1 Copyright © Cengage Learning. All rights reserved. 2. Equations and Inequalities 2.2 Applied Problems

2 Applied Problems Equations are often used to solve applied problems—that is, problems that involve applications of mathematics to other fields. Because of the unlimited variety of applied problems, it is difficult to state specific rules for finding solutions.

3 Applied Problems The following guidelines may be helpful, provided the problem can be formulated in terms of an equation in one variable. Guidelines for Solving Applied Problems: 1. If the problem is stated in writing, read it carefully several times and think about the given facts, together with the unknown quantity that is to be found.

4 Applied Problems 2. Introduce a letter to denote the unknown quantity. This is one of the most crucial steps in the solution. Phrases containing words such as what, find, how much, how far, or when should alert you to the unknown quantity. 3. If appropriate, draw a picture and label it. 4. List the known facts, together with any relationships that involve the unknown quantity. A relationship may be described by an equation in which written statements, instead of letters or numbers, appear on one or both sides of the equals sign.

5 Applied Problems 5. After analyzing the list in guideline 4, formulate an equation that describes precisely what is stated in words. 6. Solve the equation formulated in guideline Check the solutions obtained in guideline 6 by referring to the original statement of the problem. Verify that the solution agrees with the stated conditions. The use of these guidelines is illustrated in the next example.

6 Example 2 – Calculating a presale price A clothing store holding a clearance sale advertises that all prices have been discounted 20%. If a shirt is on sale for $28, what was its presale price? Solution: Since the unknown quantity is the presale price, we let x = presale price.

7 Example 2 – Solution We next note the following facts: 0.20x = discount of 20% on presale price 28 = sale price The sale price is determined as follows: (presale price) – (discount) = sale price cont’d

8 Example 2 – Solution Translating the last equation into symbols and then solving gives us x – 0.20x = x = 28 x = = 35 The presale price was $35. formulate an equation subtract 0.20x from 1x divide by 0.80 cont’d

9 Example 2 – Solution Check: If a $35 shirt is discounted 20%, then the discount (in dollars) is (0.20)(35) = 7 and the sale price is 35 – 7, or $28. cont’d

10 Applied Problems Banks and other financial institutions pay interest on investments. Usually this interest is compounded; however, if money is invested or loaned for a short period of time, simple interest may be paid, using the following formula.

11 Applied Problems The following table illustrates simple interest for three cases.

12 Example 3 – Investing money in two stocks An investment firm has $100,000 to invest for a client and decides to invest it in two stocks, A and B. The expected annual rate of return, or simple interest, for stock A is 15%, but there is some risk involved, and the client does not wish to invest more than $50,000 in this stock. The annual rate of return on the more stable stock B is anticipated to be 10%. Determine whether there is a way of investing the money so that the annual interest is (a) $12,000 (b) $13,000

13 Example 3 – Solution The annual interest is given by I = Pr, which comes from the simple interest formula I = Prt with t = 1. If we let x denote the amount invested in stock A, then 100,000 – x will be invested in stock B. This leads to the following equalities: x = amount invested in stock A at 15% 100,000 – x = amount invested in stock B at 10%

14 Example 3 – Solution 0.15x = annual interest from stock A 0.10(100,000 – x) = annual interest from stock B Adding the interest from both stocks, we obtain total annual interest = 0.15x (100,000 – x). Simplifying the right-hand side gives us total annual interest = 10, x. cont’d ()()

15 Example 3 – Solution (a) The total annual interest is $12,000 if 10, x = 12, x = 2000 x = = 40,000. Thus, $40,000 should be invested in stock A, and the remaining $60,000 should be invested in stock B. subtract 10,000 divide by 0.05 from (  ) cont’d

16 Example 3 – Solution Since the amount invested in stock A is not more than $50,000, this manner of investing the money meets the requirement of the client. Check: If $40,000 is invested in stock A and $60,000 in stock B, then the total annual interest is 40,000(0.15) + 60,000(0.10) = = 12,000. cont’d

17 Example 3 – Solution (b) The total annual interest is $13,000 if 10, x = 13, x = 3000 x = = 60,000. Thus, $60,000 should be invested in stock A and the remaining $40,000 in stock B. from ( ) subtract 10,000 divide by 0.05 cont’d

18 Example 3 – Solution This plan does not meet the client’s requirement that no more than $50,000 be invested in stock A. Hence, the firm cannot invest the client’s money in stocks A and B such that the total annual interest is $13,000. cont’d

19 Applied Problems In certain applications, it is necessary to combine two substances to obtain a prescribed mixture, as illustrated in the next example.

20 Example 4 – Mixing chemicals A chemist has 10 milliliters of a solution that contains a 30% concentration of acid. How many milliliters of pure acid must be added in order to increase the concentration to 50%? Solution: Since the unknown quantity is the amount of pure acid to add, we let x = number of mL of pure acid to be added.

21 Example 4 – Solution To help visualize the problem, let us draw a picture, as in Figure 1, and attach appropriate labels. cont’d

22 Example 4 – Solution Since we can express the amount of pure acid in the final solution as either 3 + x (from the first two beakers) or 0.50(10 + x), we obtain the equation 3 + x = 0.50(10 + x). We now solve for x: 3 + x = x 0.5x = 2 multiply factors subtract 0.5x and 3 cont’d

23 Example 4 – Solution x = = 4 Hence, 4 milliliters of pure acid should be added to the original solution. Check: If 4 milliliters of acid is added to the original solution, then the new solution contains 14 milliliters, 7 milliliters of which is pure acid. This is the desired 50% concentration. divide by 0.5 cont’d

24 Example 6 – Comparing times traveled by cars Two cities are connected by means of a highway. A car leaves city B at 1:00 P.M. and travels at a constant rate of 40 mi/hr toward city C. Thirty minutes later, another car leaves B and travels toward C at a constant rate of 55 mi/hr. If the lengths of the cars are disregarded, at what time will the second car reach the first car?

25 Example 6 – Solution Let t denote the number of hours after 1:00 P.M. traveled by the first car. Since the second car leaves B at 1:30 P.M., it has traveled hour less than the first. This leads to the following table.

26 Example 6 – Solution The schematic drawing in Figure 3 illustrates possible positions of the cars t hours after 1:00 P.M. Figure 3 cont’d

27 Example 6 – Solution The second car reaches the first car when the number of miles traveled by the two cars is equal—that is, when We now solve for t: 55t – = 40t multiply factors cont’d

28 Example 6 – Solution Thus, t is hours or, equivalently, 1 hour 50 minutes after 1:00 P.M. Consequently, the second car reaches the first at 2:50 P.M. divide by 15 subtract 40t and add cont’d

29 Example 6 – Solution Check: At 2:50 P.M. the first car has traveled for hours, and its distance from B is. At 2:50 P.M. the second car has traveled for hours and is mi from B. Hence, they are together at 2:50 P.M. cont’d

30 Example 7 – Constructing a grain-elevator hopper A grain-elevator hopper is to be constructed as shown in Figure 4, with a right circular cylinder of radius 2 feet and altitude h feet on top of a right circular cone whose altitude is one-half that of the cylinder. What value of h will make the total volume V of the hopper 500 ft 3 ? Figure 4

31 Example 7 – Solution If V cylinder and V cone denote the volumes (in ft 3 ) and h cylinder and h cone denote the heights (in feet) of the cylinder and cone, respectively, then, using the formulas for volume stated on the endpapers at the front of the text, we obtain the following: V cylinder =  r 2 h cylinder =  (2) 2 h = 4  h

32 Example 7 – Solution Since the total volume V of the hopper is to be 500 ft 3, we must have 4  h +  h =  h + 2  h =  h = 1500 h =  34.1 ft. V cylinder + V cone = V total multiply by 3 combine terms divide by 14  cont’d

33 Example 8 – Time required to do a job Two pumps are available for filling a gasoline storage tank. Pump A, used alone, can fill the tank in 3 hours, and pump B, used alone, can fill it in 4 hours. If both pumps are used simultaneously, how long will it take to fill the tank? Solution: Let t denote the number of hours needed for A and B to fill the tank if used simultaneously. It is convenient to introduce the part of the tank filled in 1 hour as follows: = part of the tank filled by A in 1 hr

34 Example 8 – Solution = part of the tank filled by B in 1 hr = part of the tank filled by A and B in 1 hr Using the fact that part filled by part filled by part filled by A in 1 hr A in 1 hr A and B in 1 hr we obtain or cont’d

35 Example 8 – Solution Taking the reciprocal of each side of the last equation gives us. Thus, if pumps A and B are used simultaneously, the tank will be filled in hours, or approximately 1 hour 43 minutes. cont’d