Copyright © Cengage Learning. All rights reserved. 1 Fundamentals Copyright © Cengage Learning. All rights reserved.

1.7 Modeling with Equations Copyright © Cengage Learning. All rights reserved.

Objectives Making and Using Models Problems About Interest Problems About Area or Length Problems About Mixtures Problems About the Time Needed to Do a Job Problems About Distance, Rate, and Time

Making and Using Models

Making and Using Models We will use the following guidelines to help us set up equations that model situations described in words.

Making and Using Models The next example illustrates how the guidelines are used to translate a “word problem” into the language of algebra.

Example 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? Solution: Identify the variable. We are asked to find the number of miles Helen has driven. So we let x = number of miles driven Translate from words to algebra. Now we translate all the information given in the problem into the language of algebra.

Example 1 – Solution Set up the model. Now we set up the model. cont’d Set up the model. Now we set up the model. 0.15x + 2(30) = 108

Example 1 – Solution Solve. Now we solve for x. 0.15x = 48 x = 320 cont’d Solve. Now we solve for x. 0.15x = 48 x = 320 Helen drove her rental car 320 miles. Subtract 60 Divide by 0.15 Calculator

Example 1 – Solution cont’d Check Your Answer: total cost = mileage cost + daily cost = 0.15(320) + 2(30) = 108

Problems About Interest

Problems About Interest When you borrow money from a bank or when a bank “borrows” your money by keeping it for you in a savings account, the borrower in each case must pay for the privilege of using the money. The fee that is paid is called interest. The most basic type of interest is simple interest, which is just an annual percentage of the total amount borrowed or deposited. The amount of a loan or deposit is called the principal P. The annual percentage paid for the use of this money is the interest rate r.

Problems About Interest We will use the variable t to stand for the number of years that the money is on deposit and the variable I to stand for the total interest earned. The following simple interest formula gives the amount of interest I earned when a principal P is deposited for t years at an interest rate r.

Problems About Interest When using this formula, remember to convert r from a percentage to a decimal. For example, in decimal form, 5% is 0.05. So at an interest rate of 5%, the interest paid on a $1000 deposit over a 3-year period is I = Prt = 1000(0.05)(3) = $150.

Example 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 % simple interest annually. If Mary’s total interest is $5025 per year, how much money is invested at each rate? Solution: Identify the variable. The problem asks for the amount she has invested at each rate. So we let x = the amount invested at 6%

Example 2 – Solution Translate from words to algebra. cont’d Translate from words to algebra. Since Mary’s total inheritance is $100,000, it follows that she invested 100,000 – x at %. We translate all the information given into the language of algebra.

Example 2 – Solution cont’d Set up the model. We use the fact that Mary’s total interest is $5025 to set up the model. 0.06x + 0.045(100,000 – x) = 5025 Solve. Now we solve for x. 0.06x + 4500 – 0.045x = 5025 0.015x + 4500 = 5025 0.015x = 525 Distributive Property Combine the x-terms Subtract 4500

Example 2 – Solution cont’d x = = 35,000 So Mary has invested $35,000 at 6% and the remaining $65,000 at %. Check your answer: total interest = 6% of $35,000 + % of $65,000 = $2100 + $2925 = $5025 Divide by 0.015

Problems About Area or Length

Problems About Area or Length When we use algebra to model a physical situation, we must sometimes use basic formulas from geometry. For example, we may need a formula for an area or a perimeter, or the formula that relates the sides of similar triangles, or the Pythagorean Theorem.

Example 3 – Dimensions of a Garden A square garden has a walkway 3 ft wide around its outer edge, as shown in Figure 1. If the area of the entire garden, including the walkway, is 18,000 ft2, what are the dimensions of the planted area? Figure 1

Example 3 – Solution Identify the variable. We are asked to find the length and width of the planted area. So we let x = the length of the planted area Translate from words to algebra. Next, translate the information from Figure 1 into the language of algebra.

Example 3 – Solution Set up the model. We now set up the model. cont’d Set up the model. We now set up the model. (x + 6)2 = 18,000 Solve. Now we solve for x. x + 6 = x = – 6 Take square roots Subtract 6

Example 3 – Solution cont’d x  128 The planted area of the garden is about 128 ft by 128 ft.

Example 4 – Dimensions of a Building Lot A rectangular building lot is 8 ft longer than it is wide and has an area of 2900 ft2. Find the dimensions of the lot. Solution: Identify the variable. We are asked to find the width and length of the lot. So let w = width of lot

Example 4 – Solution Translate from words to algebra. cont’d Translate from words to algebra. Then we translate the information given in the problem into the language of algebra (see Figure 2). Figure 2

Example 4 – Solution cont’d Set up the model. Now we set up the model. w(w + 8) = 2900 Solve. Now we solve for w. w2 + 8w = 2900 w2 + 8w – 2900 = 0 Expand Subtract 2900

Example 4 – Solution (w – 50)(w + 58) = 0 w = 50 or w = – 58 cont’d (w – 50)(w + 58) = 0 w = 50 or w = – 58 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. Factor Zero-Product Property

Problems About Mixtures

Problems About Mixtures Many real-world problems involve mixing different types of substances. For example, construction workers may mix cement, gravel, and sand; fruit juice from concentrate may involve mixing different types of juices. Problems involving mixtures and concentrations make use of the fact that if an amount x of a substance is dissolved in a solution with volume V, then the concentration C of the substance is given by

Problems About Mixtures So if 10 g of sugar is dissolved in 5 L of water, then the sugar concentration is C = 10/5 = 2 g/L. Solving a mixture problem usually requires us to analyze the amount x of the substance that is in the solution. When we solve for x in this equation, we see that x = CV. Note that in many mixture problems the concentration C is expressed as a percentage.

Example 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? Solution: Identify the variable. 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

Example 6 – Solution cont’d Translate from words to algebra. In any problem of this type—in which two different substances are to be mixed—drawing a diagram helps us to organize the given information (see Figure 4). Figure 4

Example 6 – Solution cont’d The information in the figure can be translated into the language of algebra, as follows: Set up the model. 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.

Example 6 – Solution 45 + x = 0.1(900 + x) Solve. Now we solve for x. cont’d 45 + x = 0.1(900 + x) Solve. Now we solve for x. 45 + x = 90 + 0.1x 0.9x = 45 From Figure 4 Distributive Property Subtract 0.1x and 45

Example 6 – Solution cont’d x = = 50 The manufacturer should add 50 gal of pure orange juice to the soda. 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 Amounts are equal. Divide by 0.9

Problems About the Time Needed to Do a Job

Problems About the Time Needed to Do a Job When solving a problem that involves determining how long it takes several workers to complete a job, we use the fact that if a person or machine takes H time units to complete the task, then in one time unit the fraction of the task that has been completed is 1/H.

Example 7 – Time Needed to Do a Job Because of an anticipated heavy rainstorm, the water level in a reservoir must be lowered by 1 ft. Opening spillway A lowers the level by this amount in 4 hours, whereas opening the smaller spillway B does the job in 6 hours. How long will it take to lower the water level by 1 ft if both spillways are opened? Solution: Identify the variable. 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

Example 7 – Solution cont’d Translate from words to algebra. Finding an equation relating x to the other quantities in this problem is not easy. Certainly x is not simply 4 + 6, because that would mean that together the two spillways require longer to lower the water level than either spillway alone. Instead, we look at the fraction of the job that can be done in 1 hour by each spillway.

Example 7 – Solution Set up the model. Now we set up the model. cont’d Set up the model. Now we set up the model. Solve. Now we solve for x. 3x + 2x = 12 5x = 12 Multiply by the LCD, 12x Add

Example 7 – Solution cont’d It will take hours, or 2 h 24 min, to lower the water level by 1 ft if both spillways are open. Divide by 5

Problems About Distance, Rate, and Time

Problems About Distance, Rate, and Time The next example deals with distance, rate (speed), and time. The formula to keep in mind here is 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.

Example 8 – A Distance-Speed-Time Problem A jet flew from New York to Los Angeles, a distance of 4200 km. The speed for the return trip was 100 km/h faster than the outbound speed. If the total trip took 13 hours of flying time, what was the jet’s speed from New York to Los Angeles? Solution: Identify the variable. We are asked for the speed of the jet from New York to Los Angeles. So let s = speed from New York to Los Angeles Then s + 100 = speed from Los Angeles to New York

Example 8 – Solution cont’d Translate from words to algebra. Now we organize the information in a table. We fill in the “Distance” column first, since we know that the cities are 4200 km apart.

Example 8 – Solution cont’d Then we fill in the “Speed” column, since we have expressed both speeds (rates) in terms of the variable s. Finally, we calculate the entries for the “Time” column, using Set up the model. The total trip took 13 hours, so we have the model

Example 8 – Solution cont’d Solve. Multiplying by the common denominator, , we get 4200(s + 100) + 4200s = 13s(s + 100) 8400s + 420,000 = 13s2 + 1300s 0 = 13s2 – 7100s – 420,000 Although this equation does factor, with numbers this large it is probably quicker to use the Quadratic Formula and a calculator.

Example 8 – Solution cont’d s = 600 or  – 53.8 Since s represents speed, we reject the negative answer and conclude that the jet’s speed from New York to Los Angeles was 600 km/h.