Chapter 3 Applications to Linear Functions
The 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 the
function is that must be optimized before we can do all the work we
did in earlier examples. It is often easiest to create a table to organize
the problem in order to construct the inequalities.
Setting Up Word Problems
Example
Suppose we have two factories, A and B. At factory A, a product
needs to be worked on for 3 hours in department 1 and 2 hours in
department 2 and we have 120 labor hours available. At factory B, the
same product is worked on for 4 hours in department 1 and 6 hours in
department 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 department
2. Write the inequalities that x and y must satisfy in order to make
sure we stay within the hours constraints.
b. Express any other constraints.
c. Find the object function.
Setting Up Word Problems
We will set up a table to organize the information.
Department 1
A
B
Profit
Department 2
Hours
Setting Up Word Problems
A
B
Profit
Department 1
3
4
5
Department 2
2
6
6
Hours
120
260
Setting Up Word Problems
A
B
Profit
Department 1
3
4
5
Department 2
2
6
6
Hours
120
260
From this, we get the constraints and object function. Maximize
P = 5x + 6y subject to the constraints

 3x + 2y ≤ 120
4x + 6y ≤ 260

x ≥ 0, y ≥ 0
Setting Up Word Problems
A
B
Profit
Department 1
3
4
5
Department 2
2
6
6
Hours
120
260
From this, we get the constraints and object function. Maximize
P = 5x + 6y subject to the constraints

 3x + 2y ≤ 120
4x + 6y ≤ 260

x ≥ 0, y ≥ 0
At this point the problem is at the point of the numerical examples we
just completed.
Baseball Equipment Example
Example
A company makes baseballs and baseball bats. Each ball requires 2
hours to make it and 2 hours of testing. Each bat requires 3 hours to
make it and 1 hour of testing. Each day there are 42 labor hours
available for making products and 26 hours available for testing. How
many of each type should the company produce daily to maximize its
daily output?
The Solution
Assign x as the number of baseballs and y as the number of bats. We
want to maximize our output, which is how many total items there are
to be produced. A table will help with the derivation of the
inequalities.
Make
Test
Balls
Bats
When we set up the table to this point, we can fill in the quantities we
are given in the statement of the problem.
The Solution
The Part We Need
Each ball requires 2 hours to make it and 2 hours of testing. Each bat
requires 3 hours to make it and 1 hour of testing
The Solution
The Part We Need
Each ball requires 2 hours to make it and 2 hours of testing. Each bat
requires 3 hours to make it and 1 hour of testing
Balls
Bats
Make
2
3
Test
2
1
The Solution
Now, we need to figure out where the limits on the different
constraints - do they go with the balls and bats or with the testing and
making of the products? The limits are on the time for the making and
testing, so we would add another row for these limits.
The Solution
Now, we need to figure out where the limits on the different
constraints - do they go with the balls and bats or with the testing and
making of the products? The limits are on the time for the making and
testing, so we would add another row for these limits.
The Part We Need
Each day there are 42 labor hours available for making products and
26 hours available for testing. How many of each type should the
company produce daily to maximize its daily output?
The Solution
Now, we need to figure out where the limits on the different
constraints - do they go with the balls and bats or with the testing and
making of the products? The limits are on the time for the making and
testing, so we would add another row for these limits.
The Part We Need
Each day there are 42 labor hours available for making products and
26 hours available for testing. How many of each type should the
company produce daily to maximize its daily output?
Balls (x)
Bats (y)
Limits
Make
2
3
42
Test
2
1
26
Output
1
1
The Solution
This now gives us all we need, besides the direction of the
inequalities. But that we should be able to get from the context of the
problem.
The Solution
This now gives us all we need, besides the direction of the
inequalities. But that we should be able to get from the context of the
problem.
The Part We Need
Each day there are 42 labor hours available for making products and
26 hours available for testing.
The Solution
This now gives us all we need, besides the direction of the
inequalities. But that we should be able to get from the context of the
problem.
The Part We Need
Each day there are 42 labor hours available for making products and
26 hours available for testing.
Balls (x)
Bats (y)
Limits
Make
2
3
≤
42
Test
2
1
≤
26
Output
1
1
The System
So, the problem becomes:
Our Task
Maximize P = x + y subject to the constraints

 2x + 3y ≤ 42
2x + y ≤ 26

x ≥ 0, y ≥ 0
The System
So, the problem becomes:
Our Task
Maximize P = x + y subject to the constraints

 2x + 3y ≤ 42
2x + y ≤ 26

x ≥ 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.
The Rewrite
Now we rewrite the inequalities to put them in slope-intercept form.

 y ≤ − 23 x + 14 I
y ≤ −2x + 26 II

x ≥ 0, y ≥ 0
The Graph and the Corners
30
20
10
10
20
30
The Graph and the Corners
30
20
10
10
20
30
The Graph and the Corners
30
20
I•
10
•
10
20
30
The Graph and the Corners
30
20
I•
10
•
10
20
30
The Graph and the Corners
30
II •
20
I•
10
•
10
•
20
30
The Graph and the Corners
30
II •
20
I•
10
•
10
•
20
30
The Graph and the Corners
30
II •
20
I•
10
F. S.
•
10
•
20
30
The Graph and the Corners
30
II •
20
I•
10
F. S.
•
10
•
20
30
The Points
1
The origin (0, 0)
The Points
1
The origin (0, 0)
2
y-intercept of I
The Points
1
The origin (0, 0)
2
y-intercept of I
We can directly obtain the y-intercept from the inequality.
The Points
1
The origin (0, 0)
2
y-intercept of I
We can directly obtain the y-intercept from the inequality.
(0, 14)
3
x-intercept of II
The Points
1
The origin (0, 0)
2
y-intercept of I
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 to
0.
The Points
1
The origin (0, 0)
2
y-intercept of I
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 to
0.
0 = −2x + 26 ⇒ 2x = 26 ⇒ x = 13
(13, 0)
The Points
1
The origin (0, 0)
2
y-intercept of I
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 to
0.
0 = −2x + 26 ⇒ 2x = 26 ⇒ x = 13
(13, 0)
4
I=II
The Points
1
The origin (0, 0)
2
y-intercept of I
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 to
0.
0 = −2x + 26 ⇒ 2x = 26 ⇒ x = 13
(13, 0)
4
I=II
Now some algebra
The Last Point
2
− x + 14 = −2x + 26
3
−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
(0, 0)
(0, 14)
(13, 0)
(9, 8)
P=x+y
0
14
13
17
Finding the Maximum
Point
(0, 0)
(0, 14)
(13, 0)
(9, 8)
P=x+y
0
14
13
17
Therefore, we maximize our daily output at 17 products when we
make 9 balls and 8 bats.
Bicycle Example
Example
A small manufacturing plant produces two kinds of bicycles, a
3-speed and a 10-speed, in two factories. Factory A produces 16
3-speeds and 20 10-speeds a day. Factory B produces 12 3-speeds and
20 10-speeds a day. An order is received for 96 3-speeds and 140
10-speeds. It costs $1000 per day to operate factory A and $800 per
day to operate factory B. How many days should the manufacturer
operate 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 the
number of days of operation for factory B. We now set up the table.
Why are we using these as the variables?
Solution
Let x be the number of days of operation for factory A and let y be the
number of days of operation for factory B. We now set up the table.
Why are we using these as the variables?
3-speed
10-speed
cost
A
16
20
1000
B
12
20
800
order
96
140
Solution
Let x be the number of days of operation for factory A and let y be the
number of days of operation for factory B. We now set up the table.
Why are we using these as the variables?
3-speed
10-speed
cost
A
16
20
1000
B
12
20
800
order
96
140
Are these inequalities ‘less than’ or ‘greater than’?
The System of Inequalities
Minimize 1000x + 800y subject to the constraints

 16x + 12y ≥ 96
20x + 20y ≥ 140

x ≥ 0, y ≥ 0
The System of Inequalities
Minimize 1000x + 800y subject to the constraints

 16x + 12y ≥ 96
20x + 20y ≥ 140

x ≥ 0, y ≥ 0
And when we rewrite, we get
The System of Inequalities
Minimize 1000x + 800y subject to the constraints

 16x + 12y ≥ 96
20x + 20y ≥ 140

x ≥ 0, y ≥ 0
And when we rewrite, we get
Minimize 1000x + 800y subject to the constraints

 y ≥ − 43 x + 8 I
y ≥ −x + 7 II

x ≥ 0, y ≥ 0
The Graph and the Corners
9
6
3
3
6
9
The Graph and the Corners
9
6
3
3
6
9
The Graph and the Corners
9
I•
6
3
3
•
6
9
The Graph and the Corners
9
I•
6
3
3
•
6
9
The Graph and the Corners
9
I•
II •
6
3
3
•
6
•
9
The Graph and the Corners
9
I•
II •
6
3
3
•
6
•
9
The Graph and the Corners
9
I•
F. S.
II •
6
3
3
•
6
•
9
The Graph and the Corners
9
I•
y-int I
F. S.
II •
6
I=II
3
3
•
6
x-int II
•
9
The Points
1
y-intercept I
The Points
1
y-intercept I
From the slope-intercept form of the equation, we get (0, 8).
The Points
1
y-intercept I
From the slope-intercept form of the equation, we get (0, 8).
2
x-intercept of 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.
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).
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
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.
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.
4
− x + 8 = −x + 7 ⇒ x = 3
3
and we plug this into either equation to get y = 4, giving the
point (3, 4).
Testing the Points
Point
(0, 8)
(7, 0)
(3, 4)
m = 1000x + 800y
1000(0) + 800(8) = $6400
1000(7) + 800(0) = $7000
1000(3) + 800(4) = $6200
Testing the Points
Point
(0, 8)
(7, 0)
(3, 4)
m = 1000x + 800y
1000(0) + 800(8) = $6400
1000(7) + 800(0) = $7000
1000(3) + 800(4) = $6200
The minimum value here is $6200 and this occurs when factory A is
open for 3 days and factory B is open for 4 days.
Furniture Example
Example
A furniture finishing plant finishes two kinds of tables, A and B. Table
A requires 8 minutes of staining and 9 minutes of varnishing, where
table B requires 12 minutes of staining and 6 minutes of varnishing.
The staining facility is available at most 480 minutes in a day and the
varnishing facility is available at most 360 minutes a day. The plant
has 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 the
maximum profit.
The Table
staining
varnishing
profit
A
8
9
5
B
12
6
3
minutes
480
360
The System of Equations
Maximize P = 5x + 3y subject to the constraints

8x + 12y ≤ 480



9x + 6y ≤ 360
1

 y ≥ 2x

x ≥ 0, y ≥ 0
The Rewrite

y ≤ − 23 x + 40 I



y ≤ − 32 x + 60 II
III
y ≥ 12 x



x ≥ 0, y ≥ 0
The Graph and the Corners
60
40
20
20
40
60
The Graph and the Corners
60
40
20
20
40
60
The Graph and the Corners
60
40 I •
20
20
40
•
60
The Graph and the Corners
60
40 I •
20
20
40
•
60
The Graph and the Corners
60 II •
40 I •
20
20
•
40
•
60
The Graph and the Corners
60 II •
40 I •
20
20
•
40
•
60
The Graph and the Corners
60 II •
40 I •
III
20
20
•
40
•
60
The Graph and the Corners
60 II •
40 I •
III
20
F. S.
20
•
40
•
60
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
(0, 0)
(24, 24)
(30, 15)
(0, 40)
P = 5x + 3y
0
192
195
120
The maximum profit of P=$195 occurs at when we make 30 of type A
and 15 of type B.
Truck Transport Example
Example
A truck traveling from New York to Baltimore is to be loaded with
two types of cargo. Each crate of cargo A is 4 ft3 in volume, weighs
100 lbs and earns $13 for the driver. Each crate of cargo B is 3 ft3 in
volume, weighs 200 lbs and earns $9 for the driver. The truck can
carry 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 twice
the number of crates of cargo A. Find the number of each type of
cargo that would maximize profit.
The Table
volume
weight
earnings
A
4
100
13
B
3
200
9
capacity
300
10000
The System of Inequalities
Maximize E = 13x + 9y subject to the constraints

4x + 3y ≤ 300



100x + 200y ≤ 10000
y ≤ 2x



x ≥ 0, y ≥ 0
The Rewrite

y ≤ − 43 x + 100 I



y ≤ − 12 x + 50
II
y
≤
2x
III



x ≥ 0, y ≥ 0
The Graph and the Corners
100
75
50
25
25
50
75
100
The Graph and the Corners
100
75
50
25
25
50
75
100
The Graph and the Corners
III
100I •
75
50 II •
25
25
50
•
75
•
100
The Graph and the Corners
III
100I •
75
50 II •
25
F. S.
25
50
•
75
•
100
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
(0, 0)
(20, 40)
(60, 20)
(75, 0)
E = 13x + 9y
0
620
960
975
The maximum occurs of $975 occurs when we ship 75 crates of cargo
A and 0 crates of cargo B..
Another Example
Example
A manufacturer produces two items, A and B. A maximum of 2000
units can be produced per day. The cost is $3 per unit for A and $5 per
unit for B. The daily production budget is $7500. If the manufacturer
makes a makes a profit of $1.75 per unit for A and $2.50 per unit for
B, 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.
units
cost
profit
A
1
3
1.75
B
1
5
2.50
limits
2000
7500
The System of Inequalities
Maximize P = 1.75x + 2.50y subject to the constraints

 x + y ≤ 2000
3x + 5y ≤ 7500

x ≥ 0, y ≥ 0
The Rewrite

I
 y ≤ −x + 2000
y ≤ − 35 x + 1500 II

x ≥ 0, y ≥ 0
The Graph and the Corners
2000
1500
1000
500
500
1000
1500
2000
2500
The Graph and the Corners
2000 • I
1500 •
II
1000
500
500
1000
1500
•
2000
•
2500
The Graph and the Corners
2000 • I
1500 •
II
1000
500
F. S.
500
1000
1500
•
2000
•
2500
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
(0, 0)
(0, 1500)
(1250, 750)
(2000, 0)
P = 1.75x + 2.50y
0
3750
4062.5
3500
The maximum occurs of $4062.50 occurs when we produce 1250 of
product A and 750 of product B.
Pottery Example
Example
A potter is making cups and plates. It takes her 6 minutes to make a
cup and 3 minutes to make a plate. Each cup uses 3/4 lb. of clay and
each plate uses one lb. of clay. She has 20 hours available for making
the cups and plates and has 250 lbs. of clay on hand. She makes a
profit of $2 on each cup and $1.50 on each plate. How many cups and
how many plates should she make in order to maximize her profit?
Pottery Example
Example
A potter is making cups and plates. It takes her 6 minutes to make a
cup and 3 minutes to make a plate. Each cup uses 3/4 lb. of clay and
each plate uses one lb. of clay. She has 20 hours available for making
the cups and plates and has 250 lbs. of clay on hand. She makes a
profit of $2 on each cup and $1.50 on each plate. How many cups and
how 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 one
to make it consistent. In order to avoid more fractions, we will use
minutes.
The Table
time
clay
profit
cups
6
3
4
2.00
plates
3
1
1.50
limits
1200
250
The System of Inequalities
Maximize P = 2x + 1.5y subject to the constraints

 6x + 3y ≤ 1200
3
x + y ≤ 250
 4
x ≥ 0, y ≥ 0
The Rewrite

 y ≤ −2x + 400 I
y ≤ − 34 x + 250 II

x ≥ 0, y ≥ 0
The Graph and the Corners
400 • I
300
•
II
200
100
100
•
200
•
300
400
The Graph and the Corners
400 • I
300
•
II
200
100
F. S.
100
•
200
•
300
400
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
(0, 0)
(0, 250)
(200, 0)
(120, 160)
P = 2x + 1.5y
0
375
400
480
The maximum profit of $480 occurring when 120 cups and 160 plates
are sold.
A Non-Table Example
Example
A calculator company produces a scientific calculator and a graphing
calculator. Long-term projections indicate an expected demand of at
least 100 scientific and 80 graphing calculators each day. Because of
limitations on production capacity, no more than 200 scientific and
170 graphing calculators can be made daily. To satisfy a shipping
contract, a total of at least 200 calculators much be shipped each day.
If each scientific calculator sold results in a $2 loss, but each graphing
calculator produces a $5 profit, how many of each type should be
made daily to maximize net profits?
Solution
An example like this doesn’t lend itself to a table to organize the
information.
Solution
An example like this doesn’t lend itself to a table to organize the
information.
We can begin with the assignment of variables as before, with x being
the number of scientific calculators and y being the number of
graphing calculators.
Solution
An example like this doesn’t lend itself to a table to organize the
information.
We can begin with the assignment of variables as before, with x being
the number of scientific calculators and y being the number of
graphing calculators.
Maximize P = −2x + 5y subject to the constraints

 100 ≤ x ≤ 200
80 ≤ y ≤ 170

x + y ≥ 200
The Rewrite

 100 ≤ x ≤ 200
80 ≤ x ≤ 170

y ≥ −x + 200
The Graph and the Corners
200
150
100
50
50
100
150
200
The Graph and the Corners
200
150
100
50
50
100
150
200
The Graph and the Corners
200
150
100
50
50
100
150
200
The Graph and the Corners
200 •
150
100
50
50
100
150
•
200
The Graph and the Corners
200 •
150
F. S.
100
50
50
100
150
•
200
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)
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 both
because each has at least one coordinate that comes from a horizontal
or vertical line.
Testing the Points
Point
(100, 170)
(200, 170)
(200, 80)
(100, 100)
(120, 80)
P = −2x + 5y
650
450
0
300
160
The maximum profit of $650 occurs when 100 scientific and 170
graphing calculators are sold.
Points on an Exam
Example
A student is taking an exam consisting of 10 essay questions and 50
short answer questions. He has 90 minutes to complete the exam and
knows he cannot possibly answer all questions. The essay questions
are worth 20 points and the short answer questions are worth 5 points.
An essay question takes 10 minutes to answer and a short answer
question takes 2 minutes. The student must do at least 3 essay
questions and at least 10 short answer questions. Find how many
correct of each type will maximize the exam score.
The Table
Let x be the number of essay questions and let y be the number of
short answer questions.
The Table
Let x be the number of essay questions and let y be the number of
short answer questions.
time
quantity
required
value
essay
10
10
3
20
short answer
2
50
10
5
available
90
The System of Inequalities
Maximize Score = 20x + 5y subject to the constraints

 10x + 2y ≤ 90
3 ≤ x ≤ 10

10 ≤ y ≤ 50
The Rewrite

 y ≤ −5x + 45
3 ≤ x ≤ 10

10 ≤ y ≤ 50
The Plot and the Corners
50
37.5
25
12.5
2.5
5
7.5
10
The Plot and the Corners
50
37.5
25
12.5
2.5
5
7.5
10
The Plot and the Corners
50
37.5
25
12.5
2.5
5
7.5
10
The Plot and the Corners
50
•
37.5
25
12.5
•
2.5
5
7.5
10
The Plot and the Corners
50
•
37.5
25
F. S.
12.5
•
2.5
5
7.5
10
The Points
Again, as in last example, at least one of the coordinates of each point
is either directly from a constraint represented by a horizontal or a
vertical line in the graph.
The Points
Again, as in last example, at least one of the coordinates of each point
is either directly from a constraint represented by a horizontal or a
vertical line in the graph.
1
(3, 30)
2
(7, 10)
3
(3, 10)
Testing the Points
Point
(3, 30)
(7, 10)
(3, 10)
S = 20x + 5y
210
190
110
The maximum score of 210 occurs when 3 essay and 30 short answer
questions are (correctly) answered.
Law Firm Example
Example
The Sue All Law Firm handles two types of lawsuits: medical
malpractice suits against unscrupulous heart surgeons for performing
unnecessary surgery, and suits against hard-working math professors
for failing students who do not deserve to pass. Math professor
lawsuits each require 6 person-months of preparation and the hiring of
5 expert witnesses, whereas medical lawsuits each require 10
person-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 it
cannot afford to hire more than 15 expert witnesses. It makes an
average profit of $1 million per math professor sued and $5 million
per heart surgeon sued. How many of each type of lawsuit should it
initiate in order to maximize its expected profits?
The Table
Let x be the number of lawsuits against doctors and let y be the
number of lawsuits against math professors.
The Table
Let x be the number of lawsuits against doctors and let y be the
number of lawsuits against math professors.
prep
witnesses
profit
medical
10
3
5
math
6
5
1
limits
30
15
The System of Inequalities
Maximize Profit = 5x + y subject to the constraints

 10x + 6y ≤ 30
3x + 5y ≤ 15

x ≥ 0, y ≥ 0
The Rewrite
Maximize Profit = 5x + y subject to the constraints

 y ≤ − 53 x + 5 I
y ≤ − 35 x + 3 II

x ≥ 0, y ≥ 0
The Graph and the Corners
9
6
3
3
6
9
The Graph and the Corners
9
6
3
3
6
9
The Graph and the Corners
9
6
I
3
3
6
9
The Graph and the Corners
9
6
I
3
3
II 6
9
The Graph and the Corners
9
6
I
3
F.S.
3
II 6
9
The Graph and the Corners
9
6
I
y-int II
3
I=II
F.S.
origin
3x-int I
II 6
9
The Points
The only one that requires algebra is the point where the lines
intersect.
3
5
− x+5=− x+3
3
5
−25x + 75 = −9x + 45
30 = 16x
15
=x
8
5 15
y=− ·
+5
3 8
25
=− +5
8
25 40
=− +
8
8
15
=
8
The Points
The only one that requires algebra is the point where the lines
intersect.
3
5
− x+5=− x+3
3
5
−25x + 75 = −9x + 45
30 = 16x
15
=x
8
1
2
3
4
(0, 0)
(3, 0)
15 15
8, 8
(0, 3)
5 15
y=− ·
+5
3 8
25
=− +5
8
25 40
=− +
8
8
15
=
8
Testing the Points
Point
(0, 0)
(3, 0) 15 15
8, 8
(0, 3)
P = 5x + y
0
15
45
4
3
The maximum profit of $15 million occurs when 3 medical lawsuits
and no professor lawsuits are handled.
Testing the Points
Point
(0, 0)
(3, 0) 15 15
8, 8
(0, 3)
P = 5x + y
0
15
45
4
3
The maximum profit of $15 million occurs when 3 medical lawsuits
and no professor lawsuits are handled.
How did we know the intersection of the two lines gives an
impossible point?