2. Essential Understanding and
Objectives
● Essential Understanding: Some real-world problems involve multiple linear
relationships. Linear programming accounts for all of these linear
relationships and gives the solution to the problem.
● Objectives:
● Students will be able to:
● Solving problems using linear programming
● Define constraint, linear programming, feasible region, and objective function
3. Iowa Core Curriculum
• Algebra
• A.CED.3 Represent constraints by equations or inequalities,
and by systems of equations and/or inequalities, and interpret
solutions as viable or nonviable options in a modeling context.
For example, represent inequalities describing nutritional and
cost constraints on combinations of different foods.
4. Linear Programming
Businesses use linear
programming to find out how to
maximize profit or minimize
costs. Most have constraints on
what they can use or buy.
5. Linear Programming
Linear programming is a process of finding
a maximum or minimum of a function by
using coordinates of the polygon formed
by the graph of the constraints.
6. What is a constraint?
A restriction...
A boundary…
A limitation…
7. What is the feasible region?
The feasible region is the area of the
graph in which all the constraints are met.
8. Objective Function
• The quantity you are trying to maximize or minimize is
modeled by this.
• Usually this quantity is the cost or profit
• Looks something like this C = ax + by, where a and b are real
numbers
9. Vertex Principle
• If there is a maximum or minimum value of the linear objective
function, it occurs at one or more vertices of the feasible
region.
Online Example
10. Find the minimum and maximum
value of the function f(x, y) = 3x - 2y.
We are given the constraints:
•y ≥ 2
• 1 ≤ x ≤5
•y ≤ x + 3
11. Linear Programming
• Find the minimum and maximum
values by graphing the inequalities
and finding the vertices of the polygon
formed.
• Substitute the vertices into the function
and find the largest and smallest
values.
13. Linear Programming
•The vertices of the quadrilateral
formed are:
(1, 2) (1, 4) (5, 2) (5, 8)
•Plug these points into the
objective function
f(x, y) = 3x - 2y
20. Example 1
A farmer has 25 days to plant cotton and soybeans.
The cotton can be planted at a rate of 9 acres per day,
and the soybeans can be planted at a rate of 12 acres
a day. The farmer has 275 acres available. If the
profit for soybeans is $18 per acre and the profit for
cotton is $25 per acre, how many acres of each crop
should be planted to maximize profits?
21. Step 1: Define the variables
What are the unknown values?
Let c = number of acres of cotton to
plant
Let s = number of acres of soybeans to
plant
22. Step 2: Write a System of
Inequalities
Write the constraints. What are the
limitations given in the problem?
The number of acres planted in cotton must
c 0 be greater than or equal to 0.
s 0 The number of acres planted in soybeans must
be greater than or equal to 0.
c s 275 The total number of acres planted must be less
than or equal to 275.
c s
25 The time available for planting must be less
9 12 than or equal to 25 days.
23. Step 3: Graph the
Inequalities
s
c
The purple area is the feasible region.
24. Step 4: Name the Vertices
of the Feasible
Region of the vertices of the
Find the coordinates
feasible region, the area inside the polygon.
(0,275)
(225,0)
(0,0)
(75,200)
25. Step 5: Write
an Equation to
be Maximized or
Minimized
p(c,s) = 25c + 18s
Maximum profit = $25 times the number
of acres of cotton planted + $18 times the
number of acres of soybeans planted.
26. Step 6: Substitute the
Coordinates into the
Equation the coordinates of the vertices into the
Substitute
maximum profit equation.
(c,s) 25c + 18s f(c,s)
(0,275) 25(0) + 18(275) 4950
(225,0) 25(225) + 18(0) 5625
(0,0) 25(0) + 18(0) 0
(75,200) 25(75) + 18(200) 5475
27. Step 7: Find the Maximum
(c,s) 25c + 18s f(c,s)
(0,275) 25(0) + 18(275) 4950
(225,0) 25(225) + 18(0) 5625
(0,0) 25(0) + 18(0) 0
(75,200) 25(75) + 18(200) 5475
225 acres of cotton and 0 acres of soybeans
should be planted for a maximum profit of
$5,625.
28. Example 2:
The Bethlehem Steel Mill can convert steel into
girders and rods. The mill can produce at most
100 units of steel a day. At least 20 girders and
at least 60 rods are required daily by regular
customers. If the profit on a girder is $8 and the
profit on a rod is $6, how many units of each
type of steel should the mill produce each day to
maximize the profits?
29. Step 1: Define the Variables
Let x = number of girders
Let y = number of rods
Step 2: Write a System of Inequalities
x 20 At least 20 girders are required daily.
y 60 At least 60 rods are required daily.
The mill can produce at most 100 units
x y 100 of steel a day.
30. Step 3: Graph the Inequalities
x 20, y 60, and x y 100
100
80
60
40
20
The purple region represents the feasible region.
-50 50 100
31. Step 4: Name the Vertices of the
Feasible Region
Find the coordinates of the vertices of the feasible region, the
area inside the polygon.
(20, 60)
(20, 80)
(40, 60)
32. Step 5: Write an
Equation to be
Maximized or Minimized
p(x,y) = 8x + 6y
Maximum profit = $8 times the number of
girders produced + $6 times the number of
rods produced
33. Step 6: Substitute the Coordinates
into the Equation
Substitute the coordinates of the vertices
into the maximum profit equation.
(x,y) 8x + 6y p(x,y)
(20, 60) 8(20) + 6(60) 520
(20, 80) 8(20) + 6(80) 640
(40, 60) 8(40) + 6(60) 680
34. Step 7: Find the Maximum
(x,y) 8x + 6y p(x,y)
(20, 60) 8(20) + 6(60) 520
(20, 80) 8(20) + 6(80) 640
(40, 60) 8(40) + 6(60) 680
40 girders and 60 rods of steel should be
produced for a maximum profit of $680.
