Modeling with Rational Functions

Modeling with Rational Functions
Section 12.6 Modeling with Rational Functions

Modeling the Mean of a Quantity
Computing the Mean Definition If a quantity Q is divided into n parts, the mean amount M of the quantity per part is given by The underground band Melted Zipper wants to make and sell a CD of its original songs. It costs about $1000 to record the music onto a digital audiotape Example

Example Continued Definition (DAT), $100 to rearrange the music and improve the sound quality, $350 for artwork for the cover and inside leaflet, and $350 to set up production. In addition, it will cost $2.50 for each CD manufactured. 1. What is the total cost of making 300 CDs? 2. Let C(n) be the total cost (in dollars) of making n CDs. Find an equation of C.

Example Continued Definition 3. Let P(n) be the price the band should set for each CD (in dollars) so that it breaks even by making and selling n CDs. Find an equation of P. 4. Find P(300). What does it mean in this situation? 5. Find n when P(n) = 10. What does it mean in this situation? 6. Describe the values of P(n) for large values of n.

Computing the Mean Solution 1. Compute the total fixed costs—the costs that do not depend on how many CDs are manufactured: = 1800 dollars Band must also pay $2.50 per CD manufactured If the band manufactures 300 CDs, the variable cost, is 2.50(300) = 750 dollars To find the total cost, we add the variable cost and the fixed costs: 2.50(300) = 2550 dollars

Computing the Mean Solution Continued Solution The total cost is equal to $2.50 times the number of CDs, plus the fixed cost of $1800: C(n) = 2.50 n 3. If the band makes and sells n CDs, it can break even by selling the CD for the amount found by dividing the total cost into n parts. This amount is the mean cost per CD:

Computing the Mean Solution Continued Solution If the band makes and sells 300 CDs, it must sell each CD for $8.50 to break even

Computing the Mean Solution Continued Solution 5. If the band can sell each CD for $10, then it must make 240 CDs to break even

Computing the Mean Solution Continued Solution 6. We see that as n increases, the mean cost (price) decreases (in order to break even) As the number of CDs manufactured increases, the more the fixed cost is spread out, so the smaller the fixed cost that each sale has to cover.

Computing the Mean Solution Continued Solution Continue to scroll down a table for larger and larger inputs n, the outputs P(n) approach 2.50, the variable cost in dollars per CD Use a graphing calculator graph to observe that the break-even CD price approaches $2.50 for large manufacturing runs It appears that for large inputs, the height of the graph of P gets close to 2.50.

Computing the Mean Solution Continued Solution If the band could sell tens of thousands of CDs, it could price them at a few cents above the $ cost and still break even, which is probably unrealistic.

Computing the Mean Example In Example 9 of Section 7.5, we modeled the U.S. annual consumption of bottled water. A reasonable model is B(t) = 16t t , where B(t) is U.S. bottled water consumption (in millions of gallons) in the year that is t years since 1990.

Computing the Mean Example In Exercise 9 of Homework 7.7, you modeled the U.S. population. A reasonable model is U(t) = t t , where U(t) is the U.S. population (in millions) at t years since 1990. 1. Let M(t) be the annual mean consumption of bottled water per person (in gallons per person) in the year that is t years since Find an equation of M.

Computing the Mean Example Continued Example 2. Find M(20). What does it mean in this situation? Find t when M(t) = 40. What does it mean in this situation? Annual mean consumption of bottled water per person is equal to the total annual consumption divided by the U.S. population: Solution

Computing the Mean Solution Continued Example Annual mean consumption of bottled water will be gallons per person in 2010, according to the model

3. We substitute 40 for M(t) in the equation of M and solve for t:
Modeling the Mean of a Quantity Computing the Mean Solution Continued Example 3. We substitute 40 for M(t) in the equation of M and solve for t:

Verify using a graphing calculator
Modeling the Mean of a Quantity Computing the Mean Solution Continued Example Verify using a graphing calculator In 2012 ( ) the average consumption with be 40 gallons per person.

Modeling the Percentage of a Quantity
Modeling the Percent of a Quantity Example The numbers of broadband cable subscribers and DSL subscribers in the United States are shown in the table for various years. Let B(t) be the number of broadband cable subscribers and D(t) be the number of DSL subscribers, both in millions, at t years since 1990.

Modeling the Percent of a Quantity Example Continued Example 1. Find equations of B and D. 2. Find an equation of the sum function B + D. What do the inputs and outputs of B + D mean in this situation? 3. In Exercise 30 of Homework 3.1, you modeled the number of U.S. households. A reasonable model is H(t) = 1.59t , where H(t) is the total number (in millions) of U.S. households at t years since 1990 (see table on next slide).

Modeling the Percent of a Quantity Example Continued Example Let P(t) be the percentage of U.S. households that are broadband cable or DSL subscribers. Find an equation of P. Assume that no household subscribes to both services. 4. Predict when 90% of U.S. households will be either broadband cable or DSL subscribers.

Modeling the Percent of a Quantity Solution Scattergrams of the data suggest that we use a linear function to model the number of broadband cable subscribers and a quadratic model to model the number of DSL subscribers. The linear regression equation of B and the quadratic regression equation of D are

Modeling the Percent of a Quantity Solution Continued Solution Inputs of B + D are the number of years since 1990 Outputs are the total number of subscribers (in millions)

Modeling the Percent of a Quantity Solution Continued Solution To find the percentage of households that subscribe, we divide the total number of subscribers by the number of households and multiply the result by 100:

Modeling the Percent of a Quantity Solution Continued Solution 4. Substitute 90 for P(t):

Modeling the Percent of a Quantity Solution Continued Solution Verify using a graphing calculator 90% of US household with be subscribers to either DSL or broadband cable by 2010.

Distance-Speed-Time Applications
Distance-Speed-Time Relations Property If an object is moving at a constant speed s for an amount of time t, then the distance traveled d is given by d = st and the time t is given by t = d/s A person plans to drive a steady 55 mph on an 80- mile trip. Compute the driving time. Example

Distance-Speed-Time Relations Solution The person is traveling at a constant speed, we use the equation t = d/s We substitute 80 for d and 55 for s in the equation: t = 80/55 ≈ 1.45 The driving time will be about 1.5 hours.

Distance-Speed-Time Relations Example A student at Seattle Central Community College plans to drive from Seattle, Washington, to Eugene, Oregon. The speed limit is 70 mph in Washington and 65 mph in Oregon. She will drive 164 miles in Washington, then 121 miles in Oregon. 1. If the student drives steadily at the speed limits, compute the driving time.

Distance-Speed-Time Relations Example Continued Example 2. If the student exceeds the speed limits, let T (a) be the driving time (in hours) at a mph above the speed limits. Find an equation of T . 3. Find T (0). Compare this result with the result in Problem 1. 4. If the student drives 5 mph over the speed limits, compute the driving time.

Distance-Speed-Time Relations Example Continued If the student wants the driving time to be 4 hours, how much over the speed limits would she have to drive? Student drives at a constant speed in Washington, Use t = d/s to compute the time (in hours) spent driving in Washington: Solution

Distance-Speed-Time Relations Solution Continued Compute the time (in hours) spent driving in Oregon: Total driving time is the sum of our two computed times:

Distance-Speed-Time Relations Solution Continued 2. If the student drives, say, 5 mph over the speed limits, then she will drive 5+70=75 mph in Washington and = 70 mph in Oregon If she drives a miles per hour over the speed limits, she will drive (a +70) mph in Washington and (a +65) mph in Oregon We use these expressions for speeds to find an equation of T:

Distance-Speed-Time Relations Solution Continued The driving time will be about 4.2 hours if the student drives at the speed limits We found the same result in Problem 1

Distance-Speed-Time Relations Solution Continued 4. If the student drives 5 mph over the speed limits, then a = 5: The driving time will be about 3.9 hours. 5. If the trip is to take 4 hours, then T (a) = 4:

Distance-Speed-Time Relations Solution Continued 4.If the student drives 5 mph over the speed limits, then a = 5: The driving time will be about 3.9 hours. 5. If the trip is to take 4 hours, then T (a) = 4: Must drive 3.5 mph above the speed limit to make it in 4 hours.

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