Finite Math
Test #1
Oct. 10, 2000
F Track
R. Hammack
Name: ________________________  
Score: _________

(1) Perform the indicated matrix operation, or explain why it cannot be done.

(a) [
5
1
 ][
1
4
 ]
=
  
2
3
2
0
  

[
5(1) + 1(2)
5(4) + 1(0)
 ]
=
 [
7
20
 ]
2(1) + 3(2)
2(4) + 3(0)
8
8

(b) [
0
-1
3
 ]
+
 [
0
1
2
 ]
=
4
7
1
-5
2
5
[
0
0
5
 ]
-1
9
6

(2) Sketch the solutions of the following system of inequalities.

2x1
+
x2
≤
 8
x1
+
x2
≤
 5
x1
+
2x2
≤
 8
x1
≥
 0
   
x2
≥
0
The solution is sketched on the right. The corner points are found to be (0, 0), (0, 4), (0,4), (2, 3) and (3,2). [Graphics:Images/temp_gr_2.gif]
(3)

Maximize

subject to ...

 

P = 10x1 + x2

2x1
+
x2
≤
 8
x1
+
x2
≤
 5
x1
+
2x2
≤
 8
x1
≥
 0
   
x2
≥
0

You may use any method. (However, notice that you sketched the feasible region in the previous problem. Feel free to use that information here.)

Using the corner points from the previous problem, we create a table:

Corner point
P = 10x1 + x2
(0, 0)
10(0) + 0 = 0
(0 ,4)
10(0) + 4 = 4
(2, 3)
10(2) + 3 = 23
(3, 2)
10(3) + 2 = 32
(4, 0)
10(4) + 0 = 40
From looking at the table, it becomes clear that the maximum value of P is 40, and this occurs when x1 = 40 and x2 = 0.

(4) Use Gauss-Jordan elimination to solve the following system of equations:

2x1
+
4x2
+
2x4
=
 6
x1
+
2x2
+
x3
+
2x4
=
 4

[
2
4
0
2
|
6
 ]
1
2
1
2
|
4
R1 <---> R2
[
1
2
1
2
|
4
 ]
2
4
0
2
|
6
-2R1 + R2 --> R2

[
1
2
1
2
|
4
 ]
0
0
-2
-2
|
-2
-1/2R2 --> R2
[
1
2
1
2
|
4
 ]
0
0
1
1
|
1
-R2 + R1 --> R1

[
1
2
0
1
|
3
 ]
0
0
1
1
|
1
  (Reduced)  

This corresponds to the system
x1
+
2x2
+
x4
=
 3
x3
+
x4
=
 1
...or ...
x1
=
 3 - 2x2 - x4
x3
=
 1 - x4

Thus, the solution are

x1
=
 3 - 2m -n
x2
=
 m
x3
=
 1 - n
x4
=
 n

(5) Use the simplex method to solve the following problem.

A small publishing company is considering whether to publish 3 books. Let's call them book A, book B and book C. Two machines are needed to print the books, a printer and a binder. First a book is printed on the printer, then it is fed into the binder. Each copy of book A takes 6 minutes on the printer, then 4 minutes on the binder. Each copy of book B takes 4 minutes on the printer, then 2 minutes on the binder. Each copy of book C takes 2 minutes on the printer, then 1 minute on the binder. The printer is available for 1000 minutes per week, and the binder is available for 800 minutes per week. Each copy of book A will bring a profit of $2, each copy of book B will bring a profit of $1, and each copy of book C will bring a profit of $3. How many copies of each book shuld be made to realize a maximum profit?

First, let's organize all the data into a table.

 
A
B
C
Maximum
Printer
6
4
2
1000 minutes
Binder
4
2
1
800 minutes
Profit
$2
$1
$3

Let x = copies of A.
Let y = copies of B.
Let z = copies of C.
... so we want to maximize profit...
 P = 2x +1y + 3z
Subject to....
  
Printer constraint ...
 6x +4y +2z ≤ 1000
Binder constraint...
 4x + 2y + z ≤ 800
Nonnegative constraint...
 x, y, z ≥ 0

There are two problem constrains, so there will be two slack variables. The problem is transformed into the following system of equations.

6x
+ 4y
+ 2z
+ s1
=
1000
4x
+ 2y
+ z
+ s2
=
800
-2x
-y
-3z
+ P
=
0

At this point we can see how to set up the simplex tableau:

   
x
y
z
s1
s2
P
   
s1  
6
4
2
1
0
0
|
1000
   1000/2 = 500
s2
4
2
1
0
1
0
|
800
 800/1 = 800
 
 
P
-2
-1
-3
0
0
1
|
0
  

Now it's time to do a pivot operation. We select the pivot column (3rd column) and the pivot row (1st row). Thus, z will enter and s1 will exit. The pivot element is 2, so first we need to change it to a 1 by doing 1/2R1 --> R1

               
   
   
3
2
1
1/2
0
0
|
500
    
 
4
2
1
0
1
0
|
800
  
 
 
 
-2
-1
-3
0
0
1
|
0
  

Now the pivot entry is 1. To finish the pivot operation, we need to get 0's elsewhere in the pivot coulum.
Thus we do -R1 + R2 ---> R2 and 3R1 + R3 ---> R3.

   
x
y
z
s1
s2
P
   
z  
3
2
1
1/2
0
0
|
500
    
s2
-1
0
0
-1/2
1
0
|
800
  
 
 
P
7
5
0
3/2
0
1
|
1500
  

There are no more negative indicators, so we can just read off the solution. A maximum profit of $1500 will be made by making z=500 copies of book C, 0 copies of book A and 0 copies of book B. That's the best that can be done given the constraints.