INTERMEDIATE MICROECONOMICS MATH REVIEW

INTERMEDIATE MICROECONOMICS MATH REVIEW August 31, 2008 OUTLINE 1. Functions Definition Inverse functions Convex and Concave functions 2. Derivative of Functions of One variable Definition Rules for finding a derivative Some applications to economics 3. Derivative of Multi-variable Functions Definition Partial Derivatives Total differentiation Some applications to economics 4. (Unconstrained) Optimization - Maximization and Minimization Problems First order conditions (FOC)- necessary 1

Second order conditions(soc)- sufficient Some applications to economics 5. Optimization with Linear Constraints Substitution Method Lagrange Multiplier Method Applications to economics 2

1 Functions Definition 1 A function f is a mapping from some set A (called the domain of the function) to some set B (called the range of the function) which satisfies a condition that to every element in A, f assigns a unique element from the set B. In short we write: f : A B 1.1 Inverse functions We will see inverse functions when we talk about supply and demand functions. We can express price as a function of quantity and quantity as a function of price. Here is a formal definition. Definition 2 Suppose that we are given a function f : X Y. We say that f has an inverse if there is a function g such that the domain of g is the range of f and such that f(x) = y if and only if g(y) = x for every x X and for every y Y. Example Demand function specifies what the quantity demanded will be for a given price, e.g. Q d (p) = 100 5p. Inverse demand function tells us what the price should be at a given quantity, e.g. P d (q) = 20 1 5 q. NOTE: Make sure you know how to find the inverse of a given function! 1.2 Concave and convex functions This section will be a about real-valued functions. (i.e. the range is the set real numbers) Definition 3 Assume we have a function f : A R, where A is a convex set. We say that the function f is convex if for all a 1 A, for all a 2 A and for all λ (0, 1) we have: f(λa 1 + (1 λ)a 2 ) λf(a 1 ) + (1 λ)f(a 2 ) 3

We say that the function f is concave if for all a 1 A, for all a 2 A and for all λ (0, 1) we have: Examples f(λa 1 + (1 λ)a 2 ) λf(a 1 ) + (1 λ)f(a 2 ) 1. Function f(x) = x 2 is convex. To see this take for example x 1 = 0, x 2 = 2 and λ = 1. Then check that the inequality holds. (NOTE: If you want to 4 completely prove that this function is convex, you need to show that the inequality holds for any x 1, x 2,and λ) 2. Function f(x) = x is concave. Try the same numbers as above. You can also try different numbers. 2 Derivative of Functions of One variable Let f : R R be a real-valued function. Definition 4 The derivative of f at point x is defined as f (x ) lim h 0 f(x + h) f(x) h Definition 5 f is said to be differentiable at x if the above limit exists. Definition 6 f is said to be differentiable if it is differentiable at every point in its domain. Intuitively, derivative of a real-valued function of one variable shows how much the value of the function changes with respect to a very very small change in its argument (x) at that point. Geometrically, derivative of a function at a point is equal to the slope of the tangent line passing through that point. Sign of the derivative gives information about the behavior of the function. If f > 0, then f is strictly increasing; similarly, if f < 0, then f is strictly decreasing. Second derivative of f, denoted by f, is the derivative of f. 4

2.1 Rules for finding derivatives 1. If k R, i.e. it is a constant, then 2. Multiplication by a constant k, 3. If k is a constant, then 4. 5. d(lnx) = 1 x d[kf(x)] dk = 0 = kf (x) k = k.xk 1 d(k x ) = k x.lnk for any constant k R. In particular, d(ex ) = ex.lne = e x. 6. Derivative of sum-difference d[f(x) ± g(x)] = f (x) ± g (x) 7. Derivative of multiplication of two functions (Product Rule) 8. Quotient Rule d[f(x).g(x)] d[f(x)/g(x)] = f (x).g(x) + g (x).f(x) = f (x).g(x) f(x).g (x) [g(x)] 2 9. Chain rule d[f(g(x))] = f (g(x)).g (x) or you can denote chain rule as, if y = f(x) and x = g(z) dy dz = dy. dz 5

2.2 Second derivative Definition 7 Second derivative of a function f at x is defined as: f (x ) lim h 0 f (x + h) f (x ) h NOTE: second derivative is simply a derivative of the derivative. 2.2.1 Second derivative and concave/convex functions Second derivative can be used to determine whether a real-valued function of one variable is concave or convex. Theorem 1 Suppose that f is a real-valued function of one variable and suppose that f exists. Then f is convex if and only if f 0. Theorem 2 Suppose that f is a real-valued function of one variable and suppose that f exists. Then f is concave if and only if f 0. 2.3 Application to economics In Econ1101, we defined marginal utility (MU) of a consumer as the increase in utility by consuming an additional unit of a good; marginal cost (MC) of a firm as the change in total cost if output increases by one unit; and marginal profit (MP) of a firm as the change in profit if output increases by one unit. Now, we are after how much the utility, cost or profit changes if we increase the amount consumed or produced by a very very small amount. So, to find marginal changes, we ll find the derivatives of utility, cost or profit functions. 6

3 Derivative of Multi-variable Functions Economic problems usually involve functions of more than one variable. For example, a consumer s utility generally depends on more than one type of good. Similarly, cost function of a firm depends on more than one type of input, i.e. cost of labor, cost of physical capital and cost of land. Problems with multivariables are of interest since trade-offs must be made among these variables. The dependence of one variable y on a series of other variables (x 1, x 2,..., x i,..., x n ) is denoted by y = f(x 1, x 2,..., x i,..., x n ). 3.1 Partial derivatives Definition 8 Let y = f(x 1, x 2,..., x i,..., x n ). A partial derivative of f with respect to x i at ˆx = (ˆx 1, ˆx 2,..., ˆx i,..., ˆx n ) 1 is: f(x) f(ˆx 1, ˆx 2,..., ˆx i + h,..., ˆx n ) f(ˆx 1, ˆx 2,..., ˆx i,..., ˆx n ) x i lim h 0 x=ˆx h NOTE: equivalent notations are y x i or f i (x). It is easy to calculate partial derivatives. If you take partial derivative with respect to x i, you have to treat other arguments as constants. The intuition is that we keep the values of all other variables fixed. Example 1. Suppose C(l, k) = 4l 2 + 3k 2. C l = 8l C k = 6k 2. Suppose U(x, y) = 2lnx + lny. U x = 2 x U y = 3 y 1 The boldface letters denote vectors. 7

3.2 Total differentiation If all the x s are varied by a small amount, the total effect on y will be the sum of effects. In other words, total change in y can be decomposed into changes resulting from changes of x 1, x 2,..., x i,..., x n. Definition 9 Let y = f(x) = f(x 1, x 2,..., x i,..., x n ). A total differential of f is defined as: Examples df(x) = f(x) x 1 1 + f(x) x 2 2 +... + f(x) n x n 1. Suppose that, we have a daily income function of an individual given by: I(h, w) = h w. It s a function of two variables: hours worked (h) and hourly wage (w). We know that: I h Hence the total differential will be: = w and I w = h di = I I dh + dw = w dh + h dw h w 2. This example will use the chain rule. Suppose that, firm s profit is a function of output and advertising. Let q denote output and a denote advertising (it can be for example the amount of $ spent on it). The profit function is: π(q, a) = 10q q 2 + 2qa + 10a a 2. We also know that firm s output is a function of world price p, i.e. q(p) = 50 + 2p. Initially q = 5, a = 5. Calculate the change in profit if the world price falls by 0.1 and firm decides to increase the money spent on advertising by 0.1. Answer: Compute the total differential of the profit function. It is given by: dπ = π π π q π dq + da = dp + q a q p a da = = (10 2q + 2a)2dp + (2q + 10 2a)da Since dp = 0.1 and da = 0.1 we get: dπ = (10 2 5+2 5) 2 ( 0.1)+(2 5+10 2 5) 0.1 = 20 0.1+10 0.1 = 1 8

3.3 Application to economics Again, partial derivatives are used to analyze the trade-off among different variables. We ll see the concepts such as marginal rate of substitution among two different goods; marginal rate of technical substitution among factors of production. It is used to calculate elasticity. For instance, price elasticity of demand and supply; or income elasticity of demand. 4 (Unconstrained) Optimization We are interested in problems such as consumer maximizing utility, consumer minimizing expenditure, firm maximizing profit or firm minimizing cost. We can find solutions to maximization and minimization problems by applying simple calculus techniques. Definition 10 Function f has a local maximum at x if there exists some open interval I that contains x and f(x ) f(x) for all x I. Definition 11 Function f has a global maximum at x if f(x ) f(x) for all x in the domain of f. Definition 12 If a function f has a local (global) maximum at x then x is called a local (global) maximizer of f. Next theorem says that if a differentiable function has a local maximum/minimum at a point, then the derivative at this point has so be zero. Theorem 3 First Order Necessary Condition for maximum/minimum Let f be a real-valued function of one variable. Suppose that f has a local maximum/minimum at some point x and suppose that f (x ) exists. Then f (x ) = 0. Remark: This theorem does not say that if f (x ) = 0 then f has a maximum/minimum at the point. In particular, we are not able to distinguish between maximum and minimum. It can also happen that neither one occurs. The next two theorems help us determine whether the function attains a maximum or a minimum. 9

Theorem 4 Sufficient Condition for maximum (SOC) Let f be a real-valued function of one variable. Suppose that f (x ) exists and suppose that f (x ) = 0 and f (x ) < 0. Then f has a local maximum at x. Theorem 5 Sufficient Condition for minimum(soc) Let f be a real-valued function of one variable. Suppose that f (x ) exists and suppose that f (x ) = 0 and f (x ) > 0. Then f has a local minimum at x. Examples 1. Suppose that the relationship between profits π and quantity produced q is given by π(q) = 1000q 5q 2. We can verify that q = 100 maximizes profits. FOC (necessary) dπ dq = 1000 10q = 0 implies that at q = 100 profit function attains either a maximum or a minimum. We need to see q = 100 is indeed a local maximum by checking SOC, i.e. π = 10 < 0. 2. Consider a function f(x) = 16x 2x 2. We can verify that the function has a local maximum at a point x = 4. We have that f (4) = 16 4x x=4 = 16 4 4 = 0. FOC is satisfied. We can calculate f (4) = 4 < 0, hence the sufficient condition is satisfied. Function f has a local maximum at 4. 3. This example will highlight the difference between a local and a global maximum/minimum. Suppose that we have a function f : R + R, i.e. f is a real-valued function defined on the set of non-negative real numbers. Assume f(x) = 1 3 x3 3 2 x2 + 2x + 10. We want to find local and global maximuma/minima. FOC for maximum/minimum is: f (x) = x 2 3x+2 = 0, which can be solved for x to give x = 1 or x = 2. Now we notice that f (x) = 2x 3, hence f (1) = 1 < 0 and f (2) = 1 > 0, so there is a local maximum at x = 1 and a local minimum at x = 2. We can evaluate the function at x = 1 and at x = 2: f(1) = 10 5 and f(2) = 10 4. However, 6 6 f(0) = 10 < 10 4 = f(2), so global minimum is not at x = 2 (it s actually 6 at x=0). We also have that lim x f(x) =, so the function doesn t have global maximum. 10

4.1 Unconstrained optimization with more than one variable Now assume that we have a real-valued function y = f(x 1,..., x n ), where all x s R. Theorem 6 First Order Necessary Condition for maximum/minimum Suppose that f has a local maximum/minimum at some point x = (x 1,..., x n). Suppose that f(x ) x 1,... f(x ) x n exist. Then f(x ) x 1 = f(x ) x 2 =... = f(x ) x n = 0. Intuitively, if one of the partials were greater or less than 0, then y could be increased by increasing or decreasing x i. 5 Constrained Optimization Consider the following problem: There is a consumer who gets utility by consuming pizza and beer. Suppose this consumer has 24 dollars to spend on these goods. The price of one slice of pizza is 2 dollars, the price of one beer is 3 dollar. The utility function of the consumer is given by: U(p, b) = 2ln(p) + ln(b) This consumer s problem can be written as: max U(p, b) = 2ln(p) + ln(b) x,y subject to 2p + 3b = 24 (BudgetConstraint) U is called the objective function, p and b are called choice variables. There are two ways to solve this problem. 5.1 Substitution method We can rewrite the budget constraint as b = 8 2 p and plug this into objective 3 function. Then, the consumer s problem becomes simply maximizing U(p) = 2ln(p) + ln(8 2 3 p) 11

FOC (necessary) implies du dp = 2 p 2/3 (8 2 3 p) = 0 We can verify p = 8 satisfies the above equation. We have to check second derivative is negative to show that this is indeed a maximum. (It is left as an exercise.) Now, we can find maximum b from the constraint. b = 8 2 3.8 = 8 3 5.2 Lagrange multiplier method One method for solving constraint maximization problems is the Lagrangian multiplier method. The formal problem: Suppose you want to find the values x 1,..., x n that maximize y = f(x 1,..., x n ) subject to First, we set up Lagrangian g(x 1,..., x n ) = c L = f(x 1,..., x n ) + λ[c g(x 1,..., x n )] (1) where λ is the additional variable called the Lagrange multiplier. (We ll treat λ as a variable in addition to x s.) Note that when the constraint holds L and f have the same value. So, if we restrict attention only to x s that satisfy the constraint, finding the constraint maximum value of f is equivalent to finding optimal(critical) values of L. The following are conditions for a critical point for the function L. Note that there are n + 1 equations with n + 1 unknowns. 12

From Equation 1, the FOCs for an optimum (critical) point are: = f + λ g = 0 x 1 x 1 x 1 = f + λ g = 0 x 2 x 2 x 2.. = f + λ g = 0 x n x n x n λ = [c g(x 1,..., x n )] = 0. Let s apply this method to the consumer s problem. The objective function is U(p, b) = 2ln(p) + ln(b) and the constraint function is g(p, b) = 2p + 3b and c = 24. The Lagrangian function is: FOCs of the Lagrangian are: L = 2ln(p) + ln(b) + λ[24 (2p + 3b)] λ p = 2 p 2λ = 0 b = 1 b 3λ = 0 = [24 (2p + 3b)] = 0. Note that the last equation simply says that the budget constraint in the maximization problem has to hold. The above system has 3 equations and 3 unknowns (p,b and λ ) and easy to solve. You ll see λ = 1/8, p = 8 and b = 8/3. (Do it!) Remark:The variable λ has an important economic interpretation. It tells us how much will the value of our maximized objective function increase if we relax our constraint. Suppose we will increase consumer s income from 24 to 25. λ = 1/8 says that maximum possible utility obtained by these goods will increase by approximately 1/8. 13