bmayer@chabotcollege.edu mth55_lec-35_sec_6-7_rational_equation_applications.ppt 1 bruce mayer, pe...

BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Chabot Mathematics

§6.7 Rational§6.7 RationalEqn AppsEqn Apps

Review §Review §

Any QUESTIONS About• §6.6 → Rational Equations

Any QUESTIONS About HomeWork• §6.6 → HW-22

6.6 MTH 55

§6.7 Rational Equation Applications§6.7 Rational Equation Applications

Problems Involving Work

Problems Involving Motion

Problems Involving Proportions

Problems involving Average Cost

Solve a Formula for a VariableSolve a Formula for a Variable

Formulas occur frequently as mathematical models. Many formulas contain rational expressions, and to solve such formulas for a specified letter, we proceed as when solving rational equations.

Solve Rational Eqn for a VariableSolve Rational Eqn for a Variable1. Determine the DESIRED letter (many times

formulas contain multiple variables)2. Multiply on both sides to clear fractions or

decimals, if that is needed. 3. Multiply if necessary to remove parentheses. 4. Get all terms with the letter to be solved for on one

side of the equation and all other terms on the other side, using the addition principle.

5. Factor out the unknown.6. Solve for the letter in question, using the

multiplication principle.

Example Example Solve for Letter Solve for Letter

Solve this formula for y: .aT

SOLN:aT

T ay T aaT

T ayyR

R T ay aT

RT Ray aT

Ray aT RT

aT RTy

Multiplying both sides by the LCD

Simplifying

Dividing both sides by Ra

Multiplying

Subtracting RT

Example Example Fluid Mechanics Fluid Mechanics

In a hydraulic system, a fluid is confined to two connecting chambers. The pressure in each chamber is the same and is given by finding the force exerted (F) divided by the surface area (A). Therefore, we know

Solve this Eqn for A2

Example Example Fluid Mechanics Fluid Mechanics

SOLUTION:

This formula can be used to calculate A2 whenever A1, F2, and F1 are known

22 1 2A

AA A A

2 1 1 2A F A F

Multiplying both sides by the LCD

Dividing both sides by F1

Problems Involving WorkProblems Involving Work

Rondae and Marrisa work during the summer painting houses. • Rondae can paint an average size

house in 12 days

• Marrisa requires 8 days to do the same painting job.

How long would it take them, working together, to paint an average size house?

House Painting cont.House Painting cont.

1. Familiarize. We familiarize ourselves with the problem by exploring two common, but incorrect, approaches.

a) One common, incorrect, approach is to add the two times. → 12 + 8 = 20

b) Another incorrect approach is to assume that Rondae and Marrisa each do half the painting.

– Rondae does ½ in 12 days = 6 days– Marrisa does ½ in 8 days = 4 days – 6 days + 4 days = 10 days.

House Painting cont.House Painting cont. A correct approach is to consider how much

of the painting job is finished in ONE day; i.e., consider the work RATE

It takes Rondae 12 days to finish painting a house, so his rate is 1/12 of the job per day.

It takes Marrisa 8 days to do the painting alone, so her rate is 1/8 of the job per day.

Working together, they can complete 1/8 + 1/12, or 5/24 of the job in one day.

Note That given a TIME-Rate

[Amount] = [Rate]•[TimeQuantity]

t/8t1/8Marrisa

t/12t1/12Rondae

Amount Completed

TimeRate of Work

Painter

Form a table to help organize the info:

2. Translate. The time that we want is some number t for which

12 8t t Portion of work done by

Marrisa in t daysPortion of work done by Rondae in t days

Or 1 o1 1

2 8 24t t

Portion of work done together in t days

3. Carry Out. We can choose any one of the above equations to solve:

24 4, or 4 days

4. Check. Test t = 24/5 days

24 241

1 2 3 51

12 8 5 5 5

5. State. Together, it will take Rondae & Marrisa 4 & 4/5 days to complete painting a house.

The WORK PrincipleThe WORK Principle

Suppose that A requires a units of time to complete a task and B requires b units of time to complete the same task.

Then A works at a rate of 1/a tasks per unit of time.

B works at a rate of 1/b tasks per unit of time,

Then A and B together work at a totalrate of [1/a + 1/b] per unit of time.

The WORK PrincipleThe WORK Principle

If A and B, working together, require t units of time to complete a task, then their combined rate is 1/t and the following equations hold:

Problems Involving MotionProblems Involving Motion

Because of a tail wind, a jet is able to fly 20 mph faster than another jet that is flying into the wind. In the same time that it takes the first jet to travel 90 miles the second jet travels 80 miles. How fast is each jet traveling?

r r+20

HEAD Wind TAIL Wind

HEADwind vs. TAILwindHEADwind vs. TAILwind

1. Familiarize. We try a guess. If the fast jet is traveling 300 mph because of a tail wind the slow jet plane would be traveling 300−20 or 280 mph.

• At 300 mph the fast jet would have a 90 mile travel-time of 90/300, or 3/10 hr.

• At 280 mph, the other jet would have a travel-time of 80/280 = 2/7 hr.

Now both planes spend the same amount of time traveling, So the guess is INcorrect.

2. Translate. Fill in the blanks using

[TimeQuantity]=[Distance]/[Rate]

AirCraft

Distance(miles)

Speed or Rate(miles per hour)

Time(hours)

Jet 1 80 r

Jet 2 90 r + 20

r r+20

Set up a RATE Table

[Distance]/[Rate] = [TimeQuantity]

AirCraft

Distance(miles)

Speed(miles per hour)

Time(hours)

Jet 1 80 r 80/r

Jet 2 90 r + 20 90/(r + 20)

The Times MUST be the SAME

Since the times must be the same for both planes, we have the equation

3. Carry Out. To solve the equation, we first Clear-Fractions multiplying both sides by the LCD of r(r+20)

( 20) (80 9

0r r r r

Complete the “Carry Out”

80( 20) 90r r

80 1600 90r r 1600 10r

Simplified by Clearing Fractions

Using the distributive law

Subtracting 80r from both sides

Dividing both sides by 10

Now we have a possible solution. The speed of the slow jet is 160 mph and the speed of the fast jet is 180 mph

4. Check. ReRead the problem to confirm that we were able to find the speeds. At 160 mph the jet would cover 80 miles in ½ hour and at 180 mph the other jet would cover 90 miles in ½ hour. Since the times are the same, the speeds Chk

5. State. One jet is traveling at 160 mph and the second jet is traveling at 180 mph

Formulas in EconomicsFormulas in Economics

Linear Production Cost Function

C x variable cost fixed costs ax b

• Where– b is the fixed cost in $

– a is the variable cost of producing each unit in $/unit (also called the marginal cost)

AverageCost ($/unit) C x C x

Formulas in EconomicsFormulas in Economics

Price-Demand Function: Suppose x units can be sold (demanded) at a price of p dollars per units.

• Where– m & n are SLOPE Constants in $/unit & unit/$

– d & k are INTERCEPT Constants in $ & units

p x mx d

x p np k

Formulas in Economics

Revenue Function

Revenue = (Price per unit)·(No. units sold)

xdmxxpxR

Profit FunctionProfit = (Total Revenue) – (Total Cost)

P x R x C x

Example Average Cost

Metro Entertainment Co. spent $100,000 in production costs for its off-Broadway play Pride & Prejudice. Once running, each performance costs $1000

a) Write the Cost Function for conducting z performances

b) Write the Average Cost Function for the z performances

c) How many performances, n, result in an average cost of $1400 per show

Example Example Average Cost Average Cost

SOLUTION a) Total Cost is the sum of the Fixed Cost and the Variable Cost

zzCshow

k1k100

SOLUTION b) The Average Cost Fcn

zCzC show

k1k100

Example Example Average Cost Average Cost

SOLUTION c) In this case for “n” Shows

k1$k100$showk1$

k100$k4.1$

nn k1$k100k4.1$ k100k4.0 n

k100k4.0 n250n

Thus 250 shows are needed to realize a per-show cost of $1400

Problems Involving ProportionsProblems Involving Proportions Recall that a RATIO of two quantities is

their QUOTIENT.• For example, 45% is the ratio of 45 to 100,

or 45/100.

A proportion is an equation stating that two ratios are EQUAL:

An equality of ratios, An equality of ratios, AA//BB = = CC//DD, is , is called a PROPORTION. The numbers called a PROPORTION. The numbers

within a proportion are said to be within a proportion are said to be proportionAL to each otherproportionAL to each other

Example Example Triangle Triangle ProportionsProportions Triangles ABC and XYZ are “similar”

y = 12

Now Solve for b if

x = 8, y = 12 and a = 7

• Note that “Similar” Triangles are “In Proportion” to Each other

Example Example Similar Triangles Similar Triangles

Set Up TheProportions

y = 12

84 or 10.5

[b is to 12]as

[7 is to 8]

AlternativeProportions

y = 12

84 or 10.5

[b is to 7]as

[12 is to 8]

Example Example Quantity Proportions Quantity Proportions

A sample of 186 hard drives contained 4 defective drives. How many defective drives would be expected in a group of 1302 HDDs?

Form a proportion in which the ratio of defective hard drives is expressed in 2 ways.

186 1302

defective drives

total drives

defective drives

total drives

186 5208x

Expect to find 28 defective HDDs

Whale ProportionalityWhale Proportionality

To determine the number of humpback whales in a pod, a marine biologist, using tail markings, identifies 35 members of the pod.

Several weeks later, 50 whales from the SAME pod are randomly sighted. Of the 50 sighted, 18 are from the 35 originally identified. Estimate the number of whales in the pod.

Tagged Whale ProportionsTagged Whale Proportions

1. Familarize. We need to reread the problem to look for numbers that could be used to approximate a percentage of the of the pod sighted.

Since 18 of the 35 whales that were later sighted were among those originally identified, the ratio 18/50 estimates the percentage of the pod originally identified.

HumpBack WhalesHumpBack Whales

2. Translate: Stating the Proportion

Marked whales sighted later

Total Whales sighted later

Whales originally Marked

Total Whales in pod

3. CarryOut

550 50

50 35 or 97.22

50 35 18 w

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