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Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 1 6.5 Perimeter, Value, Interest, and Mixture Problems.

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Presentation on theme: "Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 1 6.5 Perimeter, Value, Interest, and Mixture Problems."— Presentation transcript:

1 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 1 6.5 Perimeter, Value, Interest, and Mixture Problems

2 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 2 Five-Step Problem-Solving Method To solve some problems in which we want to find two quantities, it is useful to perform the following five steps: Step 1: Define each variable. For each quantity that we are trying to find, we usually define a variable to represent that unknown quantity.

3 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 3 Five-Step Problem-Solving Method Step 2: Write a system of two equations. We find a system of two equations by using the variables from step 1. We can usually write each equation either by translating the information stated in the problem into mathematics or by making a substitution into a formula. Step 3: Solve the system. We solve the system of equations from step 2.

4 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 4 Five-Step Problem-Solving Method Step 4: Describe each result. We use a complete sentence to describe the quantities we found. Step 5: Check. We reread the problem and check that the quantities we found agree with the given information.

5 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 5 Perimeter Recall from Section 4.6 that the formula of the perimeter P of a rectangle with length L and width W is P = 2L + 2W.

6 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 6 Example: Solving a Perimeter Problem Throughout history, rectangles who length is about 1.62 times their width, called golden rectangles, have been viewed as the most pleasing to the eye. Many famous structures, including the Parthenon in Athens and the Great Pyramid of Giza in Cairo, incorporate golden rectangles into their design. For Leonardo Da Vinci’s Mona Lisa, the edges of the painting form a golden rectangle.

7 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 7 Example: Solving a Perimeter Problem Suppose an artist wants a piece of canvas in the shape of a golden rectangle with a perimeter of 9 feet. Find the dimensions of the canvas.

8 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 8 Solution Step 1: Define each variable. Let W be the width (in feet) and L be the length (in feet.) Step 2: Write a system of two equations. Since the length must be 1.62 times the width, our first equation is L = 1.62W. Because the perimeter is 9 feet, we find our second equation by substituting 9 for P in the perimeter formula P = 2L + 2W:9 = 2L + 2W.

9 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 9 Solution The system is

10 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 10 Solution Step 3: Solve the system. We can solve the system by substitution. We substitute 1.62W for L in equation (2) and then solve for W:

11 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 11 Solution To find the approximate length, we substitute for W in equation (1):

12 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 12 Solution Step 4: Describe each result. The approximate width is 1.72 feet, and the approximate length is 2.78 feet. Step 5: Check. We add the lengths of the four sides: 1.72 + 2.78 + 1.72 + 2.78 = 9, which checks for a perimeter of 9 feet. We can also check that 2.78 is about 1.62 times 1.72 (try it).

13 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 13 Value Problems Recall from Section 4.6 that if n objects each have value v, then their total value T is given by T = vn When some objects are sold, we refer to the total money collected as the revenue from selling the objects.

14 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 14 Example: Solving a Value Problem An auditorium has 500 balcony seats and 2100 main-level seats. If tickets for balcony seats will cost $18 less than tickets for main-level seats, what should the price be for each type of ticket so that the total revenue from a sellout performance will be $128,800?

15 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 15 Solution Step 1: Define the variable. Let b be the price (in dollars) for balcony seats and m be the price (in dollars) for main-level seats. Step 2: Define the variable. Since tickets for balcony seats will cost $18 less than tickets for main- level seats, our first equation is

16 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 16 Solution Since the total revenue is $128,800, our second equation is The units of the expressions on both sides of the equation are dollars, which suggests that our work is correct. The system is Equation (1) Equation (2)

17 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 17 Solution Step 3: Solve the system. We can use substitution to solve the system. We substitute m – 18 for b in equation (2) and solve for m: Equation (2) Then we substitute 53 for m in equation (1) and solve for b: b = 53 – 18 = 35

18 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 18 Solution Step 4: Describe each result. Tickets for balcony seats should be priced at $35 per ticket, and tickets for main-level seats should be priced at $53 per ticket. Step 5: Check. First, we find the difference in the ticket prices: 53 – 35 = 18 dollars, which checks. Then we compute the total revenue from selling 500 of the $35 tickets and 2100 of the $53 tickets: dollars, which checks.

19 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 19 Interest Problems Money deposited in an account, such as a savings account, certificate of deposit (CD), or mutual fund is called principal. A person invests money in hopes of later getting back the principal plus additional money, called interest. The annual simple-interest rate is the percentage of the principal that equals the interest earned per year.

20 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 20 Example: Finding the Interest from an Investment 1. How much interest will a person earn from investing $6000 in a Presidential Bank Internet CD account at 2.5% annual interest and $1000 in a John Hancock Health Sciences A mutual fund account at 12% annual interest for one year? 2. A person plans to invest a total of $7000. She will invest in the two accounts described in Problem 1. How much money should she invest in each account to earn a total interest of $400 in one year?

21 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 21 Solution 1. We add the interest from both accounts to compute the total investment:

22 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 22 Solution 2. Step 1: Define each variable. Let x be the money (in dollars) invested at 2.5% annual interest, and let y be the money (in dollars) invested at 12% annual interest.

23 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 23 Solution Step 2: Write a system of two equations. Our work in Problem 1 suggests the first equation: Since the total investment is $7000, our second equation is x + y = 7000.

24 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 24 Solution The system is Equation (1) Equation (2)

25 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 25 Solution Step 3: Solve the system. We can use elimination to solve the system. To eliminate the x terms, we multiply both sides of equation (2),, by –0.025, yielding the system

26 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 26 Solution Then we add the left sides and add the right sides of the equations and solve for y:

27 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 27 Solution Next, we substitute 2368.42 for y in equation (2) and solve for x:

28 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 28 Solution Step 4: Describe the result. The person should invest $4631.58 at 2.5% annual interest and $2368.42 at 12% annual interest. Step 5: Check. First, we find the sum 4631.58 + 2368.42 = 7000, which checks. Next, we calculate the total interest earned from investing $4631.58 at 2.5% and $2368.42 at 12%: 0.025(4631.58) + 0.12(2368.42) ≈ 400, which checks.

29 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 29 Example: Solving a Mixture Problem A chemist needs 8 liters of a 20% alcohol solution, but she has only a 15% alcohol solution and a 35% alcohol solution. How many liters of each solution should she mix to make the desired 8 liters of 20% alcohol solution?

30 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 30 Solution Step 1: Define each variable. Let x be the number of liters of 15% alcohol solution, and let y be the number of liters of 35% alcohol solution.

31 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 31 Solution Step 2: Write a system of two equations. Since she wants 8 liters of the total mixture, our first equation is x + y = 8. We find our second equation from the fact that the sum of the amounts of pure alcohol in the 15% alcohol solution and the 35% alcohol solution is equal to the amount of pure alcohol in the mixture:

32 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 32 Solution The system is Equation (1) Equation (2)

33 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 33 Solution Step 3: Solve the system. We can solve the system by substitution. First, we solve equation (1) for y:

34 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 34 Solution Then, we substitute 8 – x for y in equation (2) and solve for x: Next, we substitute 6 in the equation y = 8 – x and solve for y: y = 8 – 6 = 2

35 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 35 Solution Step 4: Describe each result. Six liters of the 15% alcohol solution and 2 liters of the 35% alcohol solution are required.

36 Copyright © 2015, 2008 Pearson Education, Inc. Section 6.5, Slide 36 Solution Step 5: Check. First we compute the total amount (in liters) of pure alcohol in 6 liters of the 15% alcohol solution and 2 liters of the 35% alcohol solution: 0.15(6) + 0.35(2) = 1.6 Next, we compute the amount (in liters) of pure alcohol in 8 liters of the 20% alcohol solution: 0.20(8) = 1.6 Since the two results are equal, they check. Also, 6 + 2 = 8, which checks with the chemist wanting 8 liters of the 20% solution.


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