1. Functions 2

2. Modeling with Functions 2.6

3. Modeling Many processes studied in the physical and social sciences involve understanding how one quantity varies with respect to another. Finding a function that describes the dependence of one quantity on another is called modeling.

4. Modeling For example, a biologist observes that the number of bacteria in a certain culture increases with time. He tries to model this phenomenon by finding the precise function (or rule) that relates the bacteria population to the elapsed time.

5. Modeling In this section, we will learn how to find models that can be constructed using geometric or algebraic properties of the object under study. Once the model is found, we use it to analyze and predict properties of the object or process being studied.

6. Modeling with Functions

7. We begin with a simple real-life situation that illustrates the modeling process.

8. E.g. 1—Modeling the Volume of a Box A breakfast cereal company manufactures boxes to package their product. For aesthetic reasons, the box must have the following proportions: Width is 3 times its depth. Height is 5 times its depth.

9. E.g. 1—Modeling the Volume of a Box (a) Find a function that models the volume of the box in terms of its depth. (b) Find the volume of the box if the depth is 1.5 in.

10. E.g. 1—Modeling the Volume of a Box (c) For what depth is the volume 90 in 3 ? (d) For what depth is the volume greater than 60 in 3 ?

11. E.g. 1—Thinking About the Problem Let’s experiment with the problem. If the depth is 1 in, then the width is 3 in, and the height is 5 in. So, in this case, the volume is: V = 1 x 3 x 5 = 15 in 3

12. E.g. 1—Thinking About the Problem The table gives other values.

13. E.g. 1—Thinking About the Problem Notice that: All the boxes have the same shape. The greater the depth, the greater the volume.

14. E.g. 1—Modeling Volume of a Box To find the function that models the volume of the box, we use the following steps. 1.Express the model in words. 2.Choose the variable. 3.Set up the model. 4.Use the model. Example (a)

15. E.g. 1—Modeling Volume of a Box 1. Express the model in words. We know the volume of a rectangular box is: Volume = Depth x Width x Height Example (a)

16. E.g. 1—Modeling Volume of a Box 2. Choose the variable. There are three varying quantities—width, depth, and height. Since the function we want depends on the depth, we let: x = depth of the box Example (a)

17. E.g. 1—Modeling Volume of a Box Then, we express the other dimensions of the box in terms of x. Example (a) In WordsIn Algebra Depthx Width3x3x Height5x5x

18. E.g. 1—Modeling Volume of a Box 3. Set up the model. The model is the function V that gives the volume of the box in terms of the depth x. Volume = Depth x Width x Height V(x) = x · 3x · 5x V(x) = 15x 3 Example (a)

19. E.g. 1—Modeling Volume of a Box The function V is graphed here. Example (a)

20. E.g. 1—Modeling Volume of a Box 4. Use the model. We use the model to answer the questions in parts (b), (c), and (d), as follows.

21. E.g. 1—Modeling Volume of a Box If the depth is 1.5 in., the volume is: V(1.5) = 15(1.5) 3 in 3 = 50.625 in 3 Example (b)

22. E.g. 1—Modeling Volume of a Box We need to solve the equation V(x) = 90 or The volume is 90 in 3 when the depth is about 1.82 in. Example (c)

23. E.g. 1—Modeling Volume of a Box We can also solve this equation graphically. Example (c)

24. E.g. 1—Modeling Volume of a Box We need to solve the inequality V(x) > 60 or The volume will be greater than 60 in 3 if the depth is greater than 1.59 in. Example (d)

25. E.g. 1—Modeling Volume of a Box We can also solve this inequality graphically. Example (d)

26. Modeling with Functions The steps in Example 1 are typical of how we model with functions. They are summarized as follows.

27. Guidelines for Modeling with Functions 1.Express the model in words. 2.Choose the variable. 3.Set up the model. 4.Use the model.

28. Step 1 for Modeling with Functions Express the model in words. Identify the quantity you want to model and express it, in words, as a function of the other quantities in the problem.

29. Step 2 for Modeling with Functions Choose the variable. Identify all the variables used to express the function in Step 1. Assign a symbol, such as x, to one variable and express the other variables in terms of this symbol.

30. Step 3 for Modeling with Functions Set up the model. Express the function in the language of algebra by writing it as a function of the single variable chosen in Step 2.

31. Step 4 for Modeling with Functions Use the model. Use the function to answer the questions posed in the problem. To find a maximum or a minimum, use the algebraic or graphical methods described in Section 2-5.

32. E.g. 2—Fencing a Garden A gardener has 140 feet of fencing to fence in a rectangular vegetable garden. (a) Find a function that models the area of the garden she can fence. (b) For what range of widths is the area greater than or equal to 825 ft 2 ?

33. E.g. 2—Fencing a Garden (c) Can she fence a garden with area 1250 ft 2 ? (d) Find the dimensions of the largest area she can fence.

34. E.g. 2—Thinking About the Problem If the gardener fences a plot with width 10 ft, the length must be 60 ft, because: 10 + 10 + 60 + 60 = 140 So, the area is: A = Width x Length = 10 · 60 = 600 ft 2

35. E.g. 2—Thinking About the Problem The table shows various choices for fencing the garden. We see that, as the width increases, the fenced area increases, then decreases.

36. E.g. 2—Fencing a Garden The model we want is: A function that gives the area she can fence. Example (a)

37. E.g. 2—Fencing a Garden 1. Express the model in words. We know that the area of a rectangular garden is: Area = Width x Length Example (a)

38. E.g. 2—Fencing a Garden 2. Choose the variable. There are two varying quantities—width and length. Since the function we want depends on only one variable, we let: x = width of the garden Example (a)

39. E.g. 2—Fencing a Garden Then, we must express the length in terms of x. The perimeter is fixed at 140 ft. So, the length is determined once we choose the width. Example (a)

40. E.g. 2—Fencing a Garden If we let the length be l as shown, then 2x + 2 l = 140 So, l = 70 – x Example (a)

41. E.g. 2—Fencing a Garden We summarize these facts. Example (a) In WordsIn Algebra Widthx Length70 – x

42. E.g. 2—Fencing a Garden 3. Set up the model. The model is the function A that gives the area of the garden for any width x. Area = Width x Length A(x) = x(70 – x) A(x) = 70x – x 2 Example (a)

43. E.g. 2—Fencing a Garden 4. Use the model. Again, we use the model to answer the questions in parts (b)–(d).

44. E.g. 2—Fencing a Garden We need to solve the inequality A(x) ≥ 825 To solve graphically, we graph y = 70x – x 2 and y = 825 in the same viewing rectangle. We see that: 15 ≤ x ≤ 55 Example (b)

45. E.g. 2—Fencing a Garden We see that the graph of A(x) always lies below the line y = 1250. So, an area of 1250 ft 2 is never attained. Example (c)

46. E.g. 2—Fencing a Garden We need to find the maximum value of the function A(x) = 70x – x 2. Since this is a quadratic function with a = –1 and b = 70, the maximum occurs at: So, the maximum area that she can fence has width 35 ft and length 70 – 35 = 35 ft. Example (d)

47. E.g. 3—Maximizing Revenue from Ticket Sales A hockey team plays in an arena with a seating capacity of 15,000 spectators. With the ticket price set at $14, average attendance at recent games has been 9,500. A market survey indicates that, for each dollar the ticket price is lowered, the average attendance increases by 1,000.

48. E.g. 3—Maximizing Revenue from Ticket Sales (a)Find a function that models the revenue in terms of ticket price. (b) What ticket price is so high that no one attends—and hence no revenue is generated? (c) Find the price that maximizes revenue from ticket sales.

49. E.g. 3—Thinking About the Problem With a ticket price of $14, the revenue is: 9,500 x $14 = $133,000 If the ticket price is lowered to $13, attendance increases to 9,500 + 1000 = 10,500. So, the revenue becomes: 10,500 x $13 = $136,500

50. E.g. 3—Thinking About the Problem The table shows the revenue for several ticket prices. Note that: If the ticket price is lowered, revenue increases. If it is lowered too much, revenue decreases.

51. E.g. 3—Max. Rev. from Ticket Sales The model we want is: A function that gives the revenue for any ticket price. Example (a)

52. E.g. 3—Max. Rev. from Ticket Sales 1. Express the model in words. We know that: Revenue = Ticket price x Attendance Example (a)

53. E.g. 3—Max. Rev. from Ticket Sales 2. Choose the variable. There are two varying quantities—ticket price and attendance. Since the function we want depends on price, we let: x = ticket price Example (a)

54. E.g. 3—Max. Rev. from Ticket Sales Next, we must express the attendance in terms of x. Example (a) In WordsIn Algebra Ticket pricex Amount ticket price is lowered 14 – x Increase in attendance 1000(14 – x) Attendance9500 + 1000(14 – x) = 23,500 – 10,000x

55. E.g. 3—Max. Rev. from Ticket Sales 3. Set up the model. The model is the function R that gives the revenue for a given ticket price x. Revenue = Ticket price x Attendance R(x) = x(23,500 – 1000x) R(x) = 23,500x – 1000x 2 Example (a)

56. E.g. 3—Max. Rev. from Ticket Sales 4. Use the model. We use the model to answer the questions in parts (b) and (c).

57. E.g. 3—Max. Rev. from Ticket Sales We want to find the ticket price x for which: R(x) = 23,500x – 1000x 2 = 0 We can solve this quadratic equation algebraically or graphically. Example (b)

58. E.g. 3—Max. Rev. from Ticket Sales From this graph, we see that: R(x) = 0 when x = 0 or x = 23.5 So, according to our model, the revenue would drop to zero if the ticket price is $23.50 or higher. Of course, revenue is also zero if the ticket price is zero! Example (b)

59. E.g. 3—Max. Rev. from Ticket Sales Since R(x) = 23,500x – 1000x 2 is a quadratic function with a = –1000 and b = 23,500, the maximum occurs at: Example (c)

60. E.g. 3—Max. Rev. from Ticket Sales Thus, a ticket price of $11.75 yields the maximum revenue. At this price, the revenue is: R(11.75) = 23,500(11.75) – 1000(11.75) 2 = $138,062.50 Example (c)

61. E.g. 4—Minimizing the Metal in a Can A manufacturer makes a metal can that holds 1 L (liter) of oil. What radius minimizes the amount of metal in the can?

62. E.g. 4—Thinking About the Problem To use the least amount of metal, we must minimize the surface area of the can—the area of the top, bottom, and the sides.

63. E.g. 4—Thinking About the Problem The area of the top and bottom is 2πr 2. The area of the sides is 2πrh. So, the surface area of the can is: S = 2πr 2 + 2πrh

64. E.g. 4—Thinking About the Problem The radius and height of the can must be chosen so that the volume is exactly 1 L, or 1000 cm 3. If we want a small radius, say r = 3, then the height must be just tall enough to make the total volume 1000 cm 3. That is, we must have:

65. E.g. 4—Thinking About the Problem Now that we know the radius and height, we can find the surface area of the can: S = 2π(3) 2 + 2π(3)(35.4) ≈ 729.1 cm 3 If we want a different radius, we can find the corresponding height and surface area in a similar fashion.

66. E.g. 4—Minimizing the Metal in a Can The model we want is: A function that gives the surface area of the can.

67. E.g. 4—Minimizing the Metal in a Can 1. Express the model in words. We know that, for a cylindrical can, Surface area = Area of top and bottom + Area of sides

68. E.g. 4—Minimizing the Metal in a Can 2. Choose the variable. There are two varying quantities—radius and height. Since the function we want depends on the radius, we let: r = radius of can

69. E.g. 4—Minimizing the Metal in a Can Next, we must express the height in terms of the radius r. Since the volume of a cylindrical can is V = πr 2 h and the volume must be 1000 cm 3, we have:

70. E.g. 4—Minimizing the Metal in a Can We can now express the areas of the top, bottom, and sides in terms of r only. In WordsIn Algebra Radius of canr Height of can1000/πr 2 Area of top and bottom 2πr22πr2 Area of sides (2πrh)2πr(1000/πr 2 )

71. E.g. 4—Minimizing the Metal in a Can 3. Set up the model. The model is the function S that gives the surface area of the can as a function of the radius r. Surface area = Area of top and bottom + Area of sides

72. E.g. 4—Minimizing the Metal in a Can 4. Use the model. We use the model to find the minimum surface area of the can. We graph S and zoom in on the minimum point to find that the minimum value of S is about 554 cm 2 and occurs when the radius is about 5.4 cm.