Finite Mathematics Lesson 3

Section 2.5 Section 3.1 Section 3.2 Section 3.3

Chapter 2.5, 3.1 - 3.3

I developed an "interactive tutorial" for section 2.5. Try it out and let me know how I can improve it.  The course material is getting significantly harder.  This chapter will require much reading so don't wait until that last minute take the quiz.  Good luck. 

Course Notes

Section 2.5 Gauss-Jordan Method for Calculating Inverses  

  Click here for tutorial site.

There are many methods for find the inverse of a matrix. You learned how to find the inverse of a 2 by 2 matrix in section 2.4.  We need a method for finding the inverse of a larger "square" matrixA "square" matrix is a matrix that has the same number of columns as rows. The Gauss-Jordan is one of the easiest methods to learn.  I developed an "Automatic Matrix" tutorial to help you learn this technique. 

  Click here to go to that site.

3 Important Rules

  1. Write down the matrix , and on its right,  write an identity matrix of the same size.
  2. Perform Gauss-Jordan elimination until you have an identity matrix on the left-hand side.
  3. The matrix on the right-hand side is the inverse.

3 Important Remarks

  1. Not all square matrices have inverses.  Why?
  2. You can check your work  A-1A=I    and   AA-1=I
  3. Some problems are solved using X=A-1B  . This is explained in section 2.4.  Pay attention to examples 4, 5, and 6.

When you are ready, click here for the assignment for this section.

Section 3.1   A Linear Programming Model

The book does an excellent job developing and explaining linear programming .   Read this section thoroughly and if necessary many times.  I strongly recommend using the tables to organize your data. The exercises will walk you through each step.  If you haven't got a solutions manual yet,  this may be a good time.    The graphical approach is not very efficient in solving most problems but it does help in understanding the method.  You will learn another method for solving these problems in chapter 4 that is more algorithmic in nature.

Key Point This text may shade inequalities differently that other text books.   They shade the side of the inequality that is not feasible.  I kind of like this approach because the graph looks better after the shading is done.  Look at page 129, figure 4.

Section 3.2   Linear Programming I

Linear Programming
A Graphical Approach

Maximize the objective function 7x + 4y subject to the following constraints.

The question is :  Find the value for the objective function that is the largest.  In order to find that value,  we need to graph the inequalities.

First, we want to graph the line
3x + 2y = 36
Use x and y intercepts.

  • To find x intercept, set y = 0
    3x + 2(0) = 36
    3x = 36
    x = 12,   (12, 0)  is x intercept

  • To find y intercept, set x = 0
    3(0) + 2y = 36
    2y = 36
    y = 18,   (0, 18)  is y intercept

Next, we want to graph the line
x + 4y = 32  
Use x and y intercepts.

  • To find x intercept, set y = 0
    x + 4(0) = 32
    x = 32
    (32, 0)  is x intercept

  • To find y intercept, set x = 0
    0 + 4y = 32
    4y = 32
    y = 8,   (0, 8)  is y intercept

Page 14 describes how to shade in the proper region.  First solve each inequality for y:


inequal.gif (2554 bytes)

  • Since both inequalities have less than symbols , , the feasible solution is below each line.
  • x 0  , y 0 means the x and y values are positive.  To the right of the y axis and above the x axis . (quadrant I)
  • Note:  If the inequality has a greater than symbol,   , the feasible solution is above the line.
  • Final Note:   The book shades in the area that is not in the feasible set.  Too hard to draw.
inequal1.gif (3309 bytes)
Now that the feasible set is shaded in,  we need to find the corner points.  I count 4 corner points.
The first 3 corner points are obvious.  (0, 0), (0, 8), and (12, 0) . Use the equations to the right to find the other point.
Find the remaining corner point by setting the equations equal to each other and solve for x.
Multiply both sides by 4
We get...
adding 6x and subtracting -32 from both sides 40 = 5x
The ordered pair is (8,6) x = 8, then  
(8, 6), (0, 0), (0, 8), and (12, 0)   are the four corner points.
Finally,  test each corner point in the objective function.

Maximize the objective function 7x + 4y subject to the following constraints.

Corner Point Value
(0,0) 7(0) + 4(0) = 0 + 0 = 0
(0,8) 7(0) + 4(8) = 0 +32 = 32
(12,0) 7(12) + 4(0) = 84 + 0 = 84
(6,8) 7(6) + 4(8) = 42 + 32 = 74
Maximum  is 84
inequal3.gif (4509 bytes)

The maximum value 84 occurs at the point (6,8)

 

Section 3.3   Linear Programming II

The book does a good job here.  I will add one word problem as an example.  (I know students love word problems)

An oil company owns 2 refineries.

An order is received for 1000 barrels of high grade oil, 1000 barrels of medium grade oil, and 1800 barrels of low grade oil.  How many days should each refinery be operated to fill the order at the least cost?

Solution:
Let's start by making a table.

Refinery I Refinery II Ordered
high grade 100 barrels 200 barrels 1000 barrels
medium grade 200 barrels 100 barrels 1000 barrels
low grade 300 barrels 200 barrels 1800 barrels
Cost $10,000 $9,000 ?

Now, write the inequalities:

high grade

100x + 200y 1000

medium grade

200x + 100y 1000

low grade

300x + 200y 1800

 Write the inequalities in standard form by rewrite solving for y.

This is the first inequality in red with x-int (10,0) and y-int (0,5)

The solution lies above the line

This is the 2nd inequality in blue with x-int (5,0) and y-int (0,10)

The solution lies above the line

This is the 3rd inequality in purple with x-int (6,0) and y-int (0,9)

The solution lies above the line

The feasible solution is purple  area.

Next find the corner points.  By looking at the graph and knowing x 0 and y 0 means the solution is in the first quadrant we get (10,0) and (0,10) as corner points.  We must calculate the remaining corner points by setting the corresponding inequalities equal to each other.

-.5x + 5 = -1.5x + 9 x = 4, then y = -.5(4) + 5 = 3 (4,3)
-2x + 10 = -1.5x + 9 x = 2, then y = -2(2) + 10 = 6 (2,6)
Note :  The red and blue line intersection is outside the feasible region.
Finally, let's check the cost function  10,000x + 9,000y  we want to minimize.
(0,10) 10,000(0) + 9,000(10) = 90,000 
(4,3) 10,000(4) + 9,000(3) = 67,000 
(2,6) 10,000(2) + 9,000(6) = 74,000 
(10,0) 10,000(10) + 9,000(0) = 100,000 

The minimum cost of $67,000 is achieved at (4,3).
Open refinery I for 4 days and refinery II for 3 days.

When you are ready, click here for the assignment for this section.

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