Jim Jack (J²)
MATH 1324 – Math for Business & Economics
Admin
Class
Grading
Attendance
1.2 Linear Function and Straight Lines
Intercepts
x-intercept
y-intercept
Linear and Constant functions
A function f is a linear function if f  x   mx  b ,
m  0 , where m and b are real numbers. The
domain and range are both the set of all reals.
If m  0 , then f is the constant function f  x   b ,
in which the domain is the set of all reals, and the
range is the constant b.
A linear equation in two variables is an
equation that can be written in the standard form
Ax  By  C , A,B,C (A,B, not both =0) are
constants, x, y variables.
Graph 3x  4 y  12
Find the x- and y- intercepts
Use graphing calculator
Graph x  4
Graph y  6
Find equations for vert, horiz lines thru 7,5
Graphs of Ax  By  C
If A  0
If B  0
In the Cartesian plane, the graph of any equation
of the form Ax  By  C , where A, B and C are
real constants (A and B not both 0), is a straight
line. Every straight line in the Cartesian plane is
a graph of this type. Vertical and horizontal lines
are a special case of this equation.
x  4
y6
2 x  3 y  12
Slope of a Line
If a line passes through two distinct points
P1  x1 , y1  and P2  x2 , y2 , then its slope is given by
y rise y2  y1
m


 x1  x2 
x run x2  x1
Find the slope of a line from  3,2  to 3,4.
Find the slope of a line from  1,3 to 2,3.
Find the slope of a line from  2,3 to 3,3.
Find the slope of a line from  2,4  to  2,2 .
Slope-intercept form
The equation y  mx  b , m is the slope, 0, b  is
the y-intercept, is called the slope-intercept
form of an equation of a line.
2
Find the slope and y-intercept of y   x  3
3
Write the equation of a line with a slope
y-intercept 0,2.
2
3
and
Point-Slope form
An equation of a line with slope m that passes
through the point  x1 , y1  is y  y1  m x  x1 ,
which is called the point-slope form of an
equation of a line.
Find the equation of a line that has a slope ½
and passes through the point  4,3.
Find the equation of a line that passes through
the two points  3,2  and  4,5.
Equations of a line (summary)
Standard form
Ax  By  C
Slope-intercept form
y  mx  b
Point-slope form
y  y1  m x  x1 
y  m x  x1   y1
Horizontal line
y b
Vertical line
xa
The management of a company that manufactures
skateboards has fixed costs of $300 per day and
total costs of $4300 per day at an output of 100
pairs of skateboards per day. Assume cost C is
linearly related to the output x.
Find the slope of a line joining the points
associated with outputs of 0 and 100.
Find an equation of the line relating output to cost.
Graph the cost equation from 0  x  200
Supply and Demand
At a price of $9/box of oranges, the supply is 320k
boxes while the demand is 200k. At $8.50/box,
supply is 270k, while demand is 300k. (x,p)
Find price-supply equation p  mx  b .
Find price-demand equation p  mx  b .
Find equilibrium price and quantity.
4. Systems of Equations and Inequalities
4.1. Review: Systems of Linear Equations in
Two Variables
DFN: Given the linear system
ax  by  h
cx  dy  k
a,b,c,d,h,k real constants,  x0 , y0  is a solution
of this system if each equation is satisfied by the
pair. The set of all such ordered pairs is called
the solution set.
Graphing
If 2 adult tickets and 1 child ticket cost $32,
while 1 adult ticket and 3 child tickets cost $36,
what is the price of each?
2 x  y  32
x  3 y  36
x  2y  2
x y 5
x  2y  4
2x  4 y  8
2x  4 y  8
x  2y  4
DFN: A system is consistent if at least one
solution exists, otherwise, it is inconsistent. A
consistent system is independent if there is a
unique solution (exactly one solution),
otherwise it is dependent. Two systems are
equivalent if they have the same solution set.
Thm 1: The linear system
ax  by  h
cx  dy  k
must have either
A: exactly one solution
B: no solution
C: infinitely many solutions.
There are no other possibilities.
Solve a system with a graphing calculator
5 x  2 y  15
2 x  3 y  16
Method of substitution
1. Solve one equation for one variable
2. Sub the result into the other equation, then
solve for the other variable.
3. Solve equation from step 1 using step 2.
5x  y  4
2x  3y  5
Elimination by addition
Thm 2: A system of linear equations is
transformed into an equivalent system if
A: Two equations are interchanged
B: An equation is multiplied by a non-zero
constant (both sides).
C: A constant multiple of one equation is added
to another equation.
3x  2 y  8
2 x  5 y  1
2 x  6 y  3
x  3y  2
x 1 y  4
2
 2x 
y  8
A woman wants to use milk and orange juice to
increase the amount of calcium and vitamin A in
her daily diet. An ounce of milk contains 37 mg
of calcium and 57  g of vitamin A. An ounce of
orange juice contains 5mg of calcium and 65  g
of vitamin A. How many ounces of milk and OJ
should she drink each day to provide exactly 500
mg calcium and 1200  g vitamin A?
The quantity of a product that people are willing
to buy during some period of time depends on
its price. Generally, the higher the price, the
less demand. Similarly, the quantity of a
product that a supplier is willing to sell during
some period of time increases with the price.
At a price of $1.88/lb, supply is 16,000 lb, but
demand is only 10,600 lb. When the price drops
to $1.46/lb, supply decreases to 10,000 lb, but
demand increases to 12,700 lb.
Find price-supply equation.
Find price-demand equation
Supply/Demand @ $2.09/lb, $1.32/lb, equilibrium
4.2 Systems of Linear Equations and Augmented
Matrices
A matrix is a rectangular array of numbers.
 4 5 12 
0 1 8


  3 10 9 
 6 0  1


1  4 5 
7 0  2 


a11  1
a12  4
a21  7 a22  0
a13  5
a23  2
The dimension of a matrix is m n if it has m
rows and n columns. A square matrix is n n.
a1 x  b1 y  c1 z  d1
a2 x  b2 y  c2 z  d 2
a3 x  b3 y  c3 z  d 3
 a1 b1 c1 d1 
a b c d 
 2 2 2 2
 a3 b3 c3 d 3 
square matrix m  n
row matrix m  1
column matrix n  1
principal diagonal i  j
coefficient, constant, augmented matrix
2x  3y  5
x  2 y  3
Thm 1 (like thm2 from last unit):
An augmented matrix is transformed into a rowequivalent matrix by performing any of the
following row operations.
Ri  R j
A: two rows are interchanged
B: A row is multiplied by a non-zero constant
kRi  Ri
C: a constant multiple of one row is added to
another row. kR j  Ri  Ri
Gauss-Jordan elimination
1. Do row ops to get a one at the upper left.
2. Do row ops to get a zero below that.
3. Do row ops to get a one at bottom of second
column.
4. Do row ops to get a zero above it.
3x1  4 x2  1
x1  2 x2  7
2 x1  3 x2  6
3 x1  4 x2  1
2
2 x1  x2  4
 6 x1  3x2  12
2 x1  6 x2  3
x1  3x2  2
4.3 Gauss-Jordan Elimination
3 possibilities:
a11 a12 k1

a21 a22 k 2
Reduced Row-Echelon Form
1 0 0  1
0 1 0 1 


0 0 1  2
1 0 0
0 1 4 


1 0 4 5
0 1 3 0 


0 0 0 1
1 4 0 0  3
0 0 1 0 2 


0 0 0 1 6 
1 0 3 
0 1  1


0 0 0 
A matrix is said to be in reduced row echelon
form if
1. Each row consisting entirely of zeros is
below any row having at least one nonzero
element.
2. The left-most non-zero element in any row
is a one.
3. All other elements in the column containing
the leftmost 1 of a given row are zeros.
4. The leftmost 1 in any row is to the right of
the leftmost 1 in the row above.
Main diagonal
1s, then maybe zeros
First non-zero element is a 1 (leading 1)
Leftmost leading 1 listed first
Rows with only zeros at the bottom
All elements below main diagonal are zeros
1 0 0 1
0 2 0 3


0 0 1 5
0 1  2 
1 0 3 


1 2  2 3 
0 0 1  1


1 0  3
0 0 0 


0 1  2
2x  2 y  z  3
3x  y  z  7
x  3y  2z  0
Gauss-Jordan Elimination
1. Choose the leftmost nonzero column, and use
row operations to get a 1 at the top.
2. Use multiples of the row containing the 1 at the
top to get zeros in all the remaining places in the
column containing the 1.
3. Repeat step 1 with the submatrix formed by
mentally deleting the row used in step 2 and all
rows above this row.
4. Repeat step 2 with the entire matrix including
all the rows deleted mentally. Continue until
matrix is in reduced row echelon form.
2 x  4 y  z  4
4x  8 y  7z  2
 2 x  4 y  3z  5
3 x  6 y  9 z  15
2 x  4 y  6 z  10
 2 x  3 y  4 z  6
x1  2 x2  4 x3  x4  x5  1
2 x1  4 x2  8 x3  3x4  4 x5  2
x1  3x2  7 x3 
3 x5  2
A company that rents small moving trucks wants to
purchase 25 trucks with a combined capacity of
28000 cubic feet. Three different types of trucks
are available: a 10 foot truck with a capacity of 350
cubic feet, a 14 foot truck with a capacity of 700
cubic feet, and a 24 foot truck with a capacity of
1400 cubic feet. How many trucks of each type
should the company purchase?
4.4 Matrices: Basic Operations
Two matrices are equal if they are the same size
and all corresponding elements are equal ( aij  biji, j )
 a b c  u v w a  u b  v c  w
d e f    x y z  iff d  x e  y f  z

 

Addition
The sum of two matrices of the same size is
the matrix with elements that are the sum of the
corresponding elements of the given matrices.
a b   w x  a  w b  x 
c d    y z    c  y d  z 

 
 

 2  3 0   3 1 2
1 2  5   3 2 5 

 

  1 7
5 0  2 


0
6
1  3 8  



 2 8
Properties of Matrices
Addition is commutative A  B  B  A
Addition is associative  A  B   C  A   B  C 
A matrix with all zero elements is the zero
matrix.
0 0 
0 0 0 

0
0


0 
0 
 
0 
0 
 
0 0 0 0 
0 0 0 0 


0 0 0 0
The negative of a matrix M, denoted  M , is a
matrix with elements that are the negatives of
the elements in matrix M.
a b 
 a  b 
If M  
, then  M  
.


c d 
 c  d 
We can then define subtraction as
A  B  A   B  .
3  2  2 2
5 0    3 4 

 

a b   2  1  4 3
 c d     5 6     2 4

 
 

The product of a scalar (real number) k and a
matrix M, denoted kM, is the matrix formed by
multiplying each element of M by the number k.
 3 1 0 
 2  2 1 3  


 0  1  2
Ms. Smith and Mr. Jones are salespeople in a
new car agency that sells only two models.
August was the last month for this year’s
models, and next year’s models were introduced
in September. Gross dollar sales for each month
are given in the following matrices:
August Sales
Compact Luxury
Smith $54,000 $88,000
Jones $126,000
0
(Matrix A)
September Sales
Compact Luxury
$228,000 $368,000
$304,000 $322,000
(Matrix B)
What were the combined sales in August and
September for each salesperson by model?
What was the increase in dollar sales from August
to September?
If both salespeople receive 5% commission on gross
dollar sales, what is the commission each receives
by model for September?
Product of a row matrix with a column matrix
The product of a 1 n matrix with an n  1 matrix is
an 11 matrix given by:
a1
a2
a3
 b1 
b 
 2
 an b3   a1b1  a2b2  a3b3    anbn 

 
bn 
  5
2  3 0 2  
 2
A factory produces a slalom water ski that requires
3 labor hours in the assembly department and 1
labor hour in the finishing department. Assembly
personnel receive $9/hour while finishing personnel
receive $6/hour. What is the labor cost per ski?
9
3 1  
6 
Matrix product
If A is an m  p matrix and B is a p  n matrix, the
matrix product of A and B, denoted AB, is an m n
matrix whose ij element is obtained by the product
of the i row of A with the j column of B. The
elements are  aip b pj . If the number of columns of
p 1, n
A is not equal to the number of rows of B, the
matrix product is not defined.
m p
pn
 1 3
 2 3  1 

2
0
 2 1 2  



 1 2
 2 1
 1 0  1  1 0 1  

 2 1 2 0
 1 2
 2 1
1  1 0 1  

1
0
 2 1 2 0 



 1 2
 2 6  1 2
 1  3 3 6 



1 2  2 6 
3 6  1  3 



  5
2  3 0 2  
 2
  5
 2 2  3 0 
 
 2
2
 2 1  2 1  2 1
 1 0   1 0  1 0 

 


 1 2  1 2  1 2
2
 2 6   2 6  2 6 
 1  3   1  3  1  3 

 


Notes: Matrix multiplication is not commutative.
The principle of zero products does not hold.
2
1 1
 1  1 


2
3 6 
  1  2 


2  1 a b   6 17
5 3   c d    7

4


 

Combine the time and labor costs for slalom and
trick skis from previous example.
Wages Hours
12 13 5 1.5
 7 8  3 1  



Hours Wages
5 1.5 12 13
3 1   7 8  



4.5 Inverse of a Square Matrix
The identity element for multiplication for the set
of all square matrices of order n is the square matrix
of order n, denoted by I, with 1's along the principal
diagonal and 0's elsewhere.
1
0

0
1

1 0 0   a b
0 1 0   d e


0 0 1  g h
a b
d e

 g h
1 0 0 
0 1 0 


0 0 1
c
f

i 
c  1 0 0 
f  0 1 0 


i  0 0 1
1 0  a b c 
0 1   d e f 



1 0 0 
c

0
1
0

f  
0 0 1
a b
d e

2
0 1 
1 0 


 0  1
 1 0  


2
IM  MI  M
Multiplicative Inverse
Let M be a square matrix of order n, and I be the
identity matrix of order n. If  a matrix M 1 
M 1M  MM 1  I
then M 1 is called the multiplicative inverse of M
or more simply the inverse of M.
 2 3  a c 
1 2   b d 



1  1 1   a d
0 2  1 b e


2 3 0   c f
g
h

i 
1  1 1 1 0 0 
0 2  1 0 1 0


2 3 0 0 0 1
Inverse of a Square Matrix M
If M I  can be transformed by row operations into
I B, then the resulting matrix B is M 1. However,
if we obtain all 0's in any row left of the vertical
line, then M 1 does not exist.
Find the inverse of
 4  1
 6 2 


Find the inverse of
 2  4
 3 6 


blank A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
0
1 2 3 4 5 6 7 8 91 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2
01234567 890123 456
19 5 3 18 5 20 0 3 15 4 5
4 3 19 3 5 0 15 5
1 1  5 18 20 3 4 0



91 24 66 21 80 25 9 3 72 19 20 5
Decode
46 84 85 28 47 46 4 5 10 30 48 72 29 57 38 38 57 95
1 1 1 


using 2 1 2


2 3 1
1
 5 2
A1=  2  1 0 


 4  1  1
4.6 Matrix Equations and Systems of Linear Equations
ax  b
ax  b  c
Basic Properties of Matrices
Addition
 A  B   C  A  B  C 
A B  B  A
A0  0 A
A   A    A   A
Multiplication
 AB C  ABC 
AI  IA  A
AA1  A1 A  I
Combined
A B  C   AB  AC
B  C  A  BA  CA
A  B  A  C  B  C, CA  CB and AC  BC
Solving a matrix equation
AX=B
x  y  z 1
2y  z 1
2x  3y  1
3  1
3
A1   2  2 1 


 4  5 2 
x yz 3
2y  z 1
2x  3y  4
x  y  z  5
2y  z  2
2 x  3 y  3
If the number of equations in a system equals the
number of variables, and the coefficient matrix has
an inverse, then the system will have a unique
solution that can be found using the inverse.
Will not work if:
1) coefficient matrix is singular
2) number of variables does not equal the number of
equations.
An investment advisor currently has two types of
investments available for clients: A conservative
investment A that pays 10%/year, and a higher risk
investment B that pays 20% per year. Clients may
divide investments between the two to achieve any
total return between 10% and 20%. Higher return is
higher risk. How should each client listed invest to
achieve the desired return?
Client
1
2
3
k
Investment $20000 $50000 $10000 k1
Rturn
$2400 $7500 $1300 k 2
12%
15%
13%
x1 
x2  k1
0.1x1  0.2 x2  k 2
4.7 Leontief Input-Output Analysis
Wassily Leontief 1973 Nobel Prize
500 sectors of the economy interacting
Two industry model Electric and Water
Suppose production of $1 worth of electricity
requires $0.30 worth of electricity and $0.10 worth
of water. Production of $1 worth of water requires
$0.20 worth of electricity and $0.40 worth of water.
Outside demand is:
d1  $12 million for electricity
d 2  $8 million for water
To produce this amount of water and electricity:
Basic Input-Output problem
Given the internal demands for each industry’s
output, determine the output levels for the various
industries that will meet the given final (outside)
level demand as well as the internal demand.
x1  0.3x1  0.2 x2  d1
x2  0.1x1  0.4 x2  d 2
 x1  0.3 0.2  x1   d1 
 x   0.1 0.4  x   d 
 2   2 
 2 
X  MX  D
 0.7  0.2
 1.5 0.5 
1
I M 
and  I  M   



0
.
1
0
.
6
0
.
25
1
.
75




Solution to a Two-Industry Input-Output Problem
Given two industries C1 and C2 , with
M  C1
C2
C1
a11
a21
C2
 x1 
 d1 
a12 , X    , and D   
 x2 
d 2 
a22
where aij is the input required of Ci to produce a
dollar’s worth of output for C j , the solution to
the input-output matrix equation
X  MX  D
is
X  I  M  D
assuming that  I  M  has an inverse.
1
Three-Industry model
An economy is based on three sectors, agriculture,
energy, and manufacturing. Production of a
dollar’s worth of agriculture requires an input of
$0.20 from the agriculture sector, and $0.40 from
the energy sector. Production of a dollar’s worth
of energy requires an input of $0.20 from the
energy sector, and $0.40 from the manufacturing
sector. Production of a dollar’s worth of
manufacturing requires an input of $0.10 from the
agriculture sector, $0.10 from the energy sector,
and $0.30 from the manufacturing sector. Find
the output from each sector that is needed to
satisfy a final demand of $20billion for
agriculture, $10billion from energy, and
$30billion from manufacturing.
A
E
M
A
E
M
I M 
I  M  
1
5. Linear Inequalities and Linear Programming
5.1. Linear Inequalities in Two Variables
y  2x  3
2x  3 y  5
y  2x  3
2x  3 y  5
3x  4 y  24
3x  4 y  24
y  x  2
3x  4 y  24
Graphing of Linear Inequalities
ax  by  c
The graph of the linear inequality Ax  By  C or
Ax  By  C is either the upper half-plane or the
lower half-plane determined by the line Ax  By  C .
Procedure:
1) First graph Ax  By  C as a dashed line if
equality is not included in the original
statement or a solid line if equality is included.
2) Choose a test point anywhere in the plane not
on the line and substitute into the inequality.
3) The graph of the original inequality includes
the half-plane containing the test point if the
inequality is satisfied by that point, or the
other half-plane if the inequality is not
satisfied.
2x  3y  6
y  3
2x  5
x  3y
Interpreting a graph
Ax  By  C
A concert sales promoter wants to book a rock
group for a stadium concert. A ticket for the
playing field costs $125, while a ticket in the
stands costs $175. The group wants a total ticket
sales of at least $700,000. How many tickets of
each type must be sold?
5.2 Systems of Linear Inequalities
x y 6
2x  y  0
Solve Systems of Linear Inequalities; Corner Pts
2 x  y  22
x  y  13
2 x  5 y  50
x0
y0
A patient in a hospital on a brown rice and skim
milk diet is required to have at least 800 calories
and at least 32 grams of protein each day. Each
serving of brown rice contains 200 calories and 5
grams of protein. Each serving of skim milk
contains 80 calories and 8 grams of protein. How
many servings of each food should be eaten every
day to meet the minimum daily rqmts?
Cal
Pr
BR SM
200 80
5
8
MDR
800
32
5.3 Linear Programming in Two Dimensions:
Geometric Approach
Objective – find values of decision variables
that produce optimal value
Objective function maximized or minimized to
make that decision
Constraints – limits imposed in the decision
Feasible solutions (feasible region) – solution
set for system of linear equations.
Optimal solution – the best feasible solution
Iso-profit line – lines of constant profit
A tent manufacturer makes a standard model and an
expedition model. Each standard tent requires 1
labor-hour from the cutting department and 3 laborhours from the assembly department. Each
expedition tent requires 2 labor-hours from the
cutting department and 4 labor-hours from the
assembly department. The maximum labor hours
available per day in the cutting and assembly
departments are 32 and 84 respectively. If the
company makes a $50 profit on each standard tent,
and $80 on each expedition tent, how many tents
should be made daily to maximize profit?
Labor hours req per tent
Std Exp Max
Cut
1
2
32
Assy
3
4
84
Profit 50 80
Maximize P  50 x1  80 x2
x1  2 x2  32
Subject to
3 x1  4 x2  84
x1  0
x2  0
Fundamental Theorem of Linear Programming
If an optimal value for a linear programming
problem exists, then it occurs at one or more
corner points of the region of feasible solutions.
Procedure
1. Identify the decision variables.
2. Summarize variable information in a table.
3. Determine objective and linear obj function.
4. Write problem constraints (& non-neg).
5. Graphically determine feasible region.
6. Determine the corner points of feasible region.
7. Determine optimal solution by evaluating
objective function at corner points.
8. Interpret results in terms of original problem.
Existence of feasible solutions
1. bounded region
2. unbounded region
3. empty feasible region
Minimize and maximize z  3x1  x2 , subject to
x1  0, x2  0 , 2 x1  x2  20 , 10 x1  x2  36 , and
2 x1  5x2  36
Solving a Linear Programming Problem
1. Read problem carefully.
2. Use a table to write the objective function
and all of the constraints.
3. Sketch a graph of the region of feasible
solutions. Identify corner points.
4. Evaluate the objective function at each one
of the corner points. Choose the min/max.
Minimize and maximize z  10 x1  20 x2 , subject
to x1  0, x2  0 , 6 x1  2 x2  36 , 2 x1  4 x2  32 ,
and x2  20
Minimize and maximize P  2 x1  3x2 , subject
to x1  0, x2  0 , x1  x2  8, x1  2 x2  8, and
2 x1  x2  10
A patient in a hospital requires at least 84 units
of drug A and 120 units of drug B. Each gram
of M contains 10 units of drug A and 8 units of
drug B. Each gram of N contains 2 units of
drug A and 4 units of drug B. Now suppose that
both M and N contain an undesirable drug D, 3
units/gram in M and 1 unit/gram in N. How
many grams of M and N can be mixed to meet
the minimum daily requirements, while
minimizing intake of drug D?
M
N
MDR
A 10
B 8
D 3
2
4
1
84
120
6. Linear Programming: The Simplex Method
6.1 The Table Method:
An Introduction of the Simplex Method
Maximize P  50 x1  80 x2
x1  2 x2  32
Subject to
3 x1  4 x2  84
x1  0
x2  0
Bounded and unbounded variables
Standard Form
A linear programming problem is said to be a
standard maximization problem in standard form
if its mathematical model is of the following form:
Maximize the objective function
P  c1 x1  c2 x2    cn xn
subject to constraints of the following form:
a1 x1  a2 x2    an xn  b, b  0
with non-negative constraints
x1, x2 , xn  0
Slack variables
x1  2 x2  32
3 x1  4 x2  84
i-system 20,4 
e-system
20,4,4,8
20,8,4,8
Basic Solutions
Feasible Solutions
x1  2 x2  s1
3x1  4 x2
 32
 s2  84
Procedure: The Table Method
1. Use slack variables s1 , s2 ,, sm to convert the
i-system to e-system.
2. Form a table with m  2   m  1 / 2 rows
and m  2  columns labeled x1 , x2 , s1 , s2 ,, sm ,
then assign 0 to x1 , x2 , and 2 zeros to each comb
of variables in each row.
3. Complete each row to a soln of the e-system,
if possible. (since two values are zero, you can
easily solve two equations in two unknowns)
4. Solve the linear programming problem by
finding the maximum value of P for rows that
have no negative values.
Basic Solutions
Feasible Solutions
Fundamental Theorem of Linear Programming
If an optimal value of the objective function in
a linear programming problem exists, then it
occurs at one or more of the basic feasible
solutions.
Convert this i-system to an e-system
3 x1  2 x2  21
x1  5 x2  20
x1  0
x2  0
Find a basic solution for x1  0 and s1  0
Find a basic solution for s1  0 and s2  0
Construct a table of basic solutions and use it to
solve this linear programming problem:
Maximize P  10 x1  25 x2
3 x1  2 x2  21
Subject to
x1  5 x2  20
x1  0
x2  0
x1
0
0
0
x2
0
s1
s2
0
0
0
0
0
0
0
0
P  10 x1  25 x2
Construct a table of basic solutions and use it to
solve this linear programming problem:
Maximize P  40 x1  50 x2
x1  6 x2  72
x1  3 x2  45
Subject to 2 x1  3 x2  72
x1  0
x1
x2  0
x2
s1
s2
s3
Given a system of linear equations associated
with a linear programming problem:
The variables are divided into two (mutually
exclusive) groups: Basic variables are
selected arbitrarily with the restriction that
there are as many basic variables as
equations. The rest of the variables are
nonbasic variables.
Nonbasic variables have the value zero when
solving for the basics in a basic solution. If
a basic solution has no negative values, it is a
basic feasible solution.
Maximize P  10 x1  12 x2
x1  x2  2
Subject to
x1
0
0
0
x1  x2  3
x1  0
x2  0
x2
s1
0
0
s2
0
0
0
0
0
0
0
P  10 x1  12 x2
The Table Method (k decision var, m constraints)
1. Use slack variables s1 , s2 ,, sm  to convert
the i-system to an e-system.
2. Form the table, label the columns. Assign 0
to  x1 , x2 ,, xk , continue to all possible
combinations.
3. Complete each row to a solution of the e-sys.
4. Fill in the value of P for any row with no
negative values, then pick the maximum P.
6.2 Simplex Method: Maximization with
problem constraints of the form 
Maximize P  50 x1  80 x2
x1  2 x2  32
Subject to
3 x1  4 x2  84
x1  0
x2  0
Add slacks=>e-system; reformat objective=>init
x1  2 x2  s1
 32
3x1  4 x2
 s2
 84
 50 x1  80 x2
P0
x1, x2 , s1 , s2  0
Basic Solutions and Basic Feasible Solutions (bfs)
1. Let P always be a basic variable.
2. Note that a basic solution of this system is
still basic after P is deleted.
3. If a basic solution to the above is a bfs
without P, it is also a bfs with P.
4. A bfs can contain a negative number only
if it is the value of P.
Fundamental Theorem of Linear Programming
If an optimal value of the objective function in
a linear programming problem exists, then it
occurs at one or more of the basic feasible
solutions of the initial system.
x1  2 x2  s1
3x1  4 x2
 32
 s2
 50 x1  80 x2
 84
P0
Initial basic feasible solution, pick slacks and P
Simplex Tableau
x1
x2
s1 s2
P
s1  1
2 1 0 0 32
s2  3
4 0 1 0 84


P  50  80 0 0 1 0 
Number of variables
Select basic variables; non-basic variables
Pivot Operation
x1
x2 s1
s2
P
s1  1
2 1 0 0 32
s2  3
4 0 1 0 84


P  50  80 0 0 1 0 
Pivot Element
1.Select most negative indicator in bottom row.
This is the pivot column.
2.Divide each positive element in the pivot
column into the corresponding element in the
last column. (if no positive elements in pivot
column, stop) The pivot row is the one that
generates the smallest quotient.
3.The pivot element is at the intersection of
pivot row and column.
Pivot operation
1. Use division to put a 1 in the pivot element.
2. Use row operations to zero all other elements
in that column .
x1
x2
s1 s2
P
s1  1
2 1 0 0 32
s2  3
4 0 1 0 84


P  50  80 0 0 1 0 
Maximize P  10 x1  5x2
4 x1  x2  28
Subject to
2 x1  3 x2  24
x1  0
x2  0
Maximize P  6 x1  3x2
 2 x1  3 x2  9
Subject to
 x1  3 x2  12
x1  0
x2  0
A farmer owns a 100-acre farm and plans to plant
at most 3 crops. The seed for crops A, B, and C
cost $40, $20, and $30 per acre respectively. A
max of $3200 can be spent on seed. These crops
require 1, 2, and 1 workday per acre, and there are
a max of 160 workdays available. If the farmer
can realize profit of $100, $300, and $200 per acre
respectively, how many acres of each crop should
be planted to max profit?
Objective:
Constraints: