Math 1300: Section 5- 3 Linear Programing in Two Dimensions: Geometric Approach

Linear Programming Problem

Math 1300 Finite Mathematics
Section 5.3 Linear Programming In Two Dimensions:
Geometric Approach

Jason Aubrey

Department of Mathematics
University of Missouri

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Jason Aubrey Math 1300 Finite Mathematics


Linear programming is a mathematical process that has been
developed to help management in decision making and it has
become one of the most widely used and best-known tools of
management science.

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In general, a linear programming problem is one that is
concerned with ﬁnding the optimal value of a linear objective
function of the form
z = ax + by
(where a and b are not both 0) Where the decision variables x
and y are subject to the problem constraints.

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In general, a linear programming problem is one that is
concerned with ﬁnding the optimal value of a linear objective
function of the form
z = ax + by
(where a and b are not both 0) Where the decision variables x
and y are subject to the problem constraints.
The problem constraints are various linear inequalities and
equations. In all problems we consider, the decision variables
will also satisfy the nonnegative constraints, x ≥ 0, y ≥ 0.

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The set of points satisfying both the problem constraints and
the nonnegative constraints is called the feasible region or
feasible set for the problem.

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The set of points satisfying both the problem constraints and
the nonnegative constraints is called the feasible region or
feasible set for the problem.
Any point in the feasible region that produces the optimal value
of the objective function over the feasible region is called an
optimal solution.

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Example: Maximize P = 30x + 40y subject to

2x + y ≤ 10
x +y ≤7
x + 2y ≤ 12
x, y ≥ 0

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Theorem (Fundamental Theorem of Linear Programming)
If the optimal value of the objective function in a linear
programming problem exists, then that value must occur at one
(or more) of the corner points of the feasible region. (A corner
point is a point in the feasible set where one or more boundary
lines intersect.)

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Applying the theorem...

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1 Graph the feasible region, as discussed in Section 5-2. Be
sure to ﬁnd all corner points.

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2 Construct a corner point table listing the value of the
objective function at each corner point.

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3 Determine the optimal solution(s) from the table.

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4 For an applied problem, interpret the optimal solution(s) in
terms of the original problem.

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Back to our original problem:

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Back to our original problem:
Example: Maximize P = 30x + 40y subject to

2x + y ≤ 10
x +y ≤7
x + 2y ≤ 12
x, y ≥ 0

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6

5

4

3

2

1

0 1 2 3 4 5

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2x + y ≤ 10
6 a Boundary line:

5 (a) y = 10 − 2x
x y
4 0 10
5 0
3
Check: 2(0) + 0 ≤ 10 Yes!
2 ?

1

0 1 2 3 4 5

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x +y ≤7
6 b a Boundary line:

5 (b) y = 7 − x
x y
4 0 7
7 0
3
Check: 0 + 0 ≤ 7 Yes!
2 ?

1

0 1 2 3 4 5

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x + 2y ≤ 12
6 c b a Boundary line:

5 (c) y = 6 − 1 x
2
x y
4 0 6
3 12 0
Check: 0 + 2(0) ≤ 12 Yes!
2 ?

1

0 1 2 3 4 5

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Now we mark the feasible set and
6 c b a ﬁnd the coordinates of the corner
points.
5

4

3

2 FS
1

0 1 2 3 4 5

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Line c and the y -axis intersect at
6 c b a (0, 6)
(0, 6)
5

4

3

2 FS
1

0 1 2 3 4 5

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Line c (y = 7 − x) and Line b
c b a 1
6 (y = 6 − 2 x) intersect at (2, 5):
(0, 6)
5
(2, 5) 1
4 7−x =6− x
2
3
1
− x = −1
2
2 FS x =2
1 y =7−2=5

0 1 2 3 4 5

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Line a (y = 10 − 2x) and Line b
6 c b a (y = 7 − x) intersect at (3, 4):
(0, 6)
5
(2, 5)
10 − 2x = 7 − x
4
(3, 4) −x = −3
3
x =3
2 FS y =7−3=4
1

0 1 2 3 4 5

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Line a intersects the x-axis at
6 c b a (5, 0).
(0, 6)
5
(2, 5)
4
(3, 4)
3

2 FS
1
(5, 0)
0 1 2 3 4 5

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Don’t forget (0, 0)!
6 c b a
(0, 6)
5
(2, 5)
4
(3, 4)
3

2 FS
1
(0, 0) (5, 0)
0 1 2 3 4 5

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Now we construct our corner point ta-
6 c b a ble, and ﬁnd the maximum value of
(0, 6) the objective function P = 30x + 40y :
5 Corner Point P = 30x + 40y
(2, 5)
4 (0, 6) 240
(3, 4) (2, 5) 260
3 (3, 4) 250
(5, 0) 150
2 FS
(0, 0) 0
1
Therefore the maximum value of P
(0, 0) (5, 0)
0 1 2 3 4 5 occurs at the point (2, 5).

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Example: Minimize and maximize P = 20x + 10y subject to

2x + 3y ≥ 30
2x + y ≤ 26
−2x + 5y ≤ 34
x, y ≥ 0

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(8, 10)
10
(3, 8)
8

6

4
(12, 2)
2

0 2 4 6 8 10 12 14

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The feasible set has been drawn
(8, 10) for you. We now construct our
10
corner point table.
(3, 8)
8

6

4
(12, 2)
2

0 2 4 6 8 10 12 14

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10
corner point table.
(3, 8)
(3, 8) 140
6 (8, 10) 260
4
(12, 2) 260
(12, 2)
2

0 2 4 6 8 10 12 14

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10
corner point table.
(3, 8)
(3, 8) 140
6 (8, 10) 260
4
(12, 2) 260
Therefore the maximum value of
(12, 2)
2 P is at (8, 10) and (12, 2) and the
minimum value of P is at (3, 8).
0 2 4 6 8 10 12 14

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Example: A manufacturing plant makes two types of inﬂatable
boats, a two-person boat and a four-person boat.

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boats, a two-person boat and a four-person boat. Each
two-person boat requires 0.9 labor hours from the cutting
department and 0.8 labor-hours from the assembly department.

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Each four person boat requires 1.8 labor-hours from the cutting

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The maximum labor-hours available per month in the cutting
department and the assembly department are 864 and 672,
respectively.

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respectively. The company makes a proﬁt of $25 on each
2-person boat and $40 on each four-person boat.

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respectively. The company makes a profit of $25 on each
How many of each type should be manufactured each month to
maximize profit? What is the maximum profit?

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To solve such a problem involves:

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1 Constructing a mathematical model of the problem, and

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2 using the mathematical model to ﬁnd the solution.

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So far we have studied some of the techniques we will use for
ﬁnding the solution. Here, we introduce a 5 step procedure for
constructing a mathematical model of the problem.

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Step 1: Identify the decision variables.

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In this problem, we have

x = number of 2-person boats to manufacture
y = number of 4-person boats to manufacture

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Step 2: Summarize relevant information in table form, relating
the decision variables with the rows in the table, if possible.

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Recall: Each two-person boat requires 0.9 labor hours from the
cutting department and 0.8 labor-hours from the assembly de-
partment. Each four person boat requires 1.8 labor-hours from
the cutting department and 1.2 labor-hours from the assembly de-
partment. The maximum labor-hours available per month in the
cutting department and the assembly department are 864 and
672, respectively. The company makes a proﬁt of $25 on each

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Recall: Each two-person boat requires 0.9 labor hours from the
cutting department and 0.8 labor-hours from the assembly de-
partment. Each four person boat requires 1.8 labor-hours from
the cutting department and 1.2 labor-hours from the assembly de-
partment. The maximum labor-hours available per month in the
cutting department and the assembly department are 864 and
672, respectively. The company makes a proﬁt of $25 on each

2-person 4-person Available
Cutting 0.9 1.8 864
Assembly 0.8 1.2 672
Proﬁt $25 $40

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Cutting 0.9 1.8 864
Proﬁt $25 $40

Step 3: Determine the objective and write a linear objective
function.

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Cutting 0.9 1.8 864
Proﬁt $25 $40

function.

P = 25x + 40y

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Cutting 0.9 1.8 864
Proﬁt $25 $40

function.

P = 25x + 40y
Step 4: Write problem constraints using linear equations
and/or inequalities.

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Cutting 0.9 1.8 864
Proﬁt $25 $40

function.

P = 25x + 40y
Step 4: Write problem constraints using linear equations
and/or inequalities.

0.9x + 1.8y ≤ 864
0.8x + 1.2y ≤ 672 university-logo



P = 25x+40y
0.9x + 1.8y ≤ 864
0.8x + 1.2y ≤ 672

Step 5: Write nonnegative constraints.

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P = 25x+40y
0.9x + 1.8y ≤ 864
0.8x + 1.2y ≤ 672


x ≥0
y ≥0

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P = 25x+40y
0.9x + 1.8y ≤ 864
0.8x + 1.2y ≤ 672


x ≥0
y ≥0

We now have a mathematical model of the given problem. We
need to find the production schedule which results in maximum
profit for the company and to find that maximum profit.
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Now we follow the procedure for geometrically solving a linear
programming problem with two decision variables.
Applying the Fundamental Theorem of Linear Programming

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4 For an applied problem, interpret the optimal solution(s) in
terms of the original problem.

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500

400

300

200

100

100 200 300 400 500 600 700 800 900

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First we plot the boundary
line 0.9x + 1.8y = 864:
500 x y
0 480
400 960 0

300

200

100

100 200 300 400 500 600 700 800 900

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Test:
0.9(0) + 1.8(0) ≤ 0 Yes!
500
?
So we choose the lower half-
400
plane.
300

200

100

100 200 300 400 500 600 700 800 900

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Now plot boundary line
0.8x + 1.2y = 672:
500 x y
0 560
400 840 0

300

200

100

100 200 300 400 500 600 700 800 900

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Test:
0.8(0) + 1.2(0) ≤ 672 Yes!
500
?
So we choose the lower half-
400
plane.
300

200

100

100 200 300 400 500 600 700 800 900

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Next, we ﬁnd the intersection
point of the lines:
500
1 2
480 − x = 560 − x
400 2 3
1
x = 80
300 6
(480, 240) x = 480
200 y = 480 − (1/2)(480)
= 240
100

100 200 300 400 500 600 700 800 900

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500

400

300
(480, 240)
200 FS

100

100 200 300 400 500 600 700 800 900

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Now we solve the problem:

Corner Point P = 25x + 40y
(0, 0) 0
(0, 480) 19,200
(480, 240) 21,600
(860, 0) 21,500

We conclude that the company can make a maximum proﬁt of
$21,600 by producing 480 two-person boats and 240
four-person boats.

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Remember, there are two parts to solving an applied linear
programming problem: constructing the mathematical model
and using the geometric method to solve it.

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Example: A chicken farmer can buy a special food mix A at 20
cents per pound and a special food mix B at 40 cents per
pound. Each pound of mix A contains 3,000 units of nutrient N1
and 1,000 units of nutrient N2 ; each pound of mix B contains
4,000 units of nutrient N1 and 4,000 units of nutrient N2 . If the
minimum daily requirements for the chickens collectively are
36,000 units of nutrient N1 and 20,000 units of nutrient N2 , how
many pounds of each food mix should be used each day to
minimize daily food costs while meeting (or exceeding) the
minimum daily nutrient requirements? What is the minimum
daily cost? Construct a mathematical model and solve using
the geometric method.

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In this problem, we are asked,
how many pounds of each food mix should be used
each day to minimize daily food costs while meeting
(or exceeding) the minimum daily nutrient
requirements?

So,

x = number of pounds of mix A to use each day
y = number of pounds of mix B to use each day

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A chicken farmer can buy a special food mix A at 20 cents per
pound and a special food mix B at 40 cents per pound. Each
pound of mix A contains 3,000 units of nutrient N1 and 1,000 units
of nutrient N2 ; each pound of mix B contains 4,000 units of nutrient
N1 and 4,000 units of nutrient N2 . If the minimum daily require-
ments for the chickens collectively are 36,000 units of nutrient N1
and 20,000 units of nutrient N2 . . .

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A chicken farmer can buy a special food mix A at 20 cents per
pound and a special food mix B at 40 cents per pound. Each
pound of mix A contains 3,000 units of nutrient N1 and 1,000 units
of nutrient N2 ; each pound of mix B contains 4,000 units of nutrient
N1 and 4,000 units of nutrient N2 . If the minimum daily require-
ments for the chickens collectively are 36,000 units of nutrient N1
and 20,000 units of nutrient N2 . . .
mix A mix B Min daily req.
units of N1 3,000 4,000 36,000
units of N2 1,000 4,000 20,000
Cost per pound $0.20 $0.40

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units of N1 3,000 4,000 36,000
units of N2 1,000 4,000 20,000

function.

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units of N1 3,000 4,000 36,000
units of N2 1,000 4,000 20,000

function.
We want to minimize daily food costs so the objective
function is
C = 0.20x + 0.40y

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Step 4: Write problem constraints using linear equations and/or
inequalities.

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inequalities.

3, 000x + 4, 000y ≥ 36, 000
1, 000x + 4, 000y ≥ 20, 000

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inequalities.

3, 000x + 4, 000y ≥ 36, 000
1, 000x + 4, 000y ≥ 20, 000


x ≥0
y ≥0

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We now have a mathematical model of the given problem:

Minimize C = 0.20x + 0.40y
subject to:

3, 000x + 4, 000y ≥ 36, 000
1, 000x + 4, 000y ≥ 20, 000
x ≥0
y ≥0

We will now use the geometric method to determin how many
pounds of each food mix should be used each day to minimize
daily food costs while meeting (or exceeding) the minimum
daily nutrient requirements. We can then determine the
minimum daily cost as well.
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10
8
6
4
2

−2 0 2 4 6 8 10 12 14 16 18 20

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10
8
6
4
2

−2 0 2 4 6 8 10 12 14 16 18 20
Now plot boundary line 3, 000x + 4, 000y = 36000:
x y
0 9
12 0

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10
8
6
4
2

−2 0 2 4 6 8 10 12 14 16 18 20
Test:
3, 000(0) + 4, 000(0) ≥ 36, 000 No!
?
So we choose the upper half-plane.

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10
8
6
4
2

−2 0 2 4 6 8 10 12 14 16 18 20
Now plot boundary line 1, 000x + 4, 000y = 20, 000:
x y
0 5
20 0

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10
8
6
4
2

−2 0 2 4 6 8 10 12 14 16 18 20
Test:
1, 000(0) + 4, 000(0) ≥ 20, 000 No!
?
So we choose the upper half-plane.

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10 (0, 9)
8
6
4
2 (8, 3)
(20, 0)
−2 0 2 4 6 8 10 12 14 16 18 20
Next, we ﬁnd the intersection point of the lines:
3, 000x + 4, 000y = 36, 000
– 1, 000x + 4, 000y = 20, 000
2, 000x + 0y = 16, 000
x =8

And so, 1, 000(8) + 4, 000y = 20, 000; this implies that y =
3.
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10 (0, 9)
8
6
FS
4
2 (8, 3)
(20, 0)
−2 0 2 4 6 8 10 12 14 16 18 20
So, we have the feasible set shown above.

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10 (0, 9)
8
6
FS
4
2 (8, 3)
(20, 0)
−2 0 2 4 6 8 10 12 14 16 18 20

And now we plug the corner points into the objective function
to ﬁnd the minimum cost:
Point C = 0.20x + 0.40y
(0, 9) $3.60
(8, 3) $2.80
(20, 0) $4.00

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Point C = 0.20x + 0.40y
(0, 9) $3.60
(8, 3) $2.80
(20, 0) $4.00

We therefore conclude that the chicken farmer can feed the
chickens at a minimal cost of $2.80 per day using 8 pounds of
mix A and 3 pounds of mix B.

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Math 1300: Section 5- 3 Linear Programing in Two Dimensions: Geometric Approach

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Math 1300: Section 5- 3 Linear Programing in Two Dimensions: Geometric Approach