1. Precalculus Fifth Edition Mathematics for Calculus James Stewart Lothar Redlin Saleem Watson

2. Fundamentals 1

3. Modeling with Equations 1.6

4. Modeling with Equations Many problems in the sciences, economics, finance, medicine, and numerous other fields can be translated into algebra problems. This is one reason that algebra is so useful.

5. Modeling with Equations In this section, we use equations as mathematical models to solve real-life problems.

6. Guidelines for Modeling with Equations

7. We will use the following guidelines to help us set up equations that model situations described in words. 1.Identify the variable. 2.Express all unknown quantities in terms of the variable. 3.Set up the model. 4.Solve the equation and check your answer.

8. Guideline 1 Identify the variable. Identify the quantity that the problem asks you to find. This quantity can usually be determined by a careful reading of the question posed at the end of the problem. Then, introduce notation for the variable. Call this x or some other letter.

9. Guideline 2 Express all unknown quantities in terms of the variable. Read each sentence in the problem again, and express all the quantities mentioned in the problem in terms of the variable you defined in Step 1. To organize this information, it is sometimes helpful to draw a diagram or make a table.

10. Guideline 3 Set up the model. Find the crucial fact in the problem that gives a relationship between the expressions you listed in Step 2. Set up an equation (or model) that expresses this relationship.

11. Guideline 4 Solve the equation and check your answer. 1.Solve the equation. 2.Check your answer. 3.Express it as a sentence that answers the question posed in the problem.

12. Guidelines for Modeling with Equations The following example illustrates how these guidelines are used to translate a “word problem” into the language of algebra.

13. E.g. 1—Renting a Car A car rental company charges $30 a day and 15¢ a mile for renting a car. Helen rents a car for two days and her bill comes to $108. How many miles did she drive?

14. E.g. 1—Renting a Car We are asked to find the number of miles Helen has driven. So, we let: x = number of miles driven

15. E.g. 1—Renting a Car Then, we translate the information given in the problem into the language of algebra. In WordsIn Algebra Number of miles drivenx Mileage cost (at $0.15 per mile) 0.15x Daily cost (at $30 per day)2(30)

16. E.g. 1—Renting a Car Now, we set up the model. Mileage Cost + Daily Cost = Total Cost

17. E.g. 1—Renting a Car Helen drove her rental car 320 miles.

18. Constructing Models

19. In the examples and exercises that follow, we construct equations that model problems in many different real-life situations. Constructing Models

20. E.g. 2—Interest on an Investment Mary inherits $100,000 and invests it in two certificates of deposit. One certificate pays 6% and the other pays 4½% simple interest annually. If Mary’s total interest is $5025 per year, how much money is invested at each rate?

21. E.g. 2—Interest on an Investment The problem asks for the amount she has invested at each rate. So, we let: x = the amount invested at 6% As Mary’s total inheritance is $100,000, it follows that she invested 100,000 – x at 4½%.

22. E.g. 2—Interest on an Investment In WordsIn Algebra Amount invested at 6%x Amount invested at 4½%100,000 – x Interest earned at 6%0.06x Interest earned at 4½%0.045(100,000 – x) We translate all the information given into the language of algebra.

23. E.g. 2—Interest on an Investment We use the fact that Mary’s total interest is $5025 to set up the model. Interest at 6% + Interest at 4½% = Total Interest

24. E.g. 2—Interest on an Investment 0.06x + 0.045(100,000 – x) = 5025 0.06x + 4500 – 0.045x = 5025 (Multiply) 0.015x + 4500 = 5025 (Combine x-terms) 0.015x = 525

25. E.g. 2—Interest on an Investment Thus, So, Mary has invested $35,000 at 6% and the remaining $65,000 at 4½ %.

26. Check Your Answer Total Interest = 6% of $35,000 + 4½% of $65,000 = $2100 + $2925 = $5025

27. E.g. 3—Dimensions of a Poster A poster has a rectangular printed area 100 cm by 140 cm, and a blank strip of uniform width around the four edges. The perimeter of the poster is 1½ times the perimeter of the printed area. What is the width of the blank strip? What are the dimensions of the poster?

28. E.g. 3—Dimensions of a Poster We are asked to find the width of the blank strip. So, we let: x = the width of the blank strip

29. E.g. 3—Dimensions of a Poster Then, we translate the information in this figure into the language of algebra as follows.

30. E.g. 3—Dimensions of a Poster In WordsIn Algebra Width of blank stripx Perimeter of printed area2(100) + 2(140) = 480 Width of poster100 + 2x Length of poster140 + 2x Perimeter of poster2(100 + 2x) + 2(140 +2x)

31. E.g. 3—Dimensions of a Poster Now, we use the fact that the perimeter of the poster is 1½ times the perimeter of the printed area to set up the model. Perimeter of Poster =. Perimeter of Printed Area

32. E.g. 3—Dimensions of a Poster

33. The blank strip is 30 cm wide. So, the dimensions of the poster are: 100 + 30 + 30 = 160 cm wide by 140 + 30 + 30 = 200 cm long

34. E.g. 4—Dimensions of a Building Lot A rectangular building lot is 8 ft longer than it is wide and has an area of 2900 ft 2. Find the dimensions of the lot.

35. E.g. 4—Dimensions of a Building Lot We are asked to find the width and length of the lot. So, let: w = width of lot

36. E.g. 4—Dimensions of a Building Lot Then, we translate the information in the problem into the language of algebra. In WordsIn Algebra Width of lotw Length of lotw + 8

37. E.g. 4—Dimensions of a Building Lot Now, we set up the model. Width of Lot. Length of Lot = Area of Lot

38. E.g. 4—Dimensions of a Building Lot w(w + 8) = 2900 w 2 + 8w = 2900 (Expand) w 2 + 8w – 2900 = 0 (w – 50)(w + 58) = 0 (Factor) w = 50 or w = –58 (Zero-Product Property)

39. E.g. 4—Dimensions of a Building Lot Since the width of the lot must be a positive number, we conclude that: w = 50 ft The length of the lot is: w + 8 = 50 + 8 = 58 ft

40. E.g. 5—Height of Building Using Similar Triangles A man 6 ft tall wishes to find the height of a certain four-story building. He measures its shadow and finds it to be 28 ft long, while his own shadow is 3½ ft long. How tall is the building?

41. E.g. 5—Height of Building Using Similar Triangles The problem asks for the height of the building. So, let: h = the height of the building

42. E.g. 5—Height of Building Using Similar Triangles We use the fact that the triangles in the figure are similar. Recall that, for any pair of similar triangles, the ratios of corresponding sides are equal.

43. E.g. 5—Height of Building Using Similar Triangles Now, we translate the observations into the language of algebra. In WordsIn Algebra Height of buildingh Ratio of height to base in large triangleh/28 Ratio of height to base in small triangle6/3.5

44. E.g. 5—Height of Building Using Similar Triangles Since the large and small triangles are similar, we get: Ratio of height to base in large triangle = Ratio of height to base in small triangle The building is 48 ft tall.

45. E.g. 6—Mixtures and Concentration A manufacturer of soft drinks advertises their orange soda as “naturally flavored,” although it contains only 5% orange juice. A new federal regulation stipulates that, to be called “natural,” a drink must contain at least 10% fruit juice. How much pure orange juice must this manufacturer add to 900 gal of orange soda to conform to the new regulation?

46. E.g. 6—Mixtures and Concentration The problem asks for the amount of pure orange juice to be added. So, let: x = the amount (in gallons) of pure orange juice to be added

47. E.g. 6—Mixtures and Concentration In any problem of this type—in which two different substances are to be mixed— drawing a diagram helps us organize the given information.

48. E.g. 6—Mixtures and Concentration

49. In WordsIn Algebra Amount of orange juice to be addedx Amount of the mixture900 + x Amount of orange juice in the first vat0.05(900) = 45 Amount of orange juice in the second vat 1 ∙ x = x Amount of orange juice in the mixture0.10(900 + x) We now translate the information in the figure into the language of algebra.

50. E.g. 6—Mixtures and Concentration To set up the model, we use the fact that the total amount of orange juice in the mixture is equal to the orange juice in the first two vats. Amount of orange juice in first vat + Amount of orange juice in second vat = Amount of orange juice in mixture

51. E.g. 6—Mixtures and Concentration The manufacturer should add 50 gal of pure orange juice to the soda.

52. Check Your Answer Amount of juice before mixing = 5% of 900 gal + 50 gal pure juice = 45 gal + 50 gal = 95 gal Amount of juice after mixing = 10% of 950 gal = 95 gal

53. E.g. 7—Time Needed to Do a Job Due to an anticipated heavy rainstorm, the water level in a reservoir must be lowered by 1 ft. Opening spillway A does the job in 4 hours. Opening the smaller spillway B does it in 6 hours.

54. E.g. 7—Time Needed to Do a Job How long will it take to lower the water level by 1 ft if both spillways are opened?

55. E.g. 7—Time Needed to Do a Job We are asked to find the time needed to lower the level by 1 ft if both spillways are open. So, let: x = the time (in hours) it takes to lower the water level by 1 ft if both spillways are open

56. E.g. 7—Time Needed to Do a Job Finding an equation relating x to the other quantities in this problem is not easy. Certainly, x is not simply 4 + 6. That would mean that, together, the two spillways require longer to lower the water level than either spillway alone.

57. E.g. 7—Time Needed to Do a Job In WordsIn Algebra Time it takes to lower level 1 ft with A and B together x h Distance A lowers level in 1 h ft Distance B lowers level in 1 h ft Distance A and B together lower levels in 1 h ft Instead, we look at the fraction of the job that can be done in one hour by each spillway.

58. E.g. 7—Time Needed to Do a Job Now, we set up the model. Fraction done by A + Fraction done by B = Fraction done by both

59. E.g. 7—Time Needed to Do a Job It will take hours, or 2 h 24 min, to lower the water level by 1 ft if both spillways are open.

60. Distance, Speed, and Time The next example deals with distance, rate (speed), and time. The formula to keep in mind here is: distance = rate x time where the rate is either the constant speed or average speed of a moving object. For example, driving at 60 mi/h for 4 hours takes you a distance of 60 ∙ 4 = 240 mi.

61. E.g. 8—A Distance-Speed-Time Problem A jet flew from New York (NY) to Los Angeles (LA), a distance of 4,200 km. The speed for the return trip was 100 km/h faster than the outbound speed. If the total trip took 13 hours, what was the jet’s speed from NY to LA?

62. E.g. 8—A Distance-Speed-Time Problem We are asked for the speed of the jet from NY to LA. So, let: s = speed from NY to LA Then, s + 100 = speed from LA to NY

63. E.g. 8—A Distance-Speed-Time Problem Now, we organize the information in a table. First, we fill in the “Distance” column—as we know that the cities are 4,200 km apart.

64. E.g. 8—A Distance-Speed-Time Problem Then, we fill in the “Speed” column—as we have expressed both speeds (rates) in terms of the variable s.

65. E.g. 8—A Distance-Speed-Time Problem Finally, we calculate the entries for the “Time” column, using:

66. E.g. 8—A Distance-Speed-Time Problem The total trip took 13 hours. So, we have the model: Time from NY to LA + Time from LA to NY = Total time This gives:

67. E.g. 8—A Distance-Speed-Time Problem Multiplying by the common denominator, s(s + 100), we get: Although this equation does factor, with numbers this large, it is probably quicker to use the quadratic formula and a calculator.

68. E.g. 8—A Distance-Speed-Time Problem As s represents speed, we reject the negative answer and conclude that the jet’s speed from NY to LA was 600 km/h.

69. E.g. 9—Energy Expended in Bird Flight Ornithologists have determined that some species of birds tend to avoid flights over large bodies of water during daylight hours. Air generally rises over land and falls over water in the daytime. So, flying over water requires more energy.

70. E.g. 9—Energy Expended in Bird Flight A bird is released from point A on an island, 5 mi from B, the nearest point on a straight shoreline. It flies to a point C on the shoreline. Then, it flies along the shoreline to its nesting area D.

71. E.g. 9—Energy Expended in Bird Flight Suppose it has 170 kcal of energy reserves. It uses 10 kcal/mi flying over land. It uses 14 kcal/mi flying over water.

72. E.g. 9—Energy Expended in Bird Flight (a)Where should C be located so that the bird uses exactly 170 kcal of energy during its flight? (b)Does it have enough energy reserves to fly directly from A to D?

73. E.g. 9—Energy in Bird Flight We are asked to find the location of C. So, let: x = distance from B to C Example (a)

74. E.g. 9—Energy in Bird Flight From the figure, and from the fact that energy used = energy per mile x miles flown we determine the following. Example (a)

75. E.g. 9—Energy in Bird Flight In WordsIn Algebra Distance from B to Cx Distance flown over water (from A to C) Distance flown over land (from C to D)12 – x Energy used over water Energy used over land10(12 – x) Example (a)

76. E.g. 9—Energy in Bird Flight Now, we set up the model. Total energy used = Energy used over water + Energy used over land Example (a)

77. E.g. 9—Energy in Bird Flight Thus, To solve this equation, we eliminate the square root by first bringing all other terms to the left of the equal sign and then squaring each side. Example (a)

78. E.g. 9—Energy in Bird Flight Example (a)

79. E.g. 9—Energy in Bird Flight Example (a) This equation could be factored. However, as the numbers are so large, it is easier to use the quadratic formula and a calculator.

80. E.g. 9—Energy in Bird Flight Example (a) C should be either mi or mi from B so that the bird uses exactly 170 kcal of energy during its flight.

81. E.g. 9—Energy in Bird Flight By the Pythagorean Theorem, the length of the route directly from A to D is: Example (b)

82. E.g. 9—Energy in Bird Flight Thus, the energy the bird requires for that route is: 14 x 13 = 182 kcal This is more energy than the bird has available. So, it can’t use this route. Example (b)