Optimization with Equality ConstraintsOne important feature in the previous chapter in finding relative extrema of objective functions with one or two more choice variables is that they are independent of one another. For instance, the firm can choose any value for their first and second product without the two choices limiting each other. In a situation of constrained optimum, the firm is required to observe a restriction (i.e. production quota). Such a restriction establishes a relationships between the two choice variables and thus the relative constrained extrema will be found.Effects of a ConstraintConsider a consumer with the simple utility function:\(U = x_1 x_2 + 2 x_1\). Here, the marginal utilities are the partial derivatives:\(U_1 \equiv \frac{\partial U}{\partial x_1}\)\(U_2 \equiv \frac{\partial U}{\partial x_2}\)Without any constraint, the consumer should obviously purchase an infinite amount of both goods to achieve maximum utility. We see why a budget constraint has to be incorporated into the problem, take: \(4x_1 + 2x_2 = 60\)Thus, the choices of the goods become mutually dependent. Mathematically, the constraint narrows the domain, and hence the range of the objective function. Generally, the number of constraints is less than the number of choice variables. Finding the Stationary ValuesIf the equations are simple, we can use substitution. If complex, we can use the Lagrangian. In general, the form will be:Given the objective function, \(z = f(x, y)\)\(\)Subject to the constraint, \(g(x, y) = c\)Write the Lagrangian as, \(Z = f(x,y) + \lambda[c - g(x,y)]\)For stationary values of Z, regarded as a function of the three variables, the necessary condition is, \(Z_{\lambda} = c - g(x, y) = 0\)\(Z_x = f_x - \lambda g_x = 0\)\(Z_y = f_y - \lambda g_y = 0\)What is the Lagrange Multiplier? It measures the sensitivity of Z* to changes in the constraint. Since the three variables are endogenous, the only exogenous variable is the constraint parameter c, which can be relaxed to show the optimal solution in light of a larger budget. -----------------Class Notes: Firm TheoryHomogenous Functions & Euler's TheoremA function is homogenous of degree r if\(f(x_1, x_2, x_3 ... x_n)\) and for any t,\(f(t \vec x) = t^r f(\vec x)\)\(\vec d = (x_1, x_2, x_3, ... x_n)\)Easy case where r = 1:\(f(t \vec x) = tf(\vec x)\)Example: \(f(x_1, x_2) = x_1^\alpha x_2 ^{1 - \alpha}\)\(f(tx_1, tx_2) = (tx_1)^\alpha (tx_2)^{1 - \alpha}\)\(= t'(x_1 ^\alpha x_2 ^{1 - \alpha})\)\(= tf(\vec x)\)Constant returns to scale production functions are homogenous of degree 1. Neutrality of money:Demand for goods will not depend on level of prices, only on relative prices, and if double the money leads to double the prices...\(Q_{x_1} = f(m, \vec p)\) (demand function that states demand for good x depends on money and prices of all other goods)Thus, if \(t Q_{x_1} = f(tm, t \vec p) \)\(\therefore t^0 Q_{x_1} = Q_{x_1}\)Real demand is unaffected by a purely nominal change. Money is homogenous to the degree 0. Consider \(f(x) = y = x_1 ^{1/3} x_2 ^{1/3}\)\(f(t \vec x) = (tx_1)^{1/3} (t x_2)^{1/3}\)\(= t^{2/3} ( x_1 ^{1/3} x_2 ^{1/3})\)\(= t^{2/3}(f(\vec x))\)y is homogeneous to the degree 2/3. Fixed cost:\(y = f(\vec x) = x_1 ^\alpha x_2 ^\beta - K\), where K is fixed costsTry this:\(f(K, L) = y = z_1 K + z_2 L\)Homogenous of degree one since you can multiply the factors by t and then just factor it out to see that output will rise by the same amount. Euler's Theorem:Claim: for a function homogenous of degree r,\(f_1 x_1 + f_2 x_2 + ...f_n x_n = r f(\vec x)\)or \(\vec f_i \cdot \vec x_i = rf\) (dot product in matrices)Definition of homogeneity: \(f(t \vec x) = t^r f(\vec x)\)Differentiate with respect to t:\(\frac{df}{d(tx_1} \cdot \frac{d(tx_1)}{dt} + \frac{df}{d(tx_2)} \cdot \frac{d)tx_2)}{dt} ... = rt^{r - 1} f(\vec x)\)\(f_1 x_1 + f_2 x_2 + ...f_n x_n = r f(\vec x)\)Q.E.D.What do economists use frequently that's homogeneous to the degree 1? Production functions (Cobb-Douglas)Classic Cobb-Douglas with price theory\(y = zK^\alpha L^{1 - \alpha}\)\(\pi = y - wL - rK\)\(\pi = zK^\alpha L^{1 - \alpha} - wL - rK\)\(\frac{d \pi}{dL} = (1 - \alpha) z K^\alpha L^{- \alpha} - w = 0\)\(w = (1 - \alpha) z K^\alpha L^{- \alpha}\)\(\frac{dy}{dL} L + \frac{dy}{dK} K = y\)If labor and capital are paid their marginal products, there is no economic profit left over. Wicksell called this the "Production Exhaustion Theorem." Thus, if price theory is true and firms have CRS, capitalism actually serves the consumer. How can we actually get profits?Increasing returns to scale? \(y = zK^{2/3} L^{2/3}, r = 4/3\)\(MPK = \frac{dK}{dy} = 2/3 z K ^{-1/3} L^{2/3}\)Euler's: \(MPK \cdot K + MPL \cdot L = 4/3 y > y\)Not enough output to pay "price theory" or marginal product wages. Big problem. Reason Paul Romer will win a Nobel one day:Tension between Adam Smith's idea of division of labor (increasing returns) with ideas of efficiency (needs competition). Don't get innovation with invisible hand and perfect competition. How to have increasing returns to scale.Still, how to get monopoly profits? With decreasing returns to scale?\(y = K^\alpha L^\beta, \alpha + \beta < 1\)\(\vec {MP_i} \cdot \vec x_i = (\alpha + \beta) y\)Thus, the leftovers of decreasing returns to scale will go to the business owners. Maxima (and Minima) of 2 Variable FunctionsHow to be sure you're at the top of the mountain:Look all around in every direction, and if all of the immediate steps are lower than your current location, you're at the top. And now to prove this mathematically...To max \(f(x)\):\(f' = 0\)\(f'' < 0\)It's harder to discern that the second derivative will be negative with more variables than one (too many axies). Real trouble is with the interaction term (both production with interplay between capital and labor and consumer side):\(= f_{xX} dx^2 + 2f_{xy} dx dy + f_{yy} dy^2\)Thus, need to check that the middle term is not too positive.For well-behaved f, if \(\frac{df}{dx_i} = 0\) for both \(i = 1, 2\), a necessary condition for a max is:\(f_{11} \cdot f_{22} > f_{12} ^2\) & \(f_{11} < 0, f_{22} <0 \)In other words, the 2nd derivative must be bigger than the cross-partial, in a specific way. Example: \(y = ax_1 - \frac{b}{2} x_1 ^2 + c x_2 - \frac{d}{2} x_2 ^2\)\(a, b, c, d > 0\)\(\frac{dy}{dx_1} = a - bx_1 = 0\) F.O.C. :\(x_1 * = \frac{a}{b} ; x_2 * = \frac{c}{d}\)Now to check 2nd derivatives:\(- b \cdot (- d) > 0\)Both b and d are positive so yes, we're at a max! This was easy because there's no interaction between x1 and x2. Now to apply this to Cobb-Douglas:\(\pi = zK^\alpha L^\beta - wL - rK\)S.O.C. (second order conditions):\(\alpha (\alpha - 1) zK ^{\alpha -2} L^\beta = \pi _{KK}\)\(\beta (\beta - 1) zK ^{\alpha} L^{\beta -2} = \pi _{LL}\)What is the marginal profitability of capital if I get one more worker? Take the derivative of both:\(\alpha \beta K^{\alpha - 1} L ^{\beta - 1} = \pi _{KL}\)Now, to check that the second derivatives of capital and labor are negative, looks like the returns to capital (alpha) and labor (beta) have to be less than one:\(\alpha \beta (\alpha - 1) (\beta - 1) z^2 K^{2\alpha - 2} L ^{2\beta - 2} > \alpha ^2 \beta ^2 z^2 K ^{2\alpha - 2} L ^{2\beta - 2}\)\((\alpha - 1) (\beta - 1) > \alpha \beta\)\(\alpha \beta - \alpha - \beta + 1 > \alpha \beta\)Thus, we need \(\alpha + \beta < 1\) to be sure of a max. We NEED diminishing returns to scale (they've grown as much as they can). Thus, a constant-returns to scale firm and a price-taker, Euler's theorem will tell you that you'll earn no profits, and you don't need a determinate size (always zero firms even if you double in size). Monopoly Profit Maximization\(\pi = PQ - cQ\), where c is constant marginal costSince it is a monopoly, the price is a function of quantity: \(\pi = P(Q) Q - cQ\)\(\frac{d \pi}{dQ} = 0 = P'Q + P - c \Longrightarrow P'Q + P = c\)Thus, marginal revenue equals marginal cost. Every time the monopolist produces more stuff, he pushes down the price. Recall price elasticity of demand:\(\epsilon _D = \frac{dQ}{dP} \cdot \frac{P}{Q}\)Now multiply this by the price:\(P(\frac{dP}{dQ} \cdot \frac{Q}{P}) = c\)\(\frac{P}{c} = \frac{1}{\frac{1}{\epsilon _D} + 1} \Longrightarrow \frac{\epsilon}{1 + \epsilon}\)Where \(\epsilon < 0 \), because it's demand.Consider \(\epsilon = \frac{-1}{2}\) (i.e. relatively inelastic):\(Q = zP^{-1/2}\)So, what should your mark-up be?\(\frac{-1/2}{1 + -1/2} = -1\)What? What's up with that? Because this equation will get you the wrong answer if elasticity of demand is inelastic.