firstequationinEquation5toisolateK 3 1 1 4 4 P 4 L− K = w 4w 3 4 K =( L 4 ) ¡ ¢ p Substituting into the second equation of Equation 5 3 4 1 3 1 3 4 − P 4 4 P 4 4w 4 4 L K− = 4 L p L = r 4 1 4 3 2 ∙³ ´ ¸ = P 4 w−¡ L¢− = r Re-arranging to get L by itself¡ ¢ gives us P 3 1 2 L =( w 4 r 4 ) ∗ 4 − − Taking advantage of the symmetry of the model, we can quickly find the optimal K P 3 1 2 K =( r 4 w 4 ) ∗ 4 − − L∗ and K∗ are the firm’s factor demand equations. Optimization with Constraints The Lagrange Multiplier Method Sometimes we need to to maximize (minimize) a function that is subject to some sort of constraint. For example Maximize z = f(x, y) subject to the constraint x + y 100 ≤ Forthiskindofproblemthereisatechnique,ortrick, developed for this kind of problem known as the Lagrange Multiplier method. This method involves adding an extra variable to the problem called the lagrange multiplier, or λ. We then set up the problem as follows: 5 1. Create a new equation form the original information L = f(x, y)+λ(100 x y) or − − L = f(x, y)+λ [Zero] 2. Then follow the same steps as used in a regular maximization problem ∂L ∂x = fx λ =0 ∂L − = fy λ =0 ∂y − ∂L =100 x y =0 ∂λ − − 3. In most cases the λ will drop out with substitution. Solving these 3 equations will give you the constrained maximum solution Example 1: Suppose z = f(x, y)=xy. and the constraint is the one from above. The problem then becomes L = xy + λ(100 x y) − − Now take partial derivatives, one for each unknown, including λ ∂L ∂x = y λ =0 ∂L = x − λ =0 ∂y − ∂L =100 x y =0 ∂λ − − Starting with the first two equations, we see that x = y and λ drops out. From the third equation we can easily find that x = y =50and the constrained maximum value for z is z = xy =2500. Example 2: Maximize u =4x2 +3xy +6y2 subject to x + y =56 Set up the Lagrangian Equation: L =4x2 +3xy +6y2 + λ(56 x y) − − Take the first-order partials and set them to zero Lx =8x +3y λ =0 − Ly =3x +12y λ =0 − Lλ =56 x y =0 − − 6 From the first two equations we get 8x +3y =3x +12y x =1.8y Substitute this result into the third equation 56 1.8y y =0 − − y =20 therefore x =36 λ =348 Example 3: Cost minimization A firm produces two goods, x and y. Due to a government quota, the firm must produce subject to the constraint x + y =42.Thefirm’s cost functions is c(x, y)=8x2 xy +12y2 − The Lagrangian is L =8x2 xy +12y2 + λ(42 x y) − − − The first order conditions are Lx =16x y λ =0 − − Ly = x +24y λ =0 − − Lλ =42 x y =0 (6) − − Solving these three equations simultaneously yields x =25 y =17 λ =383 Example of duality for the consumer choice problem Example 4: Utility Maximization Consider a consumer with the utility function U = xy, who faces a budget constraint of B = Pxx + Pyy, where B, Px and Py are the budget and prices, which are given.