1
Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
TASK 2.4.1: COTTON VS. CORN
Solutions
Suppose a farmer has 150 acres available for planting corn and cotton. The cotton seeds cost $3
per acre and the corn seeds cost $5 per acre. The total labor costs for cotton will be $15 per acre
and the total labor costs for corn will be $8 per acre. The farmer expects the income from cotton
to be $80 per acre and the income from corn to be $110 per acre. The farmer can spend no more
than $540 on seeds and $1800 on labor. How many acres of corn and how many acres of cotton
should the farmer plant in order to maximize his income?
1. Create a table to organize the information above.
Cotton
Corn
Maximum allowed
Cost of seeds/acre
Cost of labor/acre
$3
$5
$540
$15
$8
$1800
Expected
income/acre
$80
$110
2. What are the variables in this situation? Assign names to these variables.
The number of acres of cotton seeds planted and the number of acres of corn
seeds planted. We will use x and y to name these variables so that we can graph
them easily.
Let x = number of acres of cotton seeds planted.
Let y = number of acres of corn seeds planted.
3. The farmer wants to maximize his income, I. Write an expression for I in terms of
the variables you define in part 2. In linear programming, we call the function to be
maximized or minimized the objective function. In this case, I is the objective
function.
I = 80x + 110y
4. Write a system of linear inequalities in x and y that describe the constraints on this
situation. Connect each inequality with the problem situation.
# 3x + 5y ! 540 cost of seeds cannot exceed $540
%
%%15x + 8y ! 1800 number of acres cannot exceed 1800
x " 0 positive number of acres of cotton
$
%
y " 0 positive number of acres of corn
%
x + y ! 150 total acres cannot exceed 150
%&
5. Rewrite each of the inequalities above in slope-intercept form (if possible) and graph
the system. The feasible region represents all points that satisfy the system. Shade
the feasible region on your graph and sketch below. Pick a point in your feasible
region and show that it satisfies all of the constraints.
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.
2
Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
y ≤ (-3/5)x +108, y ≤ (-15/8)x + 225, y ! 150 " x, x # 0, y # 0
260
240
220
200
180
160
140
120
100
80
60
40
20
-50
50
100
150
200
250
300
350
-20
-40
Pick some random points in the feasible region and have participants show how they
meet the constraints of the problem. For each point you pick, ask participants to
calculate the income earned.
6. What are the vertices of your feasible region?
There are four vertices. One is the origin, one is the intersection of the two slanted
lines and two are the intercepts of these lines.
(0, 0), (0, 108), (120, 0), (91.76, 52.94)
• Do these points satisfy the constraints?
Yes, because the inequalities are not strict inequalities, but include the
boundaries.
7. Pick some points in the feasible region and fill in the table below with the income for
the various points. Then, pick some points outside the feasible region and fill in the
table provided. When comparing the two tables, what do you notice? Explain.
Be sure that participants include all of the vertices of the region. They may choose
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.
3
Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
other points at random. We will discuss why the vertices are sufficient for
maximizing the objective function in question 9.
Points
I = 80x + 110y
(0, 0)
I=0
(0, 108)
I = 11880
(120, 0)
I = 9600
(91.76, 52.94)
I = 13164.2
Participants should try points outside the region (for example, (150, 100) gives an
income of $33,000). Picking values of x and y outside the feasible region may give a
greater income, but the constraints are not met.
8. Refer to your first table (points in the feasible region) in question 7; use the values
you obtained for income to find all points in the feasible region that could provide the
specified income. Graph these equations on a graphing calculator. What do you notice
about them? Explain.
All of the lines have slope
!8
11
. That is, they are all parallel. For example, the income
found from planting 0 acres of cotton and 108 acres of corn, $9600, produces the
9600
8
function y =
–
x . The income from planting 91.76 acres of cotton and 52.94
110
11
13164.2
8
acres of corn, $13164.20, produces the function y =
–
x . Since we are
110
11
maximizing our income, the optimal solution is the line that gives the highest yintercept while still intersecting the feasible region. This is why we only really need
to check the vertices of our region.
If dynamic geometry software is available, show the Linear Programming model
sketch. In it, participants see a feasible region with a possible income function. Drag
the income line so that it is inside the feasible region and maximizes the income.
What point of the feasible region lies on the line?
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.
4
Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
160
( )
( )
f (x ) = -
g( x ) = 140
3
!x + 1 0 8
5
15
8
LINEAR PROGRAMMING MODEL
Drag the Income function to maximize the
income. Make sure that at least one point
on the Income function is within the
feasible region.
!x + 2 2 5
Income function: y = -0.72x+89.82
120
y=f(x)
100
Income = 9880.43
C
80
60
40
Income function
20
y=g(x)
20
40
60
80
100
120
9. How many acres of cotton seed and how many acres of corn seed should the farmer
plant in order to maximize his income? Will the farmer plant all 150 acres? Why or
why not?
He should plant 91.7 acres of cotton and 52.9 acres of corn.
•
•
Why do we round down? If we round as we would mathematically, the
solution would be 92 acres of cotton and 53 acres of corn. This violates one
of our constraints since 3(92)+5(53)=541.
The farmer cannot plant all 150 acres because this is outside the feasible
region.
After participants finish this activity, split them into groups to work on the Student
Activity: Applications of Linear Programming.
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.
5
Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
Reflect and apply
A trucker hauls citrus from Florida to New York City. Below is a chart of information giving the
constraints and profit for the amount of fruit to be hauled. There is a company regulation that
prohibits the driver from carrying more crates of grapefruit than crates of oranges.
Crates of oranges (x)
Crates of
grapefruit (y)
Maximum capacity of
truck
Volume in cubic ft
4
6
Weight in lb
80
100
300
5600
Profit/crate
$2.50
$4.00
The trucker wants to maximize his profit. Below is his solution.
$
x "0
&
y "0
&&
y#x
%
& 4 x + 6y # 300
&
&'80x + 100y # 5600
50
40
P=.250x+4.00y
30
(30, 30)
!
(45, 20)
20
10
(0, 0)
20
40
60
(70, 0)
The maximum profit is $195 with 30 crates of oranges and 30 crates of grapefruit.
A disease infects a large number of orange trees, significantly decreasing the
amount of oranges available. So, the price of oranges begins to rise. The
company takes off the regulation that a trucker cannot carry more grapefruit than
oranges and the profit on oranges increases to $2.80 per crate. How does this
affect the solution to the original problem? What does the profit on oranges need
to be in order for a trucker to make a maximum profit by hauling only oranges?
If the regulation is removed, the trucker should carry a full load of grapefruit to
make the maximum profit. He can carry 50 crates and still meet his weight and
volume restrictions. When the profit on oranges reaches $2.86 per crate, it
becomes more beneficial for the trucker to carry a full load of oranges. Even
though the profit is still less per crate, the trucker can carry more crates of
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.
6
Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
oranges and still meet his weight and volume restrictions. He can carry 70 crates
of oranges. Notice that $2.86 is the solution to the equation 70x=(50)(4.00).
Math notes
In this task, participants reinforce the TEKS statements by looking at a graphical way of
solving systems of equations or inequalities to solve a problem situation. Linear
programming is a great example of mathematical modeling that is used in industry.
Looking at constraints on solution sets is an important skill that will be useful in
precalculus as students use functions to model situations.
Teaching notes
This task could be taken to the classroom as a class activity led by the teacher to
introduce linear programming. Students would need a more detailed discussion of the
system of linear inequalities in question 4. Sketching the graph of the feasible region
would also take some guidance. If you wish to use the Inequal Application on the TI-83
plus, introduce this during question 5. Or, after participants have sketched these by hand,
the facilitator can demonstrate the use of inequal to achieve the results. Students should
explore more points in the feasible region to see if they satisfy each of the constraints.
This can be done in question 7.
Technology notes
The application for the TI-83 plus can be used in place of the Geometer’s Sketchpad.
Inequal is preloaded on the Silver Edition calculators and can be downloaded free from
the Texas Instruments Educational web site.
Enter the inequalities as equations.
Place the cursor over the equal sign. Press ALPHA (green key) ZOOM to change the
equal sign to a less than or equal to sign.
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.
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Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
Press ALPHA GRAPH to change the equal sign to a greater than or equal to sign.
Arrow up to X= in the top left corner of the screen. Press ENTER.
Enter x greater than or equal to zero.
Now we need to set the window. Press WINDOW and set according to the screen shots.
It is important to change the shade resolution to 6 since we are graphing four inequalities.
Smaller resolutions make the screen too dark.
Now press GRAPH. The inequalities will be graphed and shaded.
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.
8
Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
For a clearer view of the feasible region, press ALPHA, F1(Y=). Choose 1: Ineq
Intersection.
ALPHA ZOOM allows you to trace the points of intersection. You will need to use both
the vertical and horizontal arrows to navigate to the points desired. Each time you hit a
relevant point of intersection, press STO→, ENTER. The x and y values will be stored in
lists named LINEQX and LINEQY.
To see the lists, press STAT, ENTER
We can evaluate the objective function by using the empty list. Highlight the list and
name it OBJ.
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.
9
Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
With the cursor over the list name OBJ, type in the objective function. Use quotes when
typing in the formula for OBJ—using the quotes indicates that you would like the
graphing calculator to “remember” the formula—otherwise the graphing calculator uses
the formula to calculate values and only stores the values (i.e. it “forgets” the original
function used to obtain the values).
For x, use INEQX (2nd LIST, arrow down to INEQX, ENTER). For y, use INEQY.
We can now see which values of x and y maximize the objective function.
When you are finished with the INEQUAL application, you must turn it off. Turning the
calculator off does not disengage the application.
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.
10
Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
TASK 2.4.1: COTTON VS. CORN
Suppose a farmer has 150 acres available for planting corn and cotton. The cotton seeds cost $3
per acre and the corn seeds cost $5 per acre. The total labor costs for cotton will be $15 per acre
and the total labor costs for corn will be $8 per acre. The farmer expects the income from cotton
to be $80 per acre and the income from corn to be $110 per acre. The farmer can spend no more
than $540 on seeds and $1800 on labor. How many acres of corn and how many acres of cotton
should the farmer plant in order to maximize his income?
1. Create a table to organize the information above.
2. What are the variables in this situation? Assign names to these variables.
3. The farmer wants to maximize his income, I. Write an expression for I in terms of the
variables you define in part 2. In linear programming, we call the function to be maximized
or minimized the objective function. In this case, I is the objective function.
4. Write a system of linear inequalities in x and y that describe the constraints on this situation.
Connect each inequality with the problem situation.
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.
11
Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
5. Rewrite each of the inequalities above in slope-intercept form (if possible) and graph the
system. The feasible region represents all points that satisfy the system. Shade the feasible
region on your graph and sketch below. Pick a point in your feasible region and show that it
satisfies all of the constraints.
6. What are the vertices of your feasible region?
7. Pick some points in the feasible region and fill in the table below with the income for the
various points. Then, pick some points outside the feasible region and fill in the table
provided. When comparing the two tables, what do you notice? Explain.
Points in feasible
region
I = __________________
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.
12
Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
Points outside
feasible region
I = __________________
8. Refer to your first table (points in the feasible region) in question 7; use the values you
obtained for income to find all points in the feasible region that could provide the specified
income. Graph these equations on a graphing calculator. What do you notice about them?
Explain.
9. How many acres of cotton seed and how many acres of corn seed should the farmer plant in
order to maximize his income? Will the farmer plant all 150 acres? Why or why not?
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.
13
Algebra II: Strand 2. Linear Functions; Topic 4. Applications of Linear Programming; Task 2.4.1
Reflect and apply
A trucker hauls citrus from Florida to New York City. Below is a chart of information giving the
constraints and profit for the amount of fruit to be hauled. There is a company regulation that
prohibits the driver from carrying more crates of grapefruit than crates of oranges.
Crates of oranges
(x)
Crates of
grapefruit (y)
Maximum capacity
of truck
Volume in cubic ft
4
Weight in lb
80
Profit/crate
$2.50
6
100
$4.00
300
5600
The trucker wants to maximize his profit. Below is his solution.
$
x "0
&
y "0
&&
50
y#x
%
& 4 x + 6y # 300
&
P=.250x+4.00y
&'80x + 100y # 5600
40
30
(30, 30)
!
(45, 20)
20
10
(0, 0)
20
40
60
(70, 0)
The maximum profit is $195 with 30 crates of oranges and 30 crates of grapefruit.
A disease infects a large number of orange trees, significantly decreasing the amount of oranges
available. So, the price of oranges begins to rise. The company takes off the regulation that a
trucker cannot carry more grapefruit than oranges and the profit on oranges increases to $2.80
per crate. How does this affect the solution to the original problem? What does the profit on
oranges need to be in order for a trucker to make a maximum profit by hauling only oranges?
December 20, 2004. Ensuring Teacher Quality: Algebra II, produced by the Charles A. Dana Center at The University of Texas
at Austin for the Texas Higher Education Coordinating Board.