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.

 2x1 + x2 ≤ 6 x1 + x2 ≤ 4 x1 ≥ 0 x2 ≥ 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 = x1 + x2

 2x1 + x2 ≤ 6 x1 + x2 ≤ 4 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 to solve this problem.)

Using the work from the previous problem, we make a table:

 Corner point P = x1 + x2 (0, 0) 0 + 0 = 0 (0, 4) 0 + 4 = 4 (3, 0) 3+ 0 = 3 (2, 2) 2 + 2 = 4

From the table it's clear that the maximum value of P will be 4, and that happens for either (x1, x2) = (2, 2) or (x1, x2) = (0, 4)

(4) (30 points) Use Gauss-Jordan Elimination to solve the following system of equations:
 2x1 + 2x2 + x3 = 9 2x1 + x2 + 2x3 = 11 x1 + x2 + x3 = 6

 [ 2 2 1 | 9 ] 2 1 2 | 11 1 1 1 | 6
R1 <--> R3
 [ 1 1 1 | 6 ] 2 1 2 | 11 2 2 1 | 9

-2R1 + R2 --> R2

-2R1 + R3 --> R3

 [ 1 1 1 | 6 ] 0 -1 0 | -1 0 0 -1 | -3
-R2 --> R2
 [ 1 1 1 | 6 ] 0 1 0 | 1 0 0 -1 | -3
-R2 + R1 -->R1

 [ 1 0 1 | 5 ] 0 1 0 | 1 0 0 -1 | -3
-R3 --> R3
 [ 1 0 1 | 5 ] 0 1 0 | 1 0 0 1 | 3
-R3 + R1 --> R1

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

Now you can read off the solution x1 = 2, x2 = 1, x3 = 3.

(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 pick-up 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.
 ... so we want to maximize profit... P = 15x +10y + 4z Subject to.... Weight constraint ... 20x +10y +5z ≤ 1000 Time constraint... x + 2y + z ≤ 40 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.

 20x + 10y + 5z + s1 = 1000 1x + 2y + z + s2 = 40 -15x -10y -4z + P = 0

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

 x y z s1 s2 P s1 20 10 5 1 0 0 | 1000 1000/20 = 50 s2 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 s2 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 -20R2 + R1 ---> R1 and 15R2 + R3 ---> R3

 x y z s1 s2 P s1 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.