Table of Contents Chapter 1 Linear Models page Linear Models Part 1 Linear Models Activities 1 17 Linear Models Part 1 Linear Models Activities 8 Chapter Linear Programming Linear Programming Part 1 34 Linear Programming Activities 1 43 Linear Programming Part 49 Linear Programming Activities 61 More LP 64 Chapter 3 Time Value of Money Time value of Money Part 1 67 Time value of Money Activities 1 8 Time value of Money Part 86 Time value of Money Activities 93 Time value of Money Part 3 97 Time value of Money Activities 3 108 Time value of Money Part 4 11 Time value of Money Activities 4 1 Time value of Money Part 5 15 Time value of Money Activities 5 139 Chapter 4 Functions&Trendlines Functions and Trendlines Part 1 147 Functions and Trendlines Activities 1 157 Functions and Trendlines Part 167 Functions and Trendlines Activities 179 Chapter 5 Measuring Change Measuring Change Part 1 188 Measuring Change Activities 1 10 Measuring Change Part 17 Measuring Change Activities 39 Measuring Change Part 3 43 Measuring Change Activities 3 57 Measuring Change Part 4 64 Measuring Change Activities 4 74 Excel Tutorial 1 CreditCardPart1 Appendix A Excel Tutorials 79 Excel Tutorial CreditCardPart 86 Excel Tutorial 3 BreakEvenTable 93 Excel Tutorial 4 BreakEvenGraph 98 Linear Models Appendix B Answers to Exercises 301 Linear Programming 306 Time Value of Money 311 Functions and Trendlines 318 Measuring Change 35 Joseph F. Aieta 1999-003 Page 1 Printed 8/14/003 All Rights Reserved
Linear Models Part 1 Section 1.1 Page Linear Models Part 1 Linear modeling involves using linear functions to describe the relationships among variables representing data from the real world. Common business applications of models include linear depreciation, salary computations involving commissions, cost functions, revenue functions, profit functions, utility bills, tax computations, volume discounts and other pricing methods. We will examine several types of linear models starting with linear depreciation. In order to investigate and solve business problems that can be modeled by linear functions, you need to be familiar with the algebra and analytic geometry of linear equations and their graphs. To determine if you need a review of function concepts or skills related to various forms of the equation of a line, skim through the Linear Equation Review, starting on page 11. This review contains references to several useful Excel spreadsheets. Linear Depreciation Section 1.1 Each year at tax time, company accountants report company assets that have declined in value since the previous tax year. These assets include motor vehicles, computers, machinery, and other equipment. The factors that determine the value of the asset at a particular point in time (residual value) are the original purchase price, the length of time the company plans to keep the asset (useful life) and its scrap value when it is no longer useful to the company. The depreciation is taken as a tax deduction and accountants can choose among several formulas in computing the amount of this deduction. The simplest method is called straight line depreciation or linear depreciation. This method assumes that the asset depreciates by a constant amount every year. When the residual value of the asset is plotted on the vertical axis with years on the horizontal axis, then the graph displays a series of points that lie on a line which slopes downward from left to right. Example 1.1.1 We will examine linear depreciation as it applies to a rental company that owns a collection of trucks. The purchase price of the trucks ranges from $40,000 to $100,000. The useful life of the trucks for rental purposes ranges from 3 years to 15 years. One truck was purchased for $64,000 and was sold by the rental company for $6,000 after 8 years. The spreadsheet in Figure 1.1.1 shows a table and graph. Figure 1.1.1 Joseph F. Aieta Page 8/14/003
Linear Models Part 1 Section 1.1 Page 3 The annual change in residual value, commonly referred to as depreciation, is found by taking the difference between cost and scrap value and then dividing by the useful life. In this case $64,000 $6,000 = $7, 50 so the rate of change is a reduction 8 of 7,50 dollars. As we see in Figure 1.1.1 the truck had a residual value (sometimes referred to as book value) at the end of year the first year of $64.000 - $7,50 = $56,750. The coordinates of all points on the graph are of the form (year, residual value). $64,000 $6,000 = $7,50 If we calculate the slope of the line containing the points (0, $64000) and (8, $6000), we obtain 0 8 which is equivalent to $ 64,000 $6,000. It is actually more common for accountants to say that the value depreciates by 8 $7,50 per year instead of saying that the slope of the residual value line -$7,50. The formula for residual value can be succinctly stated in function notation as V(t) = 64,000 7,50 t, where t is the number of years since the truck was purchased. This is the formula that was used to create the table of values in Figure 1.1.1. This function is entered into Excel as = 64000-750*t where t refers to the named range containing years and multiplication is explicitly indicated by the asterisk. cost scrap In general, if the annual rate of change is then cost scrap V(t) cost t useful life useful life = Table 1.1.1 shows the book value of other trucks at different points in time after they were purchased. Verify the numerical values in the last column, using the above formula with either a calculator or a spreadsheet. Years of Depreciation rate Years since Book value Original cost Scrap value Useful life per year purchase (t) V(t) $80,000 $4,000 10 -$7,600.00 8 $19,00 $50,000 $9,000 5 -$8,00.00 $33,600 $100,000 $0,000 6 -$13,333.33 6 $0,000 Table 1.1.1 Using a spreadsheet such as LinearDepreciation.xls, an accountant can enter values for the original cost, the scrap value, and the useful life and have Excel automatically calculate the annual depreciation. Appendix A on Excel Basics contains tutorials for creating tables and graphs. Joseph F. Aieta Page 3 8/14/003
Linear Models Part 1 Section 1. Page 4 Example 1.1. For linear depreciation, the slope of the residual value line can be calculated using any two points on the graph. Suppose the book value of a truck is $56,000 in year 4 and $38,000 in year 8. a) If the useful life of the truck is 10 years then what is its scrap value? b) If the truck was sold as scrap for $11,000 then what was its useful life? Solution: Find the slope from two points using the formula (8, $38,000) y = x y x y x 1 1 where one point is (4, $56,000) and a second point is $38,000 $56,000 $18,000 slope = = = 4,500 8 4 4 dollars per year Using the point-slope form y - y 0 = m(x -x 0 ), we obtain y - 56,000 = -4,500 (x - 4). Solving for y we obtain y = -4,500 x + 74,000. Using t for years instead of x and the function notation V(t) instead of y, we obtain V(t) = -4,500 t + 74,000. a) If the useful life is 10 years then V(10) = -4,500 (10) +74,000 = 9,000 so the scrap value at the end of year 10 is $9,000. b) If the scrap value is $11,000 then 11,000 = -4,500 t + 74,000. When we solve for t we get 11,000 74,000 = -4,500 t or -63,000 = -4500 t or t = 14. The useful life of this truck is 14 years. Sales Applications of Linear Models Section 1. Example 1..1 For several months a company tracks its sales and its expenses (in millions of dollars) and observes a linear trend in which expenses, y, are a function of sales, x. Sales of $40 million in one month correspond to expenses of $ million. One month earlier, expenses of $10 million were accompanied by sales of $15 million. Since expenses are a function of sales the coordinates of points on the line are ordered as (sales, expenses). The point (40, ) corresponds to sales of 40 million dollars and expenses of million dollars. Solution: a) How much did sales increase from one month to another? b) How much did selling expenses increase from one month to another? c) Assuming a linear relationship, write the formula for expenses as a function of sales. a) The increase in sales, in millions of dollars, is 40 15 = 5. b) The increase in expenses, in millions of dollars, is 10 = 1. c) First we find the slope of the line from the two points (40, ) and (15, 10) 10 1 m = = = 0.48 so expenses, y, are related to sales, x, by the linear equation y 10 = 0.48(x -15). 40 15 5 This is equivalent to y = 0.48x +.8 where x and y are in millions of dollars. Joseph F. Aieta Page 4 8/14/003
Linear Models Part 1 Section 1.3 Page 5 Example 1.. A salesperson named Harry receives a base pay of $,000 per month plus a 5% commission on sales. a) If sales in one month were $5,000 then what was Harry s total pay for that month? b) If total pay was $4,750 then what was the sales volume for that month? c) Answer questions a) and b) if the base salary is unchanged but the commission structure is changed to 5% commission on sales over $30,000. Solution: a) The formula for total pay is y =,000 + 0.05 x where x is the sales volume in dollars. For sales of 5,000 dollars, Harry s total pay in dollars is,000 + 0.05(5,000) = 3,50 b) If 4,750 =,000 + 0.05 x then 0.05x = 4,750,000 =,750 and x =,750/0.05 = 55,000. Sales of $55,000 correspond to total pay of $4,750. c) Since $5,000 is less than $30,000 total pay for the month will be just base the salary of $,000. For sales above $30,000 commission based upon x dollars in sales becomes 0.05 (x 30,000) so total pay is,000 + 0.05 (x 30,000). If Harry s total pay is $4,750 then 4,750 =,000 + 0.05 (x 30,000),750 = 0.05 (x 30,000),750/0.05 = x 30,000 x = 55,000 + 30,000 = 85,000 dollars Harry needed sales of $85,000 in order for him to make $4,750 Piecewise Linear Functions Section 1.3 Many types of applications involve functions that can be described by linear segments as shown in Figure 1.3.1. This figure is made up of three line segments, and so is called a piecewise linear function. There are even situations when it is useful to approximate a continuous curve as shown in Figure 1.3. by a collection of many linear segments. Figure 1.3.1 Figure 1.3. For the piecewise linear function in Figure 1.3.1, we will focus on the left-hand endpoints of each segment. Joseph F. Aieta Page 5 8/14/003
Linear Models Part 1 Section 1.3 Page 6 The first segment is defined for a < x < b, the second segment is defined for b < x < c the third segment is defined for c < x (which is equivalent to x > c ) _a b c Example 1.3.1 Figure 1.3.3 x If we knew (1) the y coordinates that correspond to the breakpoints at x = a, x = b, and x = c and () the slope of each segment, then we could write an equation with three parts { equation equation for segment 1 equation for segment for a < x < b for b < x < c for segment 3 for c < x Let s illustrate how to develop the equations for each segment with the following hypothetical values: The left- hand endpoint for segment 1 is (0, 0) and the slope of the first segment is 0.05. The left- hand endpoint for segment is (30,000, 750) and the slope of this second segment is 0.075. The left- hand endpoint for segment 3 is (50,000, 1,300) and the slope of this third linear segment is 0.030. In the point-slope form, the equation of a line with slope m and containing the point (x 0, y 0 ) is y - y 0 = m(x -x 0 ) Using a slight variation of this form we can define each segment as y = y 0 + m(x -x 0 ) where m is the appropriate slope and (x 0, y 0 ) are the coordinates of the point at the left of each segment. { 0.05 x for 0 < x < 30,000 f(x) = 750 + 0.075 (x - 30,000) for 30,000 < x < 50,000 1,300 + 0.030 (x -50,000) for 50,000 < x Now study the following verbal interpretation of this function and then try to imagine how such a function may appear in a practical situation. The above function f(x) can be interpreted as follows: First segment: Calculate.5% of any value of x starting at 0 and going up to x = 30,000, inclusive. Middle segment: For values of x above 30,000, but not larger than 50,000, calculate.75% of the amount above 30,000 and add that to 750. Last segment: For values of x greater than 50,000 take 3% of the amount above 50,000 and add that to 1,300. It may not be immediately obvious how we found the constant, 750, for the segment over the interval 30,000 < x < 50,000 or the constant, 1,300 corresponding to the segment in which x is greater than 50,000. As we see in Figure 1.3.3, the function f(x) is continuous with no gaps or jumps. Consequently, the right-hand endpoint of each linear segment must match the left-hand endpoint of the next segment. At the upper limit of the first segment, when x =30,000 we calculate f(30,000) = 0.05(30,000) = 750. Similarly, the value of the function at x = 50,000 can be calculated as: f(50,000)=750 + 0.075 (50,000 30,000) =750 + 0.075(0,000) = 750 + 550 = 1,300. Joseph F. Aieta Page 6 8/14/003
The calculation of graduated income taxes leads to the following mathematical model: Example 1.3.1 Linear Models Part 1 Section 1.3 Page 7 Suppose that the graduated state income tax for a single person in a state on the east coast is a piecewise linear function of Adjusted Gross Income (AGI). The total tax due is computed as follows:.50% of the first $30,000 of AGI plus.75% of the AGI between $30,000 and $50,000 plus 3.00% of any amount of AGI in excess of $50,000 Determine state income tax due as a function of Adjusted Gross Income and state all results to the nearest cent. Solution: a) Find the total state tax for Anna whose AGI is $3,000.00. b) Find the total state tax for Betsy, whose AGI is $40,000.00. c) Find the total state tax for Cecilia whose AGI is $58,000.00. d) If Darlene paid $1,16.50 in taxes then what was her AGI? Let x represent the Adjusted Gross Income. The graph of this three part income tax function is displayed in Figure 1.3.3. The tax rate is increasing as AGI changes from one bracket to the next. Consequently the slopes of the segments are increasing from left to right. The break points that define the brackets are 0, 30,000, and 50,000. The rule for calculating state income tax is given by the function: { 0.05 x for 0 < x < 30,000 f(x) = 750 + 0.075 (x - 30,000) for 30,000 < x < 50,000 1,300 + 0.030 (x -50,000) for 50,000 < x a) For an income level of $3,000.00 the first rule applies since 3,000 < 30,000. The tax due is 0.05(3,000) = 575 so Anna owes $575. b) An income level of $40,000.00 is between 30,000 and 50,000 so the middle rule applies. Betsy s state tax due is 750 + 0.075(40,000-30,000) = 750 + 0.075(10,000) = 750 + 75 = 1,05 so Betsy owes $ 1,05. A common mistake in determining the second segment is to overlook the fact that the taxpayer has already paid the tax on the first $30,000 of income, which was $750. The calculation 750 + 0.075 (40,000) is incorrect since the taxpayer would then be paying tax again on the first $30,000 of income. c) For an income level of $58,000, the rule for the third segment applies. Cecilia has already paid a tax of $1,300 on the first $50,000 of her income so her total state tax due is 1,300 + 0.03(58,000-50,000) = 1,300 + 0.03(8,000) = 1300 + 40 = 1,540 dollars. d) Darlene s tax of $1,16.50 is less than $1300 but greater than $750 so the middle rule applies. Solve 1,16.50 = 750 + 0.075(x - 30,000) for x 1,16.50 750.00 = 0.075(x - 30,000) 41.50= 0.075(x - 30,000) 41.50/0.075 = x 30,000 15,000= x 30,000 15,000 + 30,000 = x so Darlene s AGI is $45,000 Joseph F. Aieta Page 7 8/14/003
Linear Models Part 1 Section 1.3 Page 8 The utility bill calculation in Example 1.3. shows that a mathematical model may be made up of linear segments whose slopes decrease instead of increase. Example 1.3. An electric utility bills its monthly charges according to the customer s monthly usage of electricity recorded in kilowatt hours. This utility charges a particular class of commercial users according to the following rate structure: A basic monthly charge of $0.00 plus $0.80 per unit for the first 50 units plus $0.70 per unit for all units in excess of 50 a) Write the rule for this cost function, C(x), as a piece-wise linear function of x where x is the number of units recorded during the month. b) What are the total charges for a commercial customer with 30 units recorded? c) What are the total charges for commercial customer with 00 units recorded? d) What is the average charge per unit if 30 units were used in a particular month? Solution: Start by drawing a rough sketch. Figure 1.3.4 The rough sketch shows two segments with the slope of the second segment slightly less than the slope of the first segment. The left-hand endpoint of the first segment is (0, 0.00) because of the $0 basic monthly charge. The slope of the first piece is 0.80 and the slope of the second piece is 0.70. a) The second coordinate of the right-hand endpoint is 0.00 + 0.80(50) = 0.00 + 40.00 = 60.00. The breakpoint (50, 60.00) is also the left-hand endpoint for the second piece. Since the slope for this second piece is 0.70, the rule for charges that correspond to units over 50 is simply 60.00 + 0.70(x 50) C(x) = { 0.00 + 0.80 x 60.00 + 0.70(x - 50) for 0 < x < 50 for 50 < x b) The total cost of 30 units is C(30)= 0.00 + 0.80(30) = 0.00 + 4.00 = 44.00 dollars. c) The total charges for using 00 units is C(70) = 60.00 + 0.70 (00-50) = 60.00 + 0.70 (150) = $165.00. d) In a month when the commercial customer used 30 units, the average charge per unit is the total cost of 30 units C(30) $44.00 divided by 30. To the nearest cent this is = = $1. 47 per unit 30 30 Joseph F. Aieta Page 8 8/14/003
Linear Models Part 1 Section 1.3 Page 9 Another situation in which piecewise linear models may apply is in the volume discounts that wholesale suppliers offer to retailers. As a retailer s order quantity increases beyond certain break points, the wholesale supplier drops the unit price. See VolumeDiscount.xls Example 1.3.3 Fashion Designs Inc. sells suits to retail stores. FDI charges a fixed fee for shipping and handling and a volume discount that depends upon the number of suits ordered by the retailer. The shipping and handling charge is $305. The minimum order is 0 suits. The volume discount structure is as follows: $75 per suit for the first 50 suits, $5 per suit for the next 50 suits, $15 per suit for each suit in excess of 100. a) Determine the cost function. b) What are the total and average costs of ordering 90 suits (to the nearest cent)? c) How many suits were ordered if the total charge was $31,555? d) What is the minimum number of suits that must be purchased so that the average charge per suit is $00 per suit or below? Solution: A rough graph should consist of three segments with decreasing slopes as shown in Figure 1.3.5. Since the minimum number of suits that can be ordered is 0, the lowest total cost is 305 + 75(0) = 5,805 dollars. $40,000 $35,000 $30,000 $5,000 $0,000 $15,000 A simple list of declining unit cost (price per suit) and the corresponding break points will help us in deriving the three-part rule for total cost. $10,000 $5,000 units $0 0 50 100 150 00 50 Figure 1.3.5 shipping and handling rate1 breakpoint 1 rate breakpoint rate3 $305.00 $75.00 50 $5.00 100 $15.00 a) At the first breakpoint, when x = 50, the total cost for 50 suits is $305 + $75(50) = $14,055. At the second breakpoint, when x = 100, the total cost for 100 suits is $14,055 + $5(100-50) = $5,305. 305 + 75 x For 0 < x < 50 { f(x) = 14,055 + 5( x -50) for 50 < x < 100 5,305 + 15( x 100) for 100 < x Joseph F. Aieta Page 9 8/14/003
Linear Models Part 1 Section 1.3 Page 10 b) The total cost of ordering 90 suits is 14,055 + 5(90-50) =14,055 + 5(40) = $3,055.00 so the average cost of ordering 90 suits is 3,055.00 / 90 = 56.17 dollars per suit. c) A total cost is $31,555 is above $5,305 so the third equation applies and we must solve the equation 5,305 + 15(x 100) = 31,555 for x 5,305 + 15(x 100) = 31,555 15(x 100) =31,555-5,305 15(x 100) = 6,50 (x 100) = 6,50/15 x 100 = 50 x = 150 suits. d) If the average charge per suit is $00, we first need to determine which segment of the function applies. Since the average price for 100 suits is $5,305/100 = $53.05, the retailer must order more than 100 to get the average price down to $00. Therefore, we need to find x such that 5,305 + 15(x 100) divided by x is $00. Start by multipying both sides of the resulting equation 5,305 + 15(x -100) = 00 x 5,305 + 15(x 100) = 00 x 5,305 + 15 x 1,500 = 00 x 5,305 1,500 = 00 x - 15 x 1,805 = 00 x - 15 x (00 15) x = 1, 805 x = 1,805 / 75 = 170.733333 by x to get Since the solution is between 170 and 171, Fashion Designs, Inc. must sell 171 suits in order to bring its average cost per suit to a value at or below $00. Note: If the solution to the equation had been x = 170.15 then we would not round down to 170 in this context. The average cost of 170 suits is f(170) / 170 5,305 + 15(170-100) 5,305 + 15(70 ) = = 00.335 which is not at or below 00 dollars. 170 170 Joseph F. Aieta Page 10 8/14/003
Linear Equations Review Page 11 Review The function concept and the different forms for the equation of a line \ input / A numerical function f is a rule with certain inputs and outputs. You could think of a function as behaving like a machine. It produces a unique output f(x), pronounced f of x, for each input x taken from a specific collection of numbers called its domain. The input variable x is often called the independent variable. The output variable (or dependent variable) is often represented by the variable y, as in y = f(x). A function of a single input variable can be described in a variety of ways: / output \ x 4 8 10 0 f(x) 0.5 0.15 0.10 0.05 numerically: as a table with two rows or two columns In the table above f(4) = ¼ or 0.5, f(8) = 1/8 or 0.15, and so on. The rule for this reciprocal function is f(x) = 1/x. algebraically: as a formula or rule If the function f is given by the rule f ( x) = 3x x 1 then f () = 3 f ( 1) = 3( 1) 1 = 1 4 1 = 7 ( 1) 1 = 3 + -1 = 4 graphically: as a graph with the input variable on the horizontal axis and the output variable on the vertical axis Suppose the graph of the function f is the line indicated below with double headed arrows that goes through the point (0,). If we start at (0, ) and go over 3 and up we reach the point (3, 4) so f(3) =4, If we start at (3,4) and go over 3 and up we reach the point (6, 6) so f(6) =6 slope of a line change in y divided by change in x Example: The slope of the line through (1, 3) and (3, 4) is given by m = y x y1 x 1 y = x 4 ( 3) 7 m = = = 3.5 3 1 Joseph F. Aieta Page 11 8/14/003
Linear Equations Review Page 1 linear functions and linear equations This category of functions turns out to be both relatively easy to work with and useful in a wide variety of business situations. Major forms for the equations of lines slope-intercept form, general form, and point-slope form slope-intercept form y = mx + b where m is the slope and b is the y-intercept. Example: y = 10x + 00 All linear functions can be described algebraically in this form where the parameters m and b are fixed values. The labels 'm' and 'b' are traditional in algebra but these letters are really arbitrary. For example, in statistics a line is often written in the form y = b 0 +b 1 x. The roles of m and b are illustrated in the worksheet slope-intercept contained in the file LinearForms.xls general form ax + by = c Example, 4x + 5y = 10 The x-intercept is c/a provided a 0. If 4x + 5y = 10, x-intercept is (.5, 0) The y-intercept is c/b provided b 0. If 4x + 5y = 10, y-intercept is (0, ) Example The roles of a, b and c are illustrated in the worksheet general form contained in the file LinearForms.xls point-slope form y - y 0 = m(x -x 0 ) where ( x 0, y 0 ) is any point on the line. A slight variation of this form y = y 0 + m(x - x 0 ) is particularly useful for determining formulas for piece-wise linear functions in section 1.3. conversions Any linear function written in general form or point-slope form can be transformed into slope-intercept form. Note: an equation such as x = 5 represents a vertical line and is not a function The equations on the left can be solved for y and transformed into slope intercept form as a linear function of x as shown on the right. 3 3x y +4 = 0 y = x + 6y = 0 y = 0 y 5 = (x 3) y = x -1 3x + 4y = 5 3 5 y = x + 4 4 Joseph F. Aieta Page 1 8/14/003
Linear Equations Review Page 13 Open the workbook LinearForms.xls and use the Excel worksheet slope intercept in answering questions 1, and In the figure below, the markers for line 1 are diamonds and the markers for line are squares. In the Excel worksheet the parameters of a line can be changed by moving the appropriate slider bars. Slope-intercept y = mx+b slope y_intercept -10 10-10 10 10 60-4 9 line 1-4 10 130 3 8-10 10-10 10 00 10 7 110 1 6 line 1 3 5 4 line 1 line Y1 = 3-4 *x + 10 x Y1 = -4*x + 10 Y =1*x + 3 Y = 1 *x + 3-10.0 50.0-7.0 Y1 = -4*x 1 + 10-9.0 46.0-6.0 Y =1*x 0+ 3-8.0 4.0-5.0-10 -9-8 -7-6 -5-4 -3 - -1-1 0 1 3 4 5 6 7 8 9 10-7.0 38.0-4.0 - -6.0 34.0-3.0-3 -5.0 30.0 -.0-4.0 6.0-1.0-4 -3.0.0 0.0-5 -.0 18.0 1.0-6 -1.0 14.0.0-7 0.0 10.0 3.0-8 1.0 6.0 4.0-9.0.0 5.0 3.0 -.0 6.0-10 4.0-6.0 7.0 5.0-10.0 8.0 6.0-14.0 9.0 7.0-18.0 10.0 8.0 -.0 11.0 9.0-6.0 1.0 10.0-30.0 13.0 Joseph F. Aieta Page 13 8/14/003
1. a) Manipulate the parameters to illustrate two parallel lines. What are their equations? b) Manipulate the parameters to illustrate identical lines. What are their equations? Linear Equations Review Page 14 c) Perpendicular lines meet at right angles. Manipulate the parameters to illustrate two lines that are perpendicular. What are their slopes? What is the relationship between the slopes of two perpendicular lines if neither one is a vertical line? d) Manipulate the parameters to illustrate two horizontal lines. What are their slopes? e) Explain why a vertical line cannot be displayed in slope-intercept form. f) Display two different lines that intersect at (1,). What are their equations? g) If a line is given in the form y - 3 = (x - 1) then it contains the point (1, 3) and has a slope of. Display this line and state its equation in slope intercept form.. a) Verify that the point (, 1) is on the line y = 3x - 5. To locate another point on this line go over one unit over in the horizontal direction and in the vertical direction. b) Verify that the point (, 4) on the line y = -x + 8. To locate another point on this line go one unit over in the horizontal direction and in the vertical direction. c) Suppose that P (x 0, y 0 ) is any point on the line y = mx + b. To locate another point on this line start at point P and go one unit to the right in the horizontal direction and in the vertical direction. What determines whether you should go up or down? Joseph F. Aieta Page 14 8/14/003
Linear Equations Review Page 15 Use the worksheet general form in the Excel workbook LinearForms.xls to help you answer questions 3 and 4. General Form ax + by = c a 1 b 1 c 1-10 10-10 10-50 50 5*x + 6*y = 0 5 6 0 line1 a b c -10 10-10 10-50 50 10*x + 0*y = 30 10 0 30 vertical line at x = 3.0 10-10 -9-8 9 11.7 999.0 8 10.8 999.0 7 10.0 999.0 6 9. 999.0 5 8.3 999.0 4 7.5 999.0 3 6.7 999.0 5.8 999.0 1 5.0 999.0 0 4. 999.0-7 -6-5 -4-3 - -1-1 0 1 3 4 5 6 3.3 999.0 -.5 999.0-3 1.7 999.0-4 0.8 999.0-5 0.0 999.0-6 -0.8 999.0-7 -1.7 999.0-8 -.5 999.0-9 -3.3 999.0-10 -4. 999.0 7 8 9 10-5.0 999.0 3. a) Display the line x - 5y = 0. What is the slope of this line? b) Display the line x - y = 5. What is the slope of this line? What are the coordinates of its intercepts? c) Display the line x + 5y = 0. What is the slope of this line? What are the coordinates of its intercepts? d) Display a distinct line parallel to x + 5y = 0 by changing only one parameter. What is the equation of this line? e) Display the line 4x - 8y = 40 and then display a line crossing the y axis at 5 that is parallel to the first line. What is the equation of this second line? f) Display the line 5x - y = 0 and state the slope and both intercepts for this line. g) Display the line -5x + y =10 and state the slope and both intercepts for this line. h) Display the line 5x +y = 40 and state the slope and both intercepts for this line. i) Display a line with y-intercept 6 and x-intercept 5. What is its equation? Can you do this more efficiently than a trial and error approach? Joseph F. Aieta Page 15 8/14/003
Linear Equations Review Page 16 k) Display the line 0x + 50y = 100. What is the slope of this line? What are the coordinates of its intercepts? l) Write an equation, in general form, for a line with x-intercept 1 and y-intercept 15 j) State the slope, x-intercept, and y -intercept for ax + by = c where a 0 and b 0. 4. a) Under what condition(s) on the parameters a, b, c will the graph of a linear equation of the form ax + by = c be a horizontal line and what is the equation of the line? b) Give an example of the equation of a horizontal line. c) What is the slope of every horizontal line? d) Under what condition(s) on the parameters a, b. c will the graph of a linear equation of the form ax + by = c be a vertical line? e) Give an example of the equation of a vertical line. f) Does a vertical line have a slope? Explain. 5. True or False. If the statement is false then explain why. a) A line that rises to the right and is almost vertical does not have a slope. b) The slope of the x-axis is zero. c) The segment connecting P(x 1, y 1 ) and Q(x 1, y ) where y 1 y is on a vertical line. d) A line segment of negative slope rises to the left. e) The slope of the y-axis is not a number. f) A line segment that is very, very close to vertical has a slope that has a large absolute value. g) No matter how large a number we may write down, there is a line segment whose slope exceeds this number. h) A line segment contained entirely in the second quadrant necessarily has a negative slope. i) The quadrant in which a line segment lies has no necessary relation to the sign of the slope of that segment. For additional practice, see Slope Intercept.xls Joseph F. Aieta Page 16 8/14/003
Linear Models Activities 1 Linear Models Activities 1 Page 17 Open the Excel workbook LinearDepreciation.xls 1. Given that the original cost of a piece of equipment = $60,000, scrap value = $8,000, and useful life = 8 years. a) What is the slope of the line and what does it represent? b) What does the vertical intercept of this line represent? c) What is the residual value of the equipment at the end of year two? d) What is the residual value of the equipment at the end of year six?. Given that the original cost of a piece of equipment = $6,000, scrap value = $4,000 and useful life = 8 years. a) What is the slope of the line and what does it represent in this context? b) In this context, give an interpretation of the point where the line crosses the vertical axis? c) Let t represent the time variable in years. Write a simplified formula, for the residual value, V(t), of this equipment as a function of t. 3. A machine, which is purchased for $9,000, will have a scrap value of $1,000 in 5 years. a) What will the machine be worth in years? b) Find a formula for the machine s residual value, V(t), after t years. 4. Suppose the original cost of a piece of equipment is $50,000 and the equipment depreciates $6000 per year until it is sold for $8000. a) What is its useful life? b) Is this problem easier to solve with a spreadsheet or with paper and pencil? Joseph F. Aieta Page 17 Printed 8/14/003
5. The chart below plots the book value of an asset assuming straight line depreciation Linear Models Activities 1 Page 18 Straight Line Depreciation $5,000 residual value $0,000 $15,000 $10,000 $5,000 $0 0 1 3 4 5 6 7 8 9 10 11 1 years a) What was the purchase price of the asset? b) Estimate the book value of the asset after 6 years. c) What is the yearly rate of depreciation? d) After how many years will the value of the asset be at or below $1000? 6. A copy card costs $15.00. When it is inserted into a campus copy machine, the machine automatically deducts the cost of the duplicating job from the current value of the card. The scatter plot shows the value remaining on the card corresponding to the total number of copies made. remaining value $15.00 $10.00 $5.00 Copy Machine Card copies $0.00 0 100 00 300 400 a) Construct a linear model for the relationship between the remaining value of the card, and the total number of copies printed. b) What value remains on the card when 00 copies have been made? c) If the value remaining on the card is $.00, how many copies have been charged to the card? 7. The relationship between total cost and number of units made is linear. Suppose cost increases by $3 for each additional unit made and the total cost of 10 units is $40, find the equation of the relationship between total cost (y) and number of units made (x). 8. Assume that the relationship between the number of units manufactured and the manufacturing cost is linear. As the number of units increases from 100 to 00, the manufacturing cost increases from $350 to $650. Find the equation of the relationship between cost (y) and number of units made (x). Joseph F. Aieta Page 18 Printed 8/14/003
Linear Models Activities 1 Page 19 9. As sales (x) change from $100 to $400, selling expense (y) changes from $75 to $150. Assume that the given data establish a linear relationship between sales and selling expense, state this relationship in the form of a linear equation. 10. If x represents sales and y represents selling expense, then (30, ) would mean $30 in sales were accompanied by $ of selling expense. Suppose that last month s figures were (14, 10) and this month s are (30, ). a) How much did sales increase? b) How much did selling expense increase? c) Make a graph showing the points and label the change in x and the change in y. 11. a) If a taxi fare (y) is 50 cents plus 0 cents per quarter mile, write the equation relating fare to number of miles traveled, x. b) The gross weekly earnings of a salesman are $50 plus 10 percent of the retail value of the goods she sells. Write the equation for earnings, E, in terms of sales volume, V. What is the common name for the slope of this line? 1. A salesperson receives a base pay of $1,000 per month plus a.5 % commission on all sales. a) If sales in one month were $60,000 then what was the total pay for that month? b) If total pay was $3,50 then what was the sales volume for that month? c) and d) Answer questions a) and b) if the base salary was unchanged at $1000 but the commission structure was.5 % commission only on sales over $40,000. 13. Colonial Gas Company charges its customers according to their usage of gas as follows: An initial $6.00 customer charge $1.180 per unit (100 cubic feet) for the first 0 units and $0.806 pr unit for each unit over 0. a) Determine the cost function and draw its graph. b) What are the total and average charges for using 15 units? c) What are the total and average charges for using 50 units? d) How many units were used if the total charge is $73.93? Joseph F. Aieta Page 19 Printed 8/14/003
Linear Models Activities 1 Page 0 14. The tax rates shown below are from the IRS 003 tax table 003 Married Filing Jointly If Taxable Income Of the is Over But Not > The Tax is: Amount > 0 1,000 0 + 10.0% 0 1,000 47,450 1,00 + 15.0% 1,000 47,450 114,650 6,518 + 7.0% 47,450 114,650 174,700 4,66 + 30.0% 114,650 174,700 311,950 4,667 + 35.0% 174,700 311,950 --- 90,714 + 38.6% 311,950 a) How much tax would a couple pay on a Taxable Income of $70,000? b) How much tax would a couple pay on a Taxable Income of $150,000? c) What was the couple s Taxable Income if the 003 tax is $3,900? 15. Two fee schedules apply for each item listed and sold on a popular online auction Website. The first fee is called an Insertion Fee and is charged to individuals who place a regular listing. The calculation of the second fee, called the Final Value Fee is based upon the final sale price. The Insertion Fee formula for a regular listing is shown in the following table. Opening Value or Minimum Price Insertion Fee $0.01 - $9.99 $0.5 $10.00 - $4.99 $0.50 $5.00 - $49.99 $1.00 $50.00 and up $.00 a) Sketch the graph of this function with initial price on the horizontal axis and the Insertion Fee on the vertical axis. b) Explain why this graph is described as a step function. c) Refer back to the previous problem #14. Would taxpayers have a legitimate objection if a graduated tax formula was a step function of adjusted gross income (AGI)? Explain. 16. Start with a blank worksheet and create an interactive linear depreciation worksheet, like the table and graph, pictured below. It should be designed in such a way that the rate of depreciation, the table, and the graph change automatically when the entries in the shaded cells for cost, and scrap value are changed. Assume that useful life is fixed at 10 years. annual cost scrap value useful_life depreciation $80,000 $4,000 10 $7,600.00 residual year value $100,000 0 $80,000 $90,000 1 $7,400 $80,000 $64,800 $70,000 3 $57,00 4 $49,600 $60,000 5 $4,000 $50,000 6 $34,400 $40,000 7 $6,800 $30,000 8 $19,00 9 $11,600 $0,000 10 $4,000 $10,000 residual value $0 0 1 3 4 5 6 7 8 9 10 years Joseph F. Aieta Page 0 Printed 8/14/003
Linear Models Part Section 1.4 Page 1 Linear Models Part Cost, Revenue, Profit, and Break-even Section 1.4 In this course we examine many situations that involve cost, revenue and profit. When the models for cost, revenue, or profit can be described by linear functions then the strategies for solving problems are the basic algebraic and geometric techniques for manipulating linear equations and drawing the graphs of lines. Example 1.4.1 CapeCodTees is a summer business run by college students. They sell t-shirts imprinted with slogans and images associated with the resort area. The owners of the business separate their costs into fixed cost and variable cost. Fixed cost includes such things as the rental of space, equipment and other costs that are incurred before any t-shirts are bought from wholesalers to be imprinted and sold. They estimate that their fixed costs will be about $,000. Variable cost is related to the cost of the plain t- shirts that they buy and the cost of the materials for imprinting each image. They estimate that variable costs will be about $6 per t-shirt. They do not include their labor as a cost since they plan to split the profits when the season is over. They will price the t-shirts at $1.50 per shirt. They believe that the maximum number of t-shirts that they can sell in a summer with good weather is around 800. If the weather is bad then sales may be significantly less. They want to know the minimum number of t-shirts that they must sell to break even. Total cost, revenue, and profit are each functions of quantity, which we will denote by q. Profit equals revenue minus cost. This can be written as P(q) = R(q) C(q) using function notation. A rough picture of the cost function looks like Figure 1.4.1 and the revenue function looks like Figure 1.4.. Cost Revenue C(q) R(q) F q q Figure 1.4.1 Figure 1.4. Several important concepts are apparent from these rough, unlabeled sketches. The cost function cuts the vertical axis at point F units above the origin since the business has fixed costs of F dollars even if no t-shirts are bought and sold. When quantity q is zero, total cost is the fixed cost, C(0) = F. There is no revenue at q = 0 so R(0) = 0 and consequently the profit at q = 0 will be negative, P(0) = R(0) C(0) = -F and the point (0, -F) on the profit function is located F units directly below the origin (0, 0). The owners of the business want to be confident that they can eventually sell a sufficient number of t-shirts in order for revenue to exceed cost. Profit must move into positive territory, at least far enough to cover the fixed cost. It should also be apparent that the slope of the revenue line in Figure 1.4. must be greater than the slope of the cost line of Figure 1.4.1 otherwise revenue could never overtake cost no matter how many t-shirts are sold and the business would always lose money. Joseph F. Aieta Page 1 Printed 8/14/003
Linear Models Part Section 1.4 Page Cost, Revenue, and Profit As long as the slope of the revenue line (price per shirt), is greater than the slope of the cost line (variable cost per shirt) the two lines will eventually cross. If the point of intersection where R(q) = C(q) is beyond any feasible sales volume then the owners will not have any profit to share. For example, if breakeven occurs at q = 1000, and there is very low probability that the business can sell that many t-shirts in one season, then the owners will have to find a way to raise prices, lower costs, or both. If this can t be accomplished then they will not have a profitable business this season. F R(q) C(q) P(q) q Figure 1.4.3 Figure 1.4.3 shows a rough sketch of cost, revenue, and profit functions, all on the same pair of axes, where quantity is the measure on the horizontal axis and dollars is the measure on the vertical axis. We define break-even as the point where revenue = cost or where profit = 0. In function notation, break-even quantity is that value of q such that R(q) = C(q). Finding the break-even quantity is equivalent to solving P (q) = 0. The difference between the slopes of the revenue and cost functions, relative to the size of fixed cost is what ultimately determines the break-even quantity. In Words Revenue is $1.50 times the number of t-shirts sold. Total cost is fixed cost plus variable cost $000 plus $6.00 per shirt. Profit is revenue minus cost. Function notation R(q) = 1.50 q C(q) = 000 + 6.00q P(q) = R(q) C(q) P(q) = 1.50 q (000+ 6.00q) P(q) = (1.50 6.00) q 000 P(q) = 6.50 q 000 In order for the company owners to get accurate graphs of the cost, revenue, and profit functions as they relate to the quantity of t-shirts imprinted and sold, they construct tables in Excel. They also use Excel s Chart Wizard to produce the graphs of the cost, revenue, and profit functions on the same coordinate axes as shown in Figure 1.4.4. Joseph F. Aieta Page Printed 8/14/003
Linear Models Part Section 1.4 Page 3 quantity Cost Revenue Profit q C R P 0 000 0 ($,000) 5 150 31.5 ($1,838) 50 300 65 ($1,675) 75 450 937.5 ($1,513) 100 600 150 ($1,350) 15 750 156.5 ($1,188) 150 900 1875 ($1,05) 175 3050 187.5 ($863) 00 300 500 ($700) 5 3350 81.5 ($538) 50 3500 315 ($375) 75 3650 3437.5 ($13) 300 3800 3750 ($50) 35 3950 406.5 $113 350 4100 4375 $75 375 450 4687.5 $438 400 4400 5000 $600 45 4550 531.5 $763 450 4700 565 $95 Break-Even Analysis C R P $7,000 $6,000 $5,000 $4,000 $3,000 $,000 $1,000 $0 -$1,000 0 100 00 300 400 500 600 -$,000 -$3,000 quantity Figure 1.4.4 The table in Figure 1.4.4 reveals that the break-even quantity is near 300. Determining when profit is equal to zero is equivalent to finding the value of q such that P(q) = 0. We use basic algebra to find the number of t-shirts where revenue is equal to cost. We will use basic algebra to solve this equation. Since P(q) = 6.50 q 000 we must solve 6.50 q 000 = 0 6.50 q = 000 q = 000/6.50 307.7 [In later examples we will use the Excel tools Goal Seek and Solver to solve equations.] If the business sells 307 t-shirts or fewer, it will be in the red (losing money). The business begins to make a profit with the sale of the 308 th t-shirt. For cost, revenue, and profit situations, economists refer to the cost of making one more unit as the marginal cost of that unit. Similarly, the additional revenue from selling one more unit is called marginal revenue and the additional profit from making and selling one more unit is called the marginal profit. When cost and quantity have a linear relationship, the marginal cost is constant. This means that the marginal cost does not depend on the quantity produced. It will be the same for 100 units as for 00 units or for 307 units. If you were to calculate C(101) C(100), C(01) C(00), or C(308) C(307) in the t-shirt example you would get the same value each time. It is identical to the slope of the cost function. In other words, if q units have been made then the marginal cost of making q + 1 unit is the variable cost per unit, regardless of the particular value of q. Similar statements apply to linear revenue and profit functions. For linear cost functions, marginal cost equals variable cost per unit Only for linear cost functions is marginal cost a constant. Chapter 5 will help us estimate marginal cost for non-linear cost functions using calculus. Joseph F. Aieta Page 3 Printed 8/14/003
Linear Models Part Section 1.4 Page 4 Example 1.4. Comparison of Profit Lines. Suppose CapeCodTees can obtain its supplies more cheaply from another wholesale outfit and thereby reduce its variable cost to $5.50 per shirt. While the quality of the t-shirts is the same, the catch is that the new distributor wants an up-front fee of $00 to cover transportation costs for the season. CapeCodTees would add this $00 to their fixed cost and see how these changes would impact profit and break-even. Their original decision is to stay with the original supplier if they have to raise their unit price above $1.50 in order to be profitable when q = 308. Original Supplier New supplier R(q) = 1.50 q R(q) = 1.50 q (unchanged) C(q) = 000 + 6.00q C(q) = 00 + 5.50q P(q) = R(q) C(q) P(q) = R(q) C(q) P(q) = 1.50 q (000 + 6.00q) P(q) = 1.50 q (00+ 5.50q) P(q) = (1.50 6.00) q 000 P(q) = (1.50 5.50) q 00 P(q) = 6.50 q 000 P(q) = 7.00 q 00 Solve for q and get q = 000/(6.50) = 307.693 Profit turns positive with the 308 th t-shirt sold Solve for q and get q = 00/7.00 = 314.8 Profit turns positive with the 315 th t-shirt sold Under their original criteria, they would stick with the original supplier. Then they start to wonder whether or not they might make more money by the end of the season if they switched to the new supplier. They look back at previous seasons and realize that even in their poorest summer, they never sold fewer than then 500 t-shirts. At q = 500 they compare P(500) = 6.50(500) 000 = $1,50 versus 7.00(500) 00 = $1,300 and decide to go with the new supplier. Applying additional algebra, they find that they are better off with the new supplier if they can sell more than 400 t-shirts. The break-even concept can occur in other contexts that involve linear functions. An individual may have two or more offers and needs to make a decision about which option is optimal. The decision in example 1.4.3 is to select the best copy shop. Example 1.4.3 You have just received bids from two printing companies for the job of printing copies of a travel brochure for your agency. Speedy Printers charges a fee of $00 plus 75 cents per copy. Quality Printers charges a fee of $150 plus 90 cents per copy a) Plot the cost functions for both print shops on the same coordinate axes with the number of copies on the horizontal axis and the cost on the vertical axis. b) For what range of copies does Quality Printers offer the best deal? c) For what range of copies does Speedy Printers offer the best deal? Joseph F. Aieta Page 4 Printed 8/14/003
Linear Models Part Section 1.4 Page 5 Figure 1.4.5 contains a cost versus quantity table and graph from the Excel file Speedy vs Quality.xls. By creating a column for the difference between the costs of using Speedy Printers minus the cost of using Quality Printers, the agency can see how much is saved by using a particular print shop rather than the other. The agency can then decide if the savings is worthwhile in light of other considerations, such as convenience. Figure 1.4.5 When the difference d = Speedy Quality is positive then choosing Quality Printers saves the agency d dollars. When the difference d = Speedy Quality is negative then choosing Speedy saves the agency d dollars where is the absolute value function. For example, if only 00 copies are printed then choosing Quality will save the agency $0.00 but if 400 copies are printed then choosing Speedy will save the agency $10. The breakpoint occurs between 35 and 350 copies. We will use Goal Seek to find the precise break point as follows: 1. The entries in column A are numerical values from patterned data and cannot contain any formulas. See Tutorials 1 and in Appendix A for the steps to create patterned data with Excel.. Highlight cell D0. 3. From the Tools menu, select Goal Seek. 4. With D0 as the Set cell, you will search for the value that will make D0 zero by changing the numeric value in cell A0. Don t forget to click OK Goal Seek changes cell A0 to 333.33 We can also find this breakpoint with pencil and paper by solving 00 + 0.75 x = 150 + 0.90 x to get x = 50/0.15 = 333.33 We conclude that Quality Printers is more cost effective for 333 or fewer copies, but Speedy Printers is the better deal for 334 copies or more. See Tutorial 3 in Appendix A for other applications of Goal Seek. Joseph F. Aieta Page 5 Printed 8/14/003
Linear Models Part Section 1.5 Page 6 Iso-value lines in two variables Section 1.5 The origin of the prefix term iso comes from a Greek word meaning same so iso-profit means the same profit, iso-cost means the same cost, iso-dollars means the same amount of money. All iso-value lines for two variables are of the form ax + by = c. For equations in this form, we can quickly determine the two intercepts and the slope. For example, if the iso-profit line is given by 4x + 3y = 480 then the y- intercept is (0, 160), the x-intercept is (10, 0) and the slope is -4/3. Example 1.5.1 Iso-profit lines After several successful seasons, the owners of a t-shirt business decide to add caps to their sales line. They observe that some of their customers come into the shop and leave with caps and no t-shirts. Other customers buy t-shirts and no caps, and some customers buy both. In the current season the owners have been able to raise their t-shirt prices without any significant impact on the number of t-shirts sold in a season. Currently they make a profit of $8.00 for each t-shirt sold and $6.50 for each cap sold. In one season they had a gross profit of $080 to the nearest dollar. The equation of this profit line can be written in general form as 8.00x + 6.50y=,080 where x is the number of t-shirts sold and y is the number of caps sold. In a season with a profit of $3,10.00 the profit line has equation 8.00 x + 6.50 y =3,10. For a profit of $4,160 the profit line is 8.00 x + 6.50 y = 4,160. If we draw the graph of each profit line on the same coordinate axis, as shown in Figure 1.5.1, we observe that each line in this family of profit lines is parallel to each other line. 700 600 iso-profit lines The slope of this family of profit lines is -8.00 divided by 6.50 which are approximately -1.3. Each profit line has an equation of the form ax + by = c where c represents profit. The larger the value of c, the further the line is from the origin and the larger the values of the x and y intercepts. For the $080 iso-profit line, the intercepts are (0, 30) and (60, 0). For the $3,10 iso-profit line the intercepts are (0,480) and (390, 0). Similarly the $4,160 iso-profit line has intercepts (0,640) and (50, 0). caps 500 400 300 00 100 0 0 100 00 300 400 500 600 t-shirts Figure 1.5.1 Note that the line closest to the origin corresponds to the lowest profit and that the line furthest from the origin corresponds to the lowest profit. Q1. Name two other points on each iso-profit line. A. The points (65, 40) and (130,160) are on the profit line 8.00 x + 6.50 y =,080 The points (65, 400) and (130, 30) are on the profit line 8.00 x + 6.50 y = 3,10 The points (65, 560) and (130, 480) are on the profit line 8.00 x + 6.50 y = 4,160 Joseph F. Aieta Page 6 Printed 8/14/003