Finite Math 
Test #1

Oct. 9, 2000

A Track 
R. Hammack


Name: ________________________ 
Score: _________

(1) (10 points) Multiply the matrices.
[ 
1

3

]  [ 
0

1

2

]  =  
7

1

5

2

5

[

1(0)+3(5)

1(1)+3(2)

1(2)+3(5)

] =  [ 
15

7

17

] 
7(0)+1(5)

7(1)+1(2)

7(2)+1(5)

5

9

19

(2) (15 points) Sketch the solutions of the following system of inequalities.
2x_{1}

+

x_{2}

≤

6 
x_{1}

+

x_{2}

≤

4 
x_{1}

≥

0  
x_{2}

≥

0 
The solution is sketched on the right. Notice that the corner points are (0,0), (0,4), (3,0), and (2,2). 
(3) (15 points)

Maximize subject to ...

P = x_{1} + x_{2}

You may use any method. (However, notice that you sketched the feasible region in the previous problem. Feel free to use that information to solve this problem.)
Using the work from the previous problem, we make a table:
Corner point

P = x_{1} + x_{2}

(0, 0)

0 + 0 = 0

(0, 4)

0 + 4 = 4

(3, 0)

3+ 0 = 3

(2, 2)

2 + 2 = 4

2x_{1}

+

2x_{2}

+

x_{3}

=

9

2x_{1}

+

x_{2}

+

2x_{3}

=

11

x_{1}

+

x_{2}

+

x_{3}

=

6


R_{1} <> R_{3} 

2R_{1} + R_{2} > R_{2} 2R_{1} + R3 > R3 

R_{2} > R_{2} 

R_{2} + R_{1} >R_{1} 

R_{3} > R_{3} 

R_{3} + R_{1} > R_{1} 

(Reduced)

(5) (30 points) Use the simplex method to solve the following problem.
A hippie farmer wants to sell three crops  apples, beans, and corn  at the local Farmer's Market. He reckons it will take him 1 hour to harvest each bushel of apples, 2 hours to harvest each bushel of beans, and 1 hour to harvest each bushel of corn. Each bushel of apples weighs 20 pounds, each bushel of beans weighs 10 pounds, and each bushel of corn weighs 5 pounds. Apples sell for $15 per bushel, beans sell for $10 per bushel, and corn sells for $4 per bushel. He has a maximum of 40 hours in which to harvest the crops. Given that his aging pickup truck can haul at most 1000 pounds of produce, how many bushels of apples, beans, and corn should he take to the Farmer's Market to realize a maximum profit? (Assume he sells everything he brings to the market.)
First, let's organize all the data into a table.
Apples

Beans

Corn

Maximum


Weight 
20

10

5

1000 pounds

Time 
1

2

1

40 hours

Profit 
$15

$10

$4

Let x = bushels of apples. Let y = bushels of beans. Let z = bushels of corn. 

There are two problem constrains, so there will be two slack variables. The problem is transformed into the following system of equations.
20x

+ 10y

+ 5z

+ s_{1}

=

1000


1x

+ 2y

+ z

+ s_{2}

=

40










15x

10y

4z

+ P

=

0

x

y

z

s_{1}

s_{2}

P


s_{1} 
20

10

5

1

0

0



1000

1000/20 = 50  
s_{2} 
1

2

1

0

1

0



40

40/1 = 40  









P 
15

10

4

0

0

1



0

Now it's time to do a pivot operation. We select the pivot column (1st column) and the pivot row (second row). Thus, x will enter and s_{2} will exit. The pivot element is already 1 (see how nicely I made it work out?) so we just need to get 0's elsewhere in the pivot column.
Thus we do 20R_{2} + R_{1} > R_{1} and 15R_{2} + R_{3 }> R_{3}
x

y

z

s_{1}

s_{2}

P


s_{1} 
0

30

15

1

20

0



200


x 
1

2

1

0

1

0



40











P 
0

20

11

0

15

1



600

There are no more negative indicators, so we can just read off the solution. The old guy will make a $600 profit if he picks and sells x=40 bushels of apples, y=0 bushels of beans, and z=0 bushels of corn. That's the best he can do given the constraints.