1. EC361: Lecture 1 Mathematical Economics Reintroduction to Functions & Equations

2. Linear Equations and Inequalities You may have heard much of this before… You will be required to solve and understand how to setup systems of linear equations as well as inequalities. You should learn how to set up these problems in Mathematica to easily solve these systems. However, you must understand the fundamentals before Mathematica can solve your problems

3. Mathematical Economics Moving beyond geometry to answer/approach economic questions. Theoretical analysis in an economic context. Not econometrics but the two are related. Setting up theoretical models that can be solved mathematically. Terms like elasticity, marginal, and diminishing will be used frequently Examples will come from all fields of economics

4. Functions General versus Functional form Example: Cobb-Douglas Production Function

5. Linear Equations, Standard Form where a is not equal to zero. This is called the standard form of the linear equation. For example, the equation is a linear equation because it can be converted to standard form by clearing of fractions and simplifying. Probably review, but… in general, a first-degree, or linear, equation in one variable is any equation that can be written in the form

6. Equivalent Equations Two equations are equivalent if one can be transformed into the other by performing a series of operations which are one of two types: 1. The same quantity is added to or subtracted from each side of a given equation. 2. Each side of a given equation is multiplied by or divided by the same nonzero quantity. To solve a linear equation, we perform these operations on the equation to obtain simpler equivalent forms, until we obtain an equation with an obvious solution.

7. Example of Solving a Linear Equation Example: Solve

8. Example of Solving a Linear Equation Example: Solve Solution: Since the LCD of 2 and 3 is 6, we multiply both sides of the equation by 6 to clear of fractions. Cancel the 6 with the 2 to obtain a factor of 3, and cancel the 6 with the 3 to obtain a factor of 2. Distribute the 3. Combine like terms.

9. Solving a Formula for a Particular Variable Example: Solve M=Nt+Nr for N.

10. Solving a Formula for a Particular Variable Example: Solve M=Nt+Nr for N. Factor out N: Divide both sides by (t + r):

11. Linear Inequalities If the equality symbol = in a linear equation is replaced by an inequality symbol (, ≤, or ≥), the resulting expression is called a first-degree, or linear, inequality. For example is a linear inequality.

12. Solving Linear Inequalities We can perform the same operations on inequalities that we perform on equations, except that the sense of the inequality reverses if we multiply or divide both sides by a negative number. For example, if we start with the true statement -2 > -9 and multiply both sides by 3, we obtain -6 > -27. The sense of the inequality remains the same. If we multiply both sides by -3 instead, we must write 18 < 81 to have a true statement. The sense of the inequality reverses.

13. Example for Solving a Linear Inequality Solve the inequality 3(x-1) < 5(x + 2) - 5

14. Example for Solving a Linear Inequality Solve the inequality 3(x-1) < 5(x + 2) - 5 Solution: 3(x-1) < 5(x + 2) - 5 3x - 3 < 5x + 10 - 5 Distribute the 3 and the 5 3x - 3 < 5x + 5Combine like terms. -2x < 8Subtract 5x from both sides, and add 3 to both sides x > -4Notice that the sense of the inequality reverses when we divide both sides by -2.

15. Interval and Inequality Notation IntervalInequalityIntervalInequality [a,b]a ≤ x ≤ b(-∞,a]x ≤ a [a,b)a ≤ x < b(-∞,a)x < a (a,b]a < x ≤ b[b,∞)x ≥ b (a,b)a < x < b(b,∞)x > b If a < b, the double inequality a < x < b means that a < x and x < b. That is, x is between a and b. Interval notation is also used to describe sets defined by single or double inequalities, as shown in the following table.

16. Interval and Inequality Notation and Line Graphs (A) Write [-5, 2) as a double inequality and graph. (B) Write x ≥ -2 in interval notation and graph.

17. Interval and Inequality Notation and Line Graphs (A) Write [-5, 2) as a double inequality and graph. (B) Write x ≥ -2 in interval notation and graph. (A) [-5, 2) is equivalent to -5 ≤ x < 2 [ ) x -5 2 (B) x ≥ -2 is equivalent to [-2, ∞) [ x -2

18. Procedure for Solving Word Problems 1. Read the problem carefully and introduce a variable to represent an unknown quantity in the problem. 2. Identify other quantities in the problem (known or unknown) and express unknown quantities in terms of the variable you introduced in the first step. 3. Write a verbal statement using the conditions stated in the problem and then write an equivalent mathematical statement (equation or inequality.) 4. Solve the equation or inequality and answer the questions posed in the problem. 5. Check that the solution solves the original problem.

19. Example: Break-Even Analysis A recording company produces compact disk (CDs). One-time fixed costs for a particular CD are $24,000; this includes costs such as recording, album design, and promotion. Variable costs amount to $6.20 per CD and include the manufacturing, distribution, and royalty costs for each disk actually manufactured and sold to a retailer. The CD is sold to retail outlets at $8.70 each. How many CDs must be manufactured and sold for the company to break even?

20. Break-Even Analysis (continued) Solution Step 1. Let x = the number of CDs manufactured and sold. Step 2. Fixed costs = $24,000 Variable costs = $6.20x C = cost of producing x CDs = fixed costs + variable costs = $24,000 + $6.20x R = revenue (return) on sales of x CDs = $8.70x

21. Break-Even Analysis (continued) Step 3. The company breaks even if R = C, that is if $8.70x = $24,000 + $6.20x Step 4. 8.7x = 24,000 + 6.2x Subtract 6.2x from both sides 2.5x = 24,000 Divide both sides by 2.5 x = 9,600 The company must make and sell 9,600 CDs to break even.

22. Break-Even Analysis (continued) Step 5. Check: Costs= $24,000 + $6.2 ∙ 9,600 = $83,520 Revenue = $8.7 ∙ 9,600 = $83,520

23. Using Mathematica

24. x y III IIIIV (3,1) (-1,-1) The Cartesian Coordinate System (cogito ergo sum?) Two points, (-1,-1) and (3,1), are plotted. Four quadrants are as labeled.

25. Linear Equations in Two Variables A linear equation in two variables is an equation that can be written in the standard form Ax + By = C, where A, B, and C are constants (A and B not both 0), and x and y are variables. A solution of an equation in two variables is an ordered pair of real numbers that satisfy the equation. For example, (4,3) is a solution of 3x - 2y = 6. The solution set of an equation in two variables is the set of all solutions of the equation. The graph of an equation is the graph of its solution set.

26. Linear Equations in Two Variables (continued) If A is not equal to zero and B is not equal to zero, then Ax + By = C can be written as This is known as slope-intercept form. If A = 0 and B is not equal to zero, then the graph is a horizontal line If A is not equal to zero and B = 0, then the graph is a vertical line

27. Using Intercepts to Graph a Line Graph 2x - 6y = 12.

28. Using Intercepts to Graph a Line Graph 2x - 6y = 12. xy 0-2y-intercept 60x-intercept 3check point

29. Using Mathematica Graph 2x - 6y = 12 in Mathematica

30. Special Cases 1. Graph x = -7 (Vertical Line) 2. 2. Graph y = 3 (Horizontal Line)

31. Solutions x = -7 y = 4

32. Slope of a Line Slope of a line: rise run

33. Find the Slope and Intercept from the Equation of a Line Example: Find the slope and y intercept of the line whose equation is 5x - 2y = 10.

34. Find the Slope and Intercept from the Equation of a Line Example: Find the slope and y intercept of the line whose equation is 5x - 2y = 10. Solution: Solve the equation for y in terms of x. Identify the coefficient of x as the slope and the y intercept as the constant term. Therefore: the slope is 5/2 and the y intercept is -5.

35. Point-Slope Form Cross-multiply and substitute the more general x for x 2 where m is the slope and (x 1, y 1 ) is a given point. It is derived from the definition of the slope of a line: To find the equation of a line, when you only have two points. The point-slope form of the equation of a line is

36. Example Find the equation of the line through the points (-5, 7) and (4, 16).

37. Example Now use the point-slope form with m = 1 and (x 1, x 2 ) = (4,16). (We could just as well have used (-5,7)). Find the equation of the line through the points (-5, 7) and (4, 16). Solution:

38. Application Office equipment was purchased for $20,000 and will have a scrap value of $2,000 after 10 years. If its value is depreciated linearly, find the linear equation that relates value (V) in dollars to time (t) in years:

39. Application Office equipment was purchased for $20,000 and will have a scrap value of $2,000 after 10 years. If its value is depreciated linearly, find the linear equation that relates value (V) in dollars to time (t) in years: Solution: When t = 0, V = 20,000 and when t = 10, V = 2,000. Thus, we have two ordered pairs (0, 20,000) and (10, 2000). We find the slope of the line using the slope formula. The y intercept is already known (when t = 0, V = 20,000, so the y intercept is 20,000). The slope is (2000-20,000)/(10 – 0) = -1,800. Therefore, our equation is V(t) = - 1,800t + 20,000.

40. Supply and Demand In a free competitive market, the price of a product is determined by the relationship between supply and demand. The price tends to stabilize at the point of intersection of the demand and supply equations. This point of intersection is called the equilibrium point. The corresponding price is called the equilibrium price. The common value of supply and demand is called the equilibrium quantity.

41. Supply and Demand Example Use the barley market data in the following table to find: (a)A linear supply equation of the form p = mx + b (b)A linear demand equation of the form p = mx + b (c)The equilibrium point.

42. Supply and Demand Example (continued) (a) To find a supply equation in the form p = mx + b, we must first find two points of the form (x, p) on the supply line. From the table, (340, 2.22) and (370, 2.72) are two such points. The slope of the line is Now use the point-slope form to find the equation of the line: p - p 1 = m(x - x 1 ) p - 2.22 = 0.0167(x - 340) p - 2.22 = 0.0167x - 5.678 p = 0.0167x - 3.458 Price-supply equation.

43. Supply and Demand Example (continued) (b) From the table, (270, 2.22) and (250, 2.72) are two points on the demand equation. The slope is p - p 1 = m(x - x 1 ) p - 2.22 = -0.025(x -270) p - 2.22 = -0.025x + 6.75 p = -0.025x + 8.97 Price-demand equation

44. Supply and Demand Example (continued) (c) If we graph the two equations in Mathematica we obtain: The equilibrium point is approximately (298, 1.52). This means that the common value of supply and demand is 298 million bushels when the price is $1.52.

45. Mathematical Modeling Mathematical modeling is the process of using mathematics to solve real-world problems. This process can be broken down into three steps: 1. Construct the mathematical model, a problem whose solution will provide information about the real-world problem. 2. Solve the mathematical model. 3. Interpret the solution to the mathematical model in terms of the original real-world problem. In this section we will discuss one of the simplest mathematical models, a linear equation.

46. Slope as a Rate of Change If x and y are related by the equation y = mx + b, where m and b are constants with m not equal to zero, then x and y are linearly related. If (x 1, y 1 ) and (x 2, y 2 ) are two distinct points on this line, then the slope of the line is This ratio is called the rate of change of y with respect to x. Since the slope of a line is unique, the rate of change of two linearly related variables is constant. Some examples of familiar rates of change are miles per hour and revolutions per minute.

47. Linear Regression In real world applications we often encounter numerical data in the form of a table. The powerful mathematical tool, regression analysis, can be used to analyze numerical data. In general, regression analysis is a process for finding a function that best fits a set of data points. In the next example, we use a linear model obtained by using linear regression.

48. Example of Linear Regression Prices for emerald-shaped diamonds taken from an on-line trader are given in the following table. Find the linear model that best fits this data. Weight (carats)Price 0.5$1,677 0.6$2,353 0.7$2,718 0.8$3,218 0.9$3,982

49. Example of Linear Regression (continued) Solution: If we enter these values into the lists in Mathematica and use the Fit function. The 1, means that there is an intercept that might not be zero. The linear equation of best fit is y = 5475x - 1042.9.

50. Scatter Plots We can plot the data points in the previous example on a Cartesian coordinate plane, either by hand or using Mathematica. Price of emerald (thousands) Weight (tenths of a carat) We can plot the graph of our line of best fit on top of the scatter plot: y = 5475x - 1042.9