1.2 Constructing Models to Solve Problems · PDF file01.11.2014 · 1.2 – Constructing Models to Solve Problems In the previous section, we learned how to solve linear equations

1.2 – Constructing Models to Solve Problems

In the previous section, we learned how to solve linear equations. In this section, we will put those skills to use in order to solve a variety of application problems.

Literal Equations and Formulas: Literal equations are those that have many or all variables. Formulas are examples of literal equations. Quite often a formula will need to be solved for a different variable which can easily be accomplished using the techniques already learned for solving linear equations. Example: Solve the formula )1( rtPA += for t. Solution: This will involve repeated use of the properties of equality.

Divide by P Subtract 1 Multiply by reciprocal of r Use distributive property

Example: Solve the formula 325

9 += CF for C.

Solution: This is the formula for converting temperature from degrees Fahrenheit to degree Celsius.

( )329

55

932

325

9

−=

=−

+=

FC

CF

CF

Example: Solve the equation CByAx =− for x. Solution: For each step of the solution, determine which property of equality is being used.

A

ByCx

ByCAx

CByAx

+=

+==−

rrP

At

P

A

rt

rtP

A

rtP

A

rtPA

1

11

1

1

)1(

−=

−=

=−

+=

+=

Problems Involving Percent: There are many different types of percent problems, including sales tax, commission, percent increase or decrease and investment problems. Example: A cattle rancher is going to sell one of his prize bulls at an auction and would like to make $45,500.00 after paying a 9% commission to the auctioneer. For what selling price will the rancher make this amount of money? Solution: Step 1: Analyze the Problem: If the commission is 9% that means that the rancher takes home 91%. Let x = the selling price. Then the IS-OF statement is

$45,500 is 91% of what number (selling price.)

Step 2: Create an Equation: The equation is based on the IS-OF statement.

x91.0500,45 = Step 3: Solve the Equation:

000,50$

91.0500,45

==

x

x

Step 4: State the Conclusion: The selling price must be $50,000 in order for the rancher to take home $45,500. Example: Between the years 2000 and 2006, the average cost for auto insurance nationwide grew 27%, to $867.00. What was the average cost in 2000? Solution: Step 1: Analyze the Problem: This is a percent increase problem. The formula for percent increase is always the amount of increase divided by the original number times 100. Let x = the average cost in 2000. Then we can create the following equation based on percent increase. Step 2: Create an Equation:

%27)100(867 =

−x

x

Step 3: Solve the Equation: This is a rational equation. To solve, multiply both sides of the equation by the LCD, which is x.

[ ]

67.682

27.1867

27.0867

27.0867

==

=−

=

−

x

x

xx

xx

xx

Step 4: State the Conclusion: The average cost of auto insurance in 2000 was $682.67.

Geometry Problems: Geometry problems are based on one or more formulas. Example: The perimeter of a regulation singles tennis court is 210 feet and the length is 3 feet less than three times the width. What are the dimensions of the court? Solution: Step 1: Analyze the Problem: This is a geometry problem. The formula for the perimeter of a rectangle is LWP 22 += . Let the width = x. Let the length = 3x-3 Step 2: Create an Equation: Like all geometry problems, the equation will be based on the formula.

)33(22210 −+= xx Step 3: Solve the Equation:

27

8216

68210

662210

)33(22210

==

−=−+=

−+=

x

x

x

xx

xx

Therefore, 783813)27(333 =−=−=−x Step 4: State the Conclusion: The width of the tennis court is 27 feet and the length is 78 feet. Example: Given a triangle, one angle is 15 degrees less than twice the smallest angle. The third angle is 35 degrees more than the smallest angle. Find these three angle measures. Solution: Step 1: Analyze the Problem: The sum of the angles of a triangle is always equal to 180 degrees. Let x = the measure of the smallest angle. Then the other angles are 2x-15 and x+35. Step 2: Create an Equation:

180)35()152( =++−+ xxx Step 3: Solve the Equation:

40

1604

180204

180)35()152(

==

=+=++−+

x

x

x

xxx

Step 4: State the Conclusion: The angles are 40, 65, and 75 degrees.

Uniform Motion Problems: Motion problems are based on the formula D=RT. It is helpful to create a table to organize the information before creating the equation. Example: If a student drives from college to home at 40 mph, then he is 15 minutes late. However, if he makes the same trip at 50 mph, he is 12 minutes early. What is the distance between his college and home? Solution: Step 1: Analyze the Problem: This is a motion problem and it may be helpful to create a table. Motion problems are based on the formula Distance = Rate x Time or D=RT. Fill in the Rate column and the Time column and then multiply these columns to obtain the distance column.

Rate Time Distance Trip 1 40 x 40x Trip 2 50 x-27 50(x-27)

Step 2: Create an Equation: The equation is based on the fact that the distance that they both travelled is equal.

)27(5040 −= xx Step 3: Solve the Equation:

135

135010

13505040

)27(5040

==

−=−=

x

x

xx

xx

135 minutes is 2.25 hours. Substituting this into 40x gives a distance of miles90)25.2(40 = Step 4: State the Conclusion: The student lives 90 miles from campus. Example: John rowed his boat from his camp on the bayou to his crab traps. Going down the bayou he caught a falling tide that increased his normal speed by 2 mph, but coming back it decreased his normal speed by 2 mph. When going with the tide, the trip took 10 minutes and going against the tide, the trip took 30 minutes. How far is it from John’s camp to the crab traps? Solution: Step 1: Analyze the Problem: This is a motion problem that involves use of the formula rtD = . As with most motion problems, it is helpful to set up a table to organize the information.

Trip Rate Time Distance To Traps x+2 10=1/6 1/6(x+2) To Camp x-2 30=1/2 1/2(x-2)

Step 2: Create an Equation: From this table we can create the following equation which is based on the distance in both directions being equal.

)2(2

1)2(

6

1 −=+ xx

Step 3: Solve the Equation:

4

82

632

)2(2

1)2(

6

1

==

−=+

−=+

x

x

xx

xx

Step 4: State the Conclusion: The distance to the traps is 60 miles.

Mixture Problems: Example: The owner of a health foods store sells dried apples for $1.20 per quarter pound, and dried apricots for $1.80 per quarter pound. How many pounds of each must he mix together to get 20 pounds of a mixture that sells for $1.68 per quarter pound? Solution: Step 1: Analyze the Problem: This is a mixture problem. A table is helpful for organizing the information. Note that I have converted the prices per quarter lb to prices per lb.

Fruit # of lbs. Rate/lb. Total amount apples x 4.80 4.8x

apricots 20-x 7.20 7.2(20-x) mixture 20 6.72 6.72(20)

Step 2: Create an Equation: From the table we can create the equation which is based on adding the dried apple and apricots together to equal the mixture.

)20(72.6)20(2.78.4 =−+ xx


4

6.94.2

4.1342.71448.4

)20(72.6)20(2.78.4

==

=−+=−+

x

x

xx

xx

Step 4: State the Conclusion: The owner must mix 4 lbs of dried apples and 16 lbs of dried apricots.

Example: How many gallons of a 3% salt solution must be mixed with 50 gallons of a 7% solution to obtain a 5% solution? Solution: Step 1: Analyze the Problem: This is a mixture problem. A table is helpful for organizing the information.

Solution % Salt Amount Salt 1 3% x 0.03x 2 7% 50 0.07(50)

Final 5% x+50 0.05(x+50) Step 2: Create an Equation: The equation will be based on the fact that the sum of the salt in solution 1 and solution 2 must be equal to the amount of salt in the final solution.

)50(05.0)50(07.003.0 +=+ xx Step 3: Solve the Equation:

50

102.0

5.205.05.303.0

)50(05.0)50(07.003.0

==

+=++=+

x

x

xx

xx

Step 4: State the Conclusion: 50 gallons of the 3% solution must be mixed.

Work Problems: Example: With the old combine Rick’s entire wheat crop can be harvested in 72 hours but a new combine can do the same job in 48 hours. How many hours would it take to harvest the crop with both combines operating? Solution: Step 1: Analyze the Problem: This is a work problem. Once again a table is helpful for organizing the information.

Combine Work /hour hours working work old 1/72 x 1/72(x) new 1/48 x 1/48(x)

Step 2: Create an Equation: From the table we can create the following equation which is based on the old and new combines working together to complete 1 job.

14872

=+ xx


8.28

3456120

34567248

14872

==

=+

=+

x

x

xx

xx

Step 4: State the Conclusion: The time it takes the combines working together to harvest the crop (complete 1 job) is 28 hours and 48 minutes. Example: Tom can paint a fence in 5 hours. Huckleberry can paint the same fence in 7 hours. If Tom gets up early and starts painting a 8:30 AM and Huckleberry joins him 2.5 hours later, at what time will they finish painting the fence? Solution: Step 1: Analyze the Problem: This is a work problem. Once again a table is helpful for organizing the information.

Combine Work /hour hours working work Tom 1/5 x 1/5(x)

Huckleberry 1/7 x-2.5 1/7(x-2.5)


17

5.2

5=−+ xx


95.3

5.4712

355.1257

17

5.2

5

==

=−+

=−+

x

x

xx

xx

Step 4: State the Conclusion: Tom will work for 3 hours and 57 minutes and Huckleberry will work for 1 hour and 27 minutes. The fence will be completed at 12:27 PM.

1.2 – Applications

Formula Problems Example: In nursing, Clark’s Rule for medication expresses a relationship between the recommended dosages for a child and for an adult. The rule states that a child’s dosage C equals the product of the weight W of the child in pounds divided by 150 and the adult’s dosage A. Determine the formula for adult dosage and then determine the adult dosage if a child weighing 60 lbs requires a dosage of 125 mg. Solution: Step 1: Analyze the Problem: First, we need to solve this formula for A, the adult dosage. Then, we can substitute in the given values to find the indicated dosage. Step 2: Create an Equation:

AW

C ×=150

Step 3: Solve the Equation: Solve for A

AW

C

WAC

AW

C

=

=

×=

150

150150

Solve for the indicated dosage.

5.31260

)125(150

150

=

=

=

A

A

AW

C

Step 4: State the Conclusion: Therefore, the adult dosage is 312.5 mg.

Geometry Problems Example: The perimeter of a regulation singles tennis court is 210 feet and the length is 3 feet less than three times the width. What are the dimensions of the court? Solution: Step 1: Analyze the Problem: This is a geometry problem. The formula for the perimeter of a rectangle is LWP 22 += . Let the width = x. Let the length = 3x-3 Step 2: Create an Equation: Like all geometry problems, the equation will be based on the formula.

)33(22210 −+= xx Step 3: Solve the Equation:

27

8216

68210

662210

)33(22210

==

−=−+=

−+=

x

x

x

xx

xx

Therefore, 783813)27(333 =−=−=−x Step 4: State the Conclusion: The width of the tennis court is 27 feet and the length is 78 feet. Example: Given a triangle, one angle is 15 degrees less than twice the smallest angle. The third angle is 35 degrees more than the smallest angle. Find these three angle measures. Solution: Step 1: Analyze the Problem: The sum of the angles of a triangle is always equal to 180 degrees. Let x = the measure of the smallest angle. Then the other angles are 2x-15 and x+35. Step 2: Create an Equation:

180)35()152( =++−+ xxx Step 3: Solve the Equation:

40

1604

180204

180)35()152(

==

=+=++−+

x

x

x

xxx

Step 4: State the Conclusion: The angles are 40, 65, and 75 degrees.

Percent Problems Example: A cattle rancher is going to sell one of his prize bulls at an auction and would like to make $45,500.00 after paying a 9% commission to the auctioneer. For what selling price will the rancher make this amount of money? Solution: Step 1: Analyze the Problem: If the commission is 9% that means that the rancher takes home 91%. Let x = the selling price. Then the IS-OF statement is $45,500 is 91% of what number (selling price.) Step 2: Create an Equation: The equation is based on the IS-OF statement.

x91.0500,45 = Step 3: Solve the Equation:

000,50$

91.0500,45

==

x

x

Step 4: State the Conclusion: The selling price must be $50,000 in order for the rancher to take home $45,500. Example: Between the years 2000 and 2006, the average cost for auto insurance nationwide grew 27%, to $867.00. What was the average cost in 2000? Solution: Step 1: Analyze the Problem: This is a percent increase problem. The formula for percent increase is always the amount of increase divided by the original number times 100. Let x = the average cost in 2000. Then we can create the following equation based on percent increase. Step 2: Create an Equation:

%27)100(867 =

−x

x

Step 3: Solve the Equation: This is a rational equation. To solve, multiply both sides of the equation by the LCD, which is x.

[ ]

67.682

27.1867

27.0867

27.0867

==

=−

=

−

x

x

xx

xx

xx

Step 4: State the Conclusion: The average cost of auto insurance in 2000 was $682.67.

Uniform Motion Problems

Example: If a student drives from college to home at 40 mph, then he is 15 minutes late. However, if he makes the same trip at 50 mph, he is 12 minutes early. What is the distance between his college and home? Solution: Step 1: Analyze the Problem: This is a motion problem and it may be helpful to create a table. Motion problems are based on the formula Distance = Rate x Time or D=RT. Fill in the Rate column and the Time column and then multiply these columns to obtain the distance column.

Rate Time Distance Trip 1 40 x 40x Trip 2 50 x-27 50(x-27)

Step 2: Create an Equation: The equation is based on the fact that the distance that they both travelled is equal.

)27(5040 −= xx Step 3: Solve the Equation:

135

135010

13505040

)27(5040

==

−=−=

x

x

xx

xx

135 minutes is 2.25 hours. Substituting this into 40x gives a distance of

miles90)25.2(40 = Step 4: State the Conclusion: The student lives 90 miles from campus.

Example: John rowed his boat from his camp on the bayou to his crab traps. Going down the bayou he caught a falling tide that increased his normal speed by 2 mph, but coming back it decreased his normal speed by 2 mph. When going with the tide, the trip took 10 minutes and going against the tide, the trip took 30 minutes. How far is it from John’s camp to the crab traps? Solution: Step 1: Analyze the Problem: This is a motion problem that involves use of the formula rtD = . As with most motion problems, it is helpful to set up a table to organize the information.

Trip Rate Time Distance To Traps x+2 10=1/6 1/6(x+2) To Camp x-2 30=1/2 1/2(x-2)

Step 2: Create an Equation: From this table we can create the following equation which is based on the distance in both directions being equal.

)2(2

1)2(

6

1 −=+ xx


4

82

632

)2(2

1)2(

6

1

==

−=+

−=+

x

x

xx

xx

Step 4: State the Conclusion: The distance to the traps is 60 miles.

Example: A cyclist leaves his training base for a morning workout, riding at the rate of 18 mph. One and one half hours later, his support staff leaves the base in a car going 45 mph in the same direction. How long will it take the support staff to catch up with the cyclist? Solution: Step 1: Analyze the Problem: This is a motion problem and it may be helpful to create a table. Motion problems are based on the formula D=RT. Fill in the Rate column and the Time column and then multiply these columns to obtain the distance column.

Rate Time Distance Cyclist 18 x 18x

Car 45 x-1.5 45(x-1.5) Step 2: Create an Equation: The equation is based on the fact that the distance that they both travelled is equal.

)5.1(4518 −= xx Step 3: Solve the Equation:

5.2

5.6727

5.674518

)5.1(4518

==

−=−=

x

x

xx

xx

Step 4: State the Conclusion: The cyclist has been travelling for 2.5 hours; therefore it will take the support car 1 hour to catch up.

Mixture Problems Example: The owner of a health foods store sells dried apples for $1.20 per quarter pound, and dried apricots for $1.80 per quarter pound. How many pounds of each must he mix together to get 20 pounds of a mixture that sells for $1.68 per quarter pound? Solution: Step 1: Analyze the Problem: This is a mixture problem. A table is helpful for organizing the information. Note that I have converted the prices per quarter lb to prices per lb.

Fruit # of lbs. Rate/lb. Total amount apples x 4.80 4.8x

apricots 20-x 7.20 7.2(20-x) mixture 20 6.72 6.72(20)

Step 2: Create an Equation: From the table we can create the equation which is based on adding the dried apple and apricots together to equal the mixture.

)20(72.6)20(2.78.4 =−+ xx


4

6.94.2

4.1342.71448.4

)20(72.6)20(2.78.4

==

=−+=−+

x

x

xx

xx

Step 4: State the Conclusion: The owner must mix 4 lbs of dried apples and 16 lbs of dried apricots.

Example: How many gallons of a 3% salt solution must be mixed with 50 gallons of a 7% solution to obtain a 5% solution? Solution: Step 1: Analyze the Problem: This is a mixture problem. A table is helpful for organizing the information.

Solution % Salt Amount Salt 1 3% x 0.03x 2 7% 50 0.07(50)

Final 5% x+50 0.05(x+50) Step 2: Create an Equation: The equation will be based on the fact that the sum of the salt in solution 1 and solution 2 must be equal to the amount of salt in the final solution.

)50(05.0)50(07.003.0 +=+ xx Step 3: Solve the Equation:

50

102.0

5.205.05.303.0

)50(05.0)50(07.003.0

==

+=++=+

x

x

xx

xx

Step 4: State the Conclusion: 50 gallons of the 3% solution must be mixed.

Investment Problems

Example: A salesperson used his $3,500 year-end bonus to purchase some old coins, with hopes of earning 15% annual interest on the gold coins and 12% annual interest on the silver coins. If he saw a return on his investment of $480.00 the first year, how much did he invest in each type of coin? Solution: Step 1: Analyze the Problem: An investment problem is a type of mixture problem. Therefore, it is often advantageous to make a table to organize the information.

Coins Earning Rate Amount Return Gold 15% x 0.15x Silver 12% 3500-x 0.12(3500-x)

Step 2: Create an Equation: The equation is based on the fact that the sum of the yields from the two investments must be equal to his total return of $480.00

480)3500(12.015.0 =−+ xx Step 3: Solve the Equation:

000,2$

6003.0

48012.042015.0

480)3500(12.015.0

==

=−+=−+

x

x

xx

xx

Step 4: State the Conclusion: $2,000 must be invested in gold and $1500 in silver.

Example: Of the $50,000 that Jason pocketed on his last real estate deal, $20,000 went to charity. He invested part of the remainder in Dreyfus New Leaders Fund with an annual yield of 16% and the rest in Templeton Growth Fund with an annual yield of 25%. If Jason made $6,060.00 on these investments in 1 year, then how much did he invest in each fund? Solution: Step 1: Analyze the Problem: An investment problem is a type of mixture problem. Therefore, it is often advantageous to make a table to organize the information.

Investment Interest Rate Amount Yield Dreyfus 16% x 0.16x

Templeton 25% 30,000-x 0.25(30,000-x)

Step 2: Create an Equation: The equation is based on the fact that the sum of the yields from the two investments must be equal to the total yield of $6,060.00.

060,6)000,30(25.016.0 =−+ xx Step 3: Solve the Equation:

000,14000,30

000,16$09.0

1440

144009.0

606025.0500,716.0

6060)000,30(25.016.0

=−=

−−=

−=−=−+

=−+

x

x

x

x

xx

xx

Step 4: State the Conclusion: $16,000 was invested in Dreyfus New Leaders Fund and $14,000 was invested in Templeton Growth Fund.

Example: Desert Sky construction contracted Amherst High and Memorial stadium for a total cost of $4.7 million. Because the construction was not completed on time, Desert Sky paid 5% of the amount of the high school contract and 4% of the stadium contract in penalties. If the penalty was $223,000 then what was the amount of each contract? Solution: Step 1: Analyze the Problem: An investment problem is a type of mixture problem. Therefore, it is often advantageous to make a table to organize the information.

Contract Penalty Rate Amount Penalty School 5% x 0.05x Stadium 4% 4,700,000-x 0.04(4,700,000-x)

Step 2: Create an Equation: The equation is based on the fact that the sum of the penalties from the two contracts is equal to the total penalty of $223,000

0000,223)000,700,4(04.005.0 =−+ xx Step 3: Solve the Equation:

000,500,3

000,3501.0

000,223000,18801.0

000,22304.0000,18805.0

000,223)000,700,4(04.005.0

==

=+=−+

=−+

x

x

x

xx

xx

Step 4: State the Conclusion: The high school contract is $3,500,000 and the stadium contract is $1,200,000.

Number-Value Problems

Example: When a child emptied his coin bank, he had a collection of pennies, nickels, and dimes. There were 20 more pennies than dimes and the number of nickels were triple the number of dimes. If the coins had a value of $5.40, how many of each type coin were in the bank? Solution: Step 1: Analyze the Problem: Number-value problems are also a type of mixture problem and a table is recommended to organize the information.

Coins Value Amount Total Pennies 0.01 x+20 0.01(x+20) Nickels 0.05 3x 0.05(3x) Dimes 0.1 x 0.1x

Step 2: Create an Equation: The equation is based on the fact that the total value of the pennies, nickels, and dimes is $5.40.

40.51.0)3(05.0)20(01.0 =+++ xxx Step 3: Solve the Equation:

20

52026

5402026

540101520

40.51.015.02.001.0

40.51.0)3(05.0)20(01.0

==

=+=+++

=+++=+++

x

x

x

xxx

xxx

xxx

Step 4: State the Conclusion: There are 20 dimes, 40 pennies, and 60 nickels.

Work Problems Example: With the old combine Rick’s entire wheat crop can be harvested in 72 hours but a new combine can do the same job in 48 hours. How many hours would it take to harvest the crop with both combines operating? Solution: Step 1: Analyze the Problem: This is a work problem. Once again a table is helpful for organizing the information.

Combine Work /hour hours working work old 1/72 x 1/72(x) new 1/48 x 1/48(x)


14872

=+ xx


8.28

3456120

34567248

14872

==

=+

=+

x

x

xx

xx

Step 4: State the Conclusion: The time it takes the combines working together to harvest the crop (complete 1 job) is 28 hours and 48 minutes.

Example: Tom can paint a fence in 5 hours. Huckleberry can paint the same fence in 7 hours. If Tom gets up early and starts painting a 8:30 AM and Huckleberry joins him 2.5 hours later, at what time will they finish painting the fence? Solution: Step 1: Analyze the Problem: This is a work problem. Once again a table is helpful for organizing the information.

Combine Work /hour hours working work Tom 1/5 x 1/5(x)

Huckleberry 1/7 x-2.5 1/7(x-2.5)


17

5.2

5=−+ xx


95.3

5.4712

355.1257

17

5.2

5

==

=−+

=−+

x

x

xx

xx

Step 4: State the Conclusion: Tom will work for 3 hours and 57 minutes and Huckleberry will work for 1 hour and 27 minutes. The fence will be completed at 12:27 PM.

Documents

1.2 Constructing Models to Solve Problems · PDF file01.11.2014 · 1.2 – Constructing Models to Solve Problems In the previous section, we learned how to solve linear equations