chapter 3 applications to linear functions the word problems

Chapter 3 Applications to Linear FunctionsThe Word Problems

The Beginnings

When dealing with word problems, we have to do a bit more work.We need to assign variables to the problem and decide what thefunction is that must be optimized before we can do all the work wedid in earlier examples. It is often easiest to create a table to organizethe problem in order to construct the inequalities.

Setting Up Word Problems

ExampleSuppose we have two factories, A and B. At factory A, a productneeds to be worked on for 3 hours in department 1 and 2 hours indepartment 2 and we have 120 labor hours available. At factory B, thesame product is worked on for 4 hours in department 1 and 6 hours indepartment 2 and there are 260 hours available for labor. The profit is$5 per unit from department 1 and $6 per unit from department 2.Write the following:

a. Let x be the units in department 1 and y be the units in department2. Write the inequalities that x and y must satisfy in order to makesure we stay within the hours constraints.

b. Express any other constraints.

c. Find the object function.


We will set up a table to organize the information.

Department 1 Department 2 HoursABProfit


Department 1 Department 2 HoursA 3 2 120B 4 6 260Profit 5 6

From this, we get the constraints and object function. MaximizeP = 5x + 6y subject to the constraints

3x + 2y ≤ 1204x + 6y ≤ 260x ≥ 0, y ≥ 0

At this point the problem is at the point of the numerical examples wejust completed.

Baseball Equipment Example

ExampleA company makes baseballs and baseball bats. Each ball requires 2hours to make it and 2 hours of testing. Each bat requires 3 hours tomake it and 1 hour of testing. Each day there are 42 labor hoursavailable for making products and 26 hours available for testing. Howmany of each type should the company produce daily to maximize itsdaily output?

The Solution

Assign x as the number of baseballs and y as the number of bats. Wewant to maximize our output, which is how many total items there areto be produced. A table will help with the derivation of theinequalities.

Make TestBallsBats

When we set up the table to this point, we can fill in the quantities weare given in the statement of the problem.

The Solution

The Part We NeedEach ball requires 2 hours to make it and 2 hours of testing. Each batrequires 3 hours to make it and 1 hour of testing

Make TestBalls 2 2Bats 3 1

The Solution

Now, we need to figure out where the limits on the differentconstraints - do they go with the balls and bats or with the testing andmaking of the products? The limits are on the time for the making andtesting, so we would add another row for these limits.

The Part We NeedEach day there are 42 labor hours available for making products and26 hours available for testing. How many of each type should thecompany produce daily to maximize its daily output?

Make Test OutputBalls (x) 2 2 1Bats (y) 3 1 1Limits 42 26

The Solution

This now gives us all we need, besides the direction of theinequalities. But that we should be able to get from the context of theproblem.

The Part We NeedEach day there are 42 labor hours available for making products and26 hours available for testing.

Make Test OutputBalls (x) 2 2 1Bats (y) 3 1 1

≤ ≤Limits 42 26

The System

So, the problem becomes:

Our TaskMaximize P = x + y subject to the constraints

2x + 3y ≤ 422x + y ≤ 26x ≥ 0, y ≥ 0

Notice we added in the restrictions on x and y being non-negative.This comes from the context of the problem ...

The Rewrite

Now we rewrite the inequalities to put them in slope-intercept form.

y ≤ −2

3 x + 14 Iy ≤ −2x + 26 IIx ≥ 0, y ≥ 0

The Rewrite

Now we rewrite the inequalities to put them in slope-intercept form.y ≤ −2

3 x + 14 Iy ≤ −2x + 26 IIx ≥ 0, y ≥ 0

The Graph and the Corners

10

20

30

10 20 30

•

•

I

•

•

II

F. S.

The Points

1 The origin (0, 0)

2 y-intercept of IWe can directly obtain the y-intercept from the inequality.(0, 14)

3 x-intercept of IIWe can solve by setting the equation associated with II equal to0.

0 = −2x + 26⇒ 2x = 26⇒ x = 13

(13, 0)4 I=II

Now some algebra

The Points

1 The origin (0, 0)2 y-intercept of I

We can directly obtain the y-intercept from the inequality.(0, 14)


0 = −2x + 26⇒ 2x = 26⇒ x = 13

(13, 0)4 I=II

Now some algebra

The Points


We can directly obtain the y-intercept from the inequality.

(0, 14)3 x-intercept of II

We can solve by setting the equation associated with II equal to0.

0 = −2x + 26⇒ 2x = 26⇒ x = 13

(13, 0)4 I=II

Now some algebra

The Points



3 x-intercept of II

We can solve by setting the equation associated with II equal to0.

0 = −2x + 26⇒ 2x = 26⇒ x = 13

(13, 0)4 I=II

Now some algebra

The Points




0 = −2x + 26⇒ 2x = 26⇒ x = 13

(13, 0)4 I=II

Now some algebra

The Points




0 = −2x + 26⇒ 2x = 26⇒ x = 13

(13, 0)

4 I=IINow some algebra

The Points




0 = −2x + 26⇒ 2x = 26⇒ x = 13

(13, 0)4 I=II

Now some algebra

The Last Point

−23

x + 14 = −2x + 26

−2x + 42 = −6x + 78

4x = 36

x = 9

and then

y = −2x + 26⇒ y = −2(9) + 26⇒ y = 8

giving the point (9, 8).

Finding the Maximum

Point P = x + y(0, 0) 0(0, 14) 14(13, 0) 13(9, 8) 17

Therefore, we maximize our daily output at 17 products when wemake 9 balls and 8 bats.

Bicycle Example

ExampleA small manufacturing plant produces two kinds of bicycles, a3-speed and a 10-speed, in two factories. Factory A produces 163-speeds and 20 10-speeds a day. Factory B produces 12 3-speeds and20 10-speeds a day. An order is received for 96 3-speeds and 14010-speeds. It costs $1000 per day to operate factory A and $800 perday to operate factory B. How many days should the manufactureroperate each factory to fill the order with the minimum cost?

Solution

Let x be the number of days of operation for factory A and let y be thenumber of days of operation for factory B. We now set up the table.Why are we using these as the variables?

A B order3-speed 16 12 9610-speed 20 20 140cost 1000 800

Are these inequalities ‘less than’ or ‘greater than’?

The System of Inequalities

Minimize 1000x + 800y subject to the constraints16x + 12y ≥ 9620x + 20y ≥ 140x ≥ 0, y ≥ 0

And when we rewrite, we get

Minimize 1000x + 800y subject to the constraintsy ≥ −4

3 x + 8 Iy ≥ −x + 7 IIx ≥ 0, y ≥ 0


3

6

9

3 6 9

•

•

I

•

•

IIF. S.

y-int I

x-int II

I=II


3

6

9

3 6 9

•

•

I

•

•

II

F. S.y-int I

x-int II

I=II


3

6

9

3 6 9

•

•

I

•

•

IIF. S.

y-int I

x-int II

I=II

The Points

1 y-intercept I

From the slope-intercept form of the equation, we get (0, 8).2 x-intercept of II

We set y = 0 in the equation y = −x + 7.

0 = −x + 7⇒ x = 7

So the point is (7, 0).3 I=II

Setting the equations equal to each other gives our x coordinate.

−43

x + 8 = −x + 7⇒ x = 3

and we plug this into either equation to get y = 4, giving thepoint (3, 4).

The Points

1 y-intercept IFrom the slope-intercept form of the equation, we get (0, 8).

2 x-intercept of IIWe set y = 0 in the equation y = −x + 7.

0 = −x + 7⇒ x = 7



−43

x + 8 = −x + 7⇒ x = 3


The Points


2 x-intercept of II

We set y = 0 in the equation y = −x + 7.

0 = −x + 7⇒ x = 7



−43

x + 8 = −x + 7⇒ x = 3


The Points



0 = −x + 7⇒ x = 7



−43

x + 8 = −x + 7⇒ x = 3


The Points



0 = −x + 7⇒ x = 7

So the point is (7, 0).

3 I=IISetting the equations equal to each other gives our x coordinate.

−43

x + 8 = −x + 7⇒ x = 3


The Points



0 = −x + 7⇒ x = 7



−43

x + 8 = −x + 7⇒ x = 3


Testing the Points

Point m = 1000x + 800y(0, 8) 1000(0) + 800(8) = $6400(7, 0) 1000(7) + 800(0) = $7000(3, 4) 1000(3) + 800(4) = $6200

The minimum value here is $6200 and this occurs when factory A isopen for 3 days and factory B is open for 4 days.

Furniture Example

ExampleA furniture finishing plant finishes two kinds of tables, A and B. TableA requires 8 minutes of staining and 9 minutes of varnishing, wheretable B requires 12 minutes of staining and 6 minutes of varnishing.The staining facility is available at most 480 minutes in a day and thevarnishing facility is available at most 360 minutes a day. The planthas to finish at least as many table Bs as half the number of table As.The profit on each table A is $5 and $3 on each table B. Find themaximum profit.

The Table

A B minutesstaining 8 12 480varnishing 9 6 360profit 5 3

The System of Equations

Maximize P = 5x + 3y subject to the constraints8x + 12y ≤ 4809x + 6y ≤ 360y ≥ 1

2 xx ≥ 0, y ≥ 0

The Rewrite

y ≤ −2

3 x + 40 Iy ≤ −3

2 x + 60 IIy ≥ 1

2 x IIIx ≥ 0, y ≥ 0


20

40

60

20 40 60

•

•

I

•

•

II

III

F. S.

The Points

1 The origin (0, 0)2 The intersection of I and II (24, 24)3 The intersection of II and III (30, 15)4 The y-intercept of I (0, 40)

Testing the Points

Point P = 5x + 3y(0, 0) 0(24, 24) 192(30, 15) 195(0, 40) 120

The maximum profit of P=$195 occurs at when we make 30 of type Aand 15 of type B.

Truck Transport Example

ExampleA truck traveling from New York to Baltimore is to be loaded withtwo types of cargo. Each crate of cargo A is 4 ft3 in volume, weighs100 lbs and earns $13 for the driver. Each crate of cargo B is 3 ft3 involume, weighs 200 lbs and earns $9 for the driver. The truck cancarry no more than 300 ft3 of crates and no more than 10, 000 lbs.The number of crates of cargo B must be less than or equal to twicethe number of crates of cargo A. Find the number of each type ofcargo that would maximize profit.

The Table

A B capacityvolume 4 3 300weight 100 200 10000earnings 13 9


Maximize E = 13x + 9y subject to the constraints4x + 3y ≤ 300100x + 200y ≤ 10000y ≤ 2xx ≥ 0, y ≥ 0

The Rewrite

y ≤ −4

3 x + 100 Iy ≤ −1

2 x + 50 IIy ≤ 2x IIIx ≥ 0, y ≥ 0


25

50

75

100

25 50 75 100

•

•

I

•

•

II

III

F. S.

The Points

1 The origin (0, 0)2 The intersection of II and III (20, 40)3 The intersection of I and II (60, 20)4 The x-intercept of I (75, 0)

Checking the Points

Point E = 13x + 9y(0, 0) 0(20, 40) 620(60, 20) 960(75, 0) 975

The maximum occurs of $975 occurs when we ship 75 crates of cargoA and 0 crates of cargo B..

Another Example

ExampleA manufacturer produces two items, A and B. A maximum of 2000units can be produced per day. The cost is $3 per unit for A and $5 perunit for B. The daily production budget is $7500. If the manufacturermakes a makes a profit of $1.75 per unit for A and $2.50 per unit forB, how many units of each should be produced to maximize profit?

The Table

Let x be the number of type A and y be the number of type B.

A B limitsunits 1 1 2000cost 3 5 7500profit 1.75 2.50


Maximize P = 1.75x + 2.50y subject to the constraintsx + y ≤ 20003x + 5y ≤ 7500x ≥ 0, y ≥ 0

The Rewrite

y ≤ −x + 2000 Iy ≤ −3

5 x + 1500 IIx ≥ 0, y ≥ 0


500

1000

1500

2000

500 1000 1500 2000 2500

•

•

I

•

•

II

F. S.

The Points

1 The origin (0, 0)2 The y-intercept of II (0, 1500)3 The intersection of I and II (1250, 750)4 The x-intercept of I (2000, 0)

Checking the Points

Point P = 1.75x + 2.50y(0, 0) 0

(0, 1500) 3750(1250, 750) 4062.5(2000, 0) 3500

The maximum occurs of $4062.50 occurs when we produce 1250 ofproduct A and 750 of product B.

Pottery Example

ExampleA potter is making cups and plates. It takes her 6 minutes to make acup and 3 minutes to make a plate. Each cup uses 3/4 lb. of clay andeach plate uses one lb. of clay. She has 20 hours available for makingthe cups and plates and has 250 lbs. of clay on hand. She makes aprofit of $2 on each cup and $1.50 on each plate. How many cups andhow many plates should she make in order to maximize her profit?

Note that time is given in hours and minutes, but we need to pick oneto make it consistent. In order to avoid more fractions, we will useminutes.

The Table

cups plates limitstime 6 3 1200clay 3

4 1 250profit 2.00 1.50


Maximize P = 2x + 1.5y subject to the constraints6x + 3y ≤ 120034 x + y ≤ 250x ≥ 0, y ≥ 0

The Rewrite

y ≤ −2x + 400 Iy ≤ −3

4 x + 250 IIx ≥ 0, y ≥ 0


100

200

300

400

100 200 300 400

•

•

I

•

•

II

F. S.

The Points

1 The origin (0, 0)2 The y-intercept of II (0, 250)3 The x-intercept of I (200, 0)4 The intersection of I and II (120, 160)

Testing the Points

Point P = 2x + 1.5y(0, 0) 0

(0, 250) 375(200, 0) 400(120, 160) 480

The maximum profit of $480 occurring when 120 cups and 160 platesare sold.

A Non-Table Example

ExampleA calculator company produces a scientific calculator and a graphingcalculator. Long-term projections indicate an expected demand of atleast 100 scientific and 80 graphing calculators each day. Because oflimitations on production capacity, no more than 200 scientific and170 graphing calculators can be made daily. To satisfy a shippingcontract, a total of at least 200 calculators much be shipped each day.If each scientific calculator sold results in a $2 loss, but each graphingcalculator produces a $5 profit, how many of each type should bemade daily to maximize net profits?

Solution

An example like this doesn’t lend itself to a table to organize theinformation.

We can begin with the assignment of variables as before, with x beingthe number of scientific calculators and y being the number ofgraphing calculators.

Maximize P = −2x + 5y subject to the constraints100 ≤ x ≤ 20080 ≤ y ≤ 170x + y ≥ 200

The Rewrite

100 ≤ x ≤ 20080 ≤ x ≤ 170y ≥ −x + 200


50

100

150

200

50 100 150 200

•

•

F. S.

The Points

1 Corners from horizontal and vertical constraints:(100, 170)(200, 170)(200, 80)

2 The line intersects with horizontal and vertical constraints:(100, 100)(120, 80)

Note: each of these points we either know one coordinate or bothbecause each has at least one coordinate that comes from a horizontalor vertical line.

Testing the Points

Point P = −2x + 5y(100, 170) 650(200, 170) 450(200, 80) 0(100, 100) 300(120, 80) 160

The maximum profit of $650 occurs when 100 scientific and 170graphing calculators are sold.

Points on an Exam

ExampleA student is taking an exam consisting of 10 essay questions and 50short answer questions. He has 90 minutes to complete the exam andknows he cannot possibly answer all questions. The essay questionsare worth 20 points and the short answer questions are worth 5 points.An essay question takes 10 minutes to answer and a short answerquestion takes 2 minutes. The student must do at least 3 essayquestions and at least 10 short answer questions. Find how manycorrect of each type will maximize the exam score.

The Table

Let x be the number of essay questions and let y be the number ofshort answer questions.

essay short answer availabletime 10 2 90quantity 10 50required 3 10value 20 5


Maximize Score = 20x + 5y subject to the constraints10x + 2y ≤ 903 ≤ x ≤ 1010 ≤ y ≤ 50

The Rewrite

y ≤ −5x + 453 ≤ x ≤ 1010 ≤ y ≤ 50

The Plot and the Corners

12.5

25

37.5

50

2.5 5 7.5 10

•

•

F. S.

The Points

Again, as in last example, at least one of the coordinates of each pointis either directly from a constraint represented by a horizontal or avertical line in the graph.

1 (3, 30)2 (7, 10)3 (3, 10)

Testing the Points

Point S = 20x + 5y(3, 30) 210(7, 10) 190(3, 10) 110

The maximum score of 210 occurs when 3 essay and 30 short answerquestions are (correctly) answered.

Law Firm Example

ExampleThe Sue All Law Firm handles two types of lawsuits: medicalmalpractice suits against unscrupulous heart surgeons for performingunnecessary surgery, and suits against hard-working math professorsfor failing students who do not deserve to pass. Math professorlawsuits each require 6 person-months of preparation and the hiring of5 expert witnesses, whereas medical lawsuits each require 10person-months of preparation and the hiring of 3 expert witnesses.The firm has a total of 30 person-months to work with and feels that itcannot afford to hire more than 15 expert witnesses. It makes anaverage profit of $1 million per math professor sued and $5 millionper heart surgeon sued. How many of each type of lawsuit should itinitiate in order to maximize its expected profits?

The Table

Let x be the number of lawsuits against doctors and let y be thenumber of lawsuits against math professors.

medical math limitsprep 10 6 30witnesses 3 5 15profit 5 1


Maximize Profit = 5x + y subject to the constraints10x + 6y ≤ 303x + 5y ≤ 15x ≥ 0, y ≥ 0

The Rewrite

Maximize Profit = 5x + y subject to the constraintsy ≤ −5

3 x + 5 Iy ≤ −3

5 x + 3 IIx ≥ 0, y ≥ 0


3

6

9

3 6 9

I

II

F.S.

x-int I

y-int II

origin

I=II

The Points

The only one that requires algebra is the point where the linesintersect.

−53

x + 5 = −35

x + 3

−25x + 75 = −9x + 45

30 = 16x158

= x

y = −53· 15

8+ 5

= −258

+ 5

= −258

+408

=158

1 (0, 0)2 (3, 0)3(15

8 ,158

)4 (0, 3)

Testing the Points

Point P = 5x + y(0, 0) 0(3, 0) 15( 158 ,

158

) 454

(0, 3) 3

The maximum profit of $15 million occurs when 3 medical lawsuitsand no professor lawsuits are handled.

How did we know the intersection of the two lines gives animpossible point?

chapter 3 applications to linear functions the word problems

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