Transcript Chapter 12

Chapter 12
Section 12.1
The Geometry of Linear
Programming
Linear Programming
A linear programming problem solves what the maximum or minimum value for an
linear objective function is inside or on the border of a feasible solution (sometimes
called constraints). One of the results of theoretical mathematics tells us that the
maximum and minimum value of a linear objective function must occur at one of the
corner points.
x=0
Maximize T = 6x - 2y
Subject to:
Corner
Point
 x0
 y0


 x4
3x  2 y  18
T = 6x – 2y
(0,0)
6(0)-2(0)=0-0=0
(0,9)
6(0)-2(9)=0-18=-18 MIN
(4,0)
6(4)-2(0)=24-0=24 MAX
(4,3)
6(4)-2(3)=24-6=18
3 x  2 y  18
2 y  3 x  18
2y
2

y
3 x
2
3 x
2
 182
9
x  4

3 x  2 y  18
3( 4)  2 y  18
12  2 y  18
2y  6
0,0
y3
-1
x=4
12
11
10
9
0,9
8
y  32 x  9
7
6
5
4
4,3
3
2
1
1
-1
2
4,0
3
4
5
6
y=0
Using Resources Efficiently
In order to make the best use of resources that a company or individual has a linear
programming approach is sometimes used. Consider the following example.
A person who enjoys woodworking makes patio tables and chairs in their spare
time. The can only do woodworking for 10 hours a week. They can only afford $100
in wood supplies a week. They spend 2 hours making a table and 5 hours making a
chair. A table uses $40 in supplies while chairs only use $10. A table sells for a profit
of $30 and a chair for a profit $20. Write this as a linear programming problem.
1. Let x = number of tables made and y = number of chairs made each week.
2. Then x > 0 and y > 0
3. The number of hours work on tables and chairs is 2x+5y that means 2x+5y<10
4. The amount of money he spends is 40x+10y that means 40x+10y<100
5. The amount of money he gets is 30x+20y that means T = 30x+20y
Maximize T = 30x+20y subject to:
x  0
y  0


2 x  5 y  10
40x  10 y  100
In order to solve the linear programming problem
to the right we do the following 3 steps:
Maximize T = 30x+20y
subject to:
1. Graph the region given by the constraints and
locate the corner points.
x  0
y  0


2 x  5 y  10
40x  10 y  100
2. Plug all corner points into the objective
function.
3. Select which corner point gives the maximum
or minimum value depending on what is wanted.
12
11
2 x  5 y  10
10
5 y  2 x  10
40x  10 y  100
10 y  40x  100
y
y  4 x  10
9
8
7
5
 4 x  10 y  20
4
0,2 3
40x  10 y  100
36x  80
80 20
x

36 9
 209 , 109 
2
0,01
1
-1
x2
Find the point of intersection
of the two lines.
6
-1
2
5
 52 ,0
2
3
4
5
6
2  209  5 y  10
40
9
 5y 
5y 
50
9
y  109
Maximum
90
9
Locate all other corner
points and fill in values
for objective function:
Corner
Point
0,0
0,2
 52 ,0
 209 , 109 
T = 30x + 20y
T  30  0  20  0  0
T  30  0  20  2  40
T  30 52  20 0  75
T  30 209  20 109  800
9  77.8
Example
Roger runs his own side business of detailing cars and vans during the evenings.
He only has enough time to detail 10 vehicles each week. Each car he details uses
11 dollars worth of cleaning supplies and each van uses 16 dollars of cleaning
supplies. He only can spend at most 125 dollars each week on cleaning supplies.
Each car he details earns a profit of 28 dollars and each van a profit of 32 dollars.
Formulate the above situation as a Linear Programming problem.
Let:
x = Number of cars detailed in 1 week
y = Number of vans detailed in 1 week
x  y  10
11x  16y  125
x  0 and y  0
T  28x  32 y
He only has enough time to
detail 10 vehicles each week.
He only can spend at most
125 dollars each week on
cleaning supplies.
Negative numbers do not
make sense here.
A profit of 15 dollars and each
van a profit of 18 dollars.
This is written in the
following way as a linear
programming problem:
Maximize: T=28x+32y
subject to:
x  0
y  0


 x  y  10
11x  16 y  125
Maximize the value T for
T=28x+32y
subject to the constraints:
To solve the linear the linear programming
problem to the right (from previous slide).
1. Graph the region described by the inequalities
(i.e. this is the feasible region).
2. Find all the corner points (i.e. where the
boundary lines intersect).
3. Plug each corner point into the objective
function to see what is maximum.
x0
11x  16 y  125
x  y  10
12
16 y  11x  125
y   x  10
11
y
11
16
x  125
16
10
Finding Corner Points
9
8
0, 
125
16 7
x  y  10
11x  16 y  125
x
0
11x  16 y  125
16 y  125
16x  16 y  160
 11x  16 y  125
y  125
16
6
5
7,3
4
3
10,0
1
-1
0,0-1
 35
5x
2
1
2
3
4
5
6
7
8
9
10
11
x  y  10
12
y0
x7
y3
11x  16y  125
x  0
y  0


 x  y  10
11x  16 y  125
Maximum
Corner
Points
T=28x+32y
0,0
28  0  32  0  0
10,0
28 10  32  0  280
0, 125

16
28 0  32 125
16  250
7,3
28  7  32  3  292