Economics Session Overview: Optimizing Economic Functions with Derivatives

Lecturer: Dr. Monica Lambon-Quayefio Contact Information: mplambon-quayefio@ug.edu.gh College of Education School of Continuing and Distance Education 2014/2015 2016/2017

Session Overview Economics basically studies the science of choice and the most common criterion for making a particular choice among alternatives involves the goal of maximizing something (such as a firm s profit, consumer s utility) or of minimizing something ( such as minimizing the cost of production for a firm). This session introduces students to the concept of optimizing economic functions by determining their maximum or minimum values using derivatives. Objectives At the end of the session, the student will Be able to determine whether or not a given function is to have a maximum or a minimum point Be able to determine the stationary point of a function Be able to classify stationary points as maximum or minimum points. Slide 2

Session Outline The key topics to be covered in the session are as follows: First derivative Test for Concavity Finding stationary points and distinguishing between maximum and minimum points Optimization of Economic Functions Slide 3

Reading List Sydsaeter, K. and P. Hammond, Essential Mathematics for Economic Analysis, 2nd Edition, Prentice Hall, 2006- Chapter 8 Dowling, E. T., Introduction to Mathematical Economics , 3rdEdition, Shaum s Outline Series, McGraw-Hill Inc., 2001.- Chapter 4 Chiang, A. C., Fundamental Methods of Mathematical Economics , McGraw Hill Book Co., New York, 1984.- Chapter 9 Slide 4

Topic One THE CONCEPT OF CONCAVITY AND ITS TEST Slide 5

Concave upward and downward Definition: (a) Ifthe graph of f lies above all of its tangentson an interval, thenfis called concave upward on that interval. (b) Ifthe graph of f lies below all of its tangentson an interval, thenfis called concave downward on that interval. Concave upward Concave downward

Inflection Points Definition: A point P on a curvey = f(x) is called an inflection point if fis continuous there and the curve changes from concave upward to concave downward or from concave downward to concave upward at P. Inflection points

What doesf say aboutf ? Concavity test: (a) If f (x) > 0 for all x of an interval, then the graph of f is concave upward on the interval. (b) If f (x) < 0 for all x of an interval, then the graph of f is concave downward on the interval. Example: Find the intervals of concavity of the function f (x) = x3 1.5x2 6x + 5. Solution: f (x) = 3x2 3x 6 f (x) = 6x - 3 f (x) > 0 for x > 0.5 , thus it is concave upward on (0.5, ) . f (x) < 0 for x < 0.5 , thus it is concave downward on (- , 0.5) . Thus, the graph has an inflection point at x = 0.5 since concavity changes from upward to downward at this point .

Topic One STATIONARY POINTS OF FUNCTIONS Slide 9

Definitions Given that a function f defined is defined on an interval for any two numbers x1 and x2 ,then f is said to be an increasing function if f(x1) < f(x2) when x1 < x2 and a decreasing function if f(x1) > f(x2) when x1< x2 X2 f(x) X1 X2 X1 10

Increasing and Decreasing Functions Derivative Test for Increasing / Decreasing Function: (a) If f (x) > 0 on the interval (x1 ,x2 ) , then f is increasing on that interval. (b) If f (x) < 0 on an interval (x1 ,x2 ) , then f is decreasing on that interval. Example: Find where the function f (x) = x3 1.5x2 6x + 5 is increasing and where it is decreasing. Solution: f (x) = 3x2 3x 6 = 3(x + 1)(x - 2) f (x) > 0 for x < -1 and x > 2 ; thus the function is increasing on (- , -1) and (2, ) . f (x) < 0 for -1 < x < 2 ; thus the function is decreasing on (-1, 2) .

Stationarity of Functions The stationary points of a function is defined as the points of the function where the gradient is zero. Stationary points can either be a maximum or minimum. A maximum point x dy = = 0 dx x A minimum point Slide 12

To find the stationary points of any function, find the first derivative of the function, equate to zero and then solve. Examples Find the coordinates of the stationary points on the curve x x x y 9 3 = = 3 2 Solution: ?? ?? = 3?2 -6x-9 ?? ?? to zero and solve for x. Equate the first derivative 3?2- 6x-9=0 Mathematical and Nonmathematical Economics Divide through by 3 to obtain: ?2- 2x-3=0 Factorize to obtain: (x-3)(x+1)=0 Solve for x to obtain: x=3, -1

To solve for the y values, substitute the x values into the given function. 3 2 = = = = = = = = = = x y 3 ) 3 ( 27 1 ) 3 ( 3 27 + + 3 ) 3 ( 9 = = 5 27 = = 9 27 3 2 = = x y 1 ( ) 1 ( 3 ) 1 ( 9 ) 1 The stationary points are therefore: (3, -27) and ( -1, 5)

Topic Two MAXIMUM AND MINIMUM POINTS Slide 15

Mathematical Economics Versus Econometrics Once stationary points have been determined, it is sometimes important to be able to determine the nature of a stationary point ( whether it is a max or a min ). There are several ways of doing this. e.g. On the left of a maximum, the gradient is positive On the right of a maximum, the gradient is negative

So, for a max the gradients are + + 0 At the max On the left of the max On the right of the max The opposite is true for a minimum + + 0 Calculating the gradients on the left and right of a stationary point tells us whether the point is a max or a min.

Stationary Points: Example Find the coordinates of the stationary points of the curve 1 4 + + = = x x y 2 Determine whether or not these points are maximum and minimum

= + ) 1 ( 2 4 1 y x x dy = = 2 x 4 dx 4 dy = = x 2 = = = = x 0 2 0 dx Substitute in (1): 2 = = = = ) 2 ( 4 + + dy y 3 y ) 2 ( 1 On the left of x = 2 e.g. at x = 1, = = = = 0 ) 1 ( 2 4 2 dx dy On the right of x = 2 e.g. at x = 3, = = = = 0 ) 3 ( 2 4 2 dx ) 3 + + , 2 ( We have is a min 0

Determining Stationarity using Second Order Derivatives Another method used in determining maximum or minimum points is the use of second order derivative. The second derivative uses the notion of gradient of a gradient. This is denoted by and it is simply found by 2 dx 2 d y Differentiating a function twice. Slide 20

Determining Stationarity using Second Order Derivatives Suppose f is continuous near a. (a) If f (a) = 0 and f (a) > 0 then f has a minimum at a. (b) If f (c) = 0 and f (c) < 0 then f has a local maximum at c. Example: Find the extrema ( ie. Maximum/ minimum) of the function f (x) = x3 1.5x2 6x + 5. Solution: f (x) = 3x2 3x 6 = 3(x + 1)(x - 2) , f (x) = 6x - 3 f (-1) = 6*(-1) 3 = -9 < 0, so x = -1 is a maximum f (2) = 6*2 3 = 9 > 0, so x = 2 is a minimum so f (x) =0 at x=-1 and x=2 Slide 21

Summary of what y and y say about the curve First derivative: y y y Curve is rising/ increasing. is positive is negative Curve is falling/ decreasing. Possible local maximum or minimum. is zero Second derivative: y is positive Curve is concave up. y Curve is concave down. is negative y is zero Possible inflection point (where concavity changes).

Trial Questions Find the stationary points and determine the nature of these points of the functions below: 1. 2. ( ) 4 2 3 3 2 = = + + 2 + + y x x x 3 9 10 = + 3 2 f x x x x Slide 23

Topic Two OPTIMIZING ECONOMIC FUNCTIONS Slide 24

Session Problem Set Many economic theories begin with the assumption that an economic agent is seeking to find the optimal value of some function. These functions to be optimized are usually termed objective functions. Examples of such functions include: Consumers seeking to maximize utility Firms seeking to maximize profit Firms seeking to minimize production costs Governments seeking to maximize the use of scare natural resources Governments and agencies seeking to minimize levels of pollution

Single Variable Functions In general, single variable functions are functions whose dependent variable is one. Examples of such functions include the following: A firm s total revenue function which is a function of the quantity produced only: ? ? = 32? ?2 A firm s total cost function which is a function of the quantity produced only: TC(q)=3?3+?2-6q+4

Optimizing Functions The process of optimization simply involves determining the values of the function of interest which yields the maximum or minimum values as the case may be. Simple example: Manager of a firm wishes to maximize profits . = (q ) f Here, the firm seeks to determine the quantity level that will yield the highest profit for the firm. * = f(q) * occur at q* Maximum profits of Quantity q*

Optimization using Derivatives The process of optimization makes use of the first and second order derivatives studied in the previous sessions. To optimize any economic function, we follow the steps below: Determine the stationary points using first order derivatives Determine the nature of the stationary points using the second order derivatives Evaluate the objective function at the stationary points

First Order Condition for Max/Min For a function of one variable to attain its maximum value at some point, the derivative at that point must be zero ie. ?? ??= 0 This condition is described as the necessary condition for a maximum or a minimum point. However, the mere fact that the first derivative is zero does not necessarily mean that the function is at its maximum or minimum at that point. There is therefore the need to probe further to determine whether the point is a max or a min point. The second order derivative test helps to ensure that the point that makes ?? ??= 0 is indeed a maximum or minimum point. This second order derivative test is described as the sufficient condition for a maximum or minimum point to exist.

Example: Profit Maximization Suppose that the relationship between profit and output is = 1,000q - 5q2 The first order condition for a maximum is d /dq = 1,000 - 10q = 0 q* = 100 This means that at q=100, the profit function is stationary. Meaning at this quantity level, the profit function could either be at its minimum or maximum level. We therefore find the second derivative to ascertain the nature of the function at q* ??2 = -10 < 0. This means, q=100 is indeed a maximum point. This means that for the firm to attain its maximum profit level, it has to produce at q=100. ?2?

Trial Question Given the following total revenue (TR) and total cost (TC)functions below as functions of quantity, find the levels of q at which the firm maximizes its profit (?). TR= 1400q 6?2 and the TC=1500+80q.