The Collin Freight Company has an order for three products t

The Collin Freight Company has an order for three products to be delivered to a destination. Product I requires 10 cubic feet, weighs 10 pounds, and has a value of $100. Product II requires 8 cubic feet, weighs 20 pounds, and has a value of $20. Product III requires 20 cubic feet, weighs 40 pounds, and has a value of $200. If the carrier can carry 6,000 cubic feet, 11,000 pounds, and is insured for $36.900, how many of each product can be carried? a. Identify the variables of the problem using x, y and z. b. Fill out the following table from the information given: c. Write the augmented matrix for the system of equations. d. Solve the system of equations using Gauss-Jordan Elimination. (Attach work) e. In at least one complete sentence, answer the question: How many of each product can be carried?

Solution

(a). The variables of the problem are the number of the products I,II and III. Let these be x , y and z respectively.

(b)

x

y

z

Total

Volume (cu.ft.)

10

8

20

38

Weight (lb.)

10

20

40

70

Value ($)

100

20

200

320

(c) The 3 equations which can be arrived at from the conditions about the maximum weight and volume which can be carried by the carrier and the limit on insurance are as under:

10x+8y+20z 6000. On dividing both the sides by 2, we get 5x+4y+10z 3000…(1).

10x+20y+40z 11000. On dividing both the sides by 10, we get x +2y+4z 1100…(2).

100x +20y+200z 36900. On dividing both the sides by 20, we get 5x+y+10z 1845…(3)

The augmented matrix of the system is A =

5

4

10

3000

1

2

4

1100

5

1

10

1845

( we can create the augmented matrix of the system from the original inequalities also, but smaller numbers are easier to deal with).

(d) We can reduce A to its RREF, using Gauss-Jordan elimination as under:

Multiply the 1st row by 1/5

Add -1 times the 1st row to the 2nd row              

Add -5 times the 1st row to the 3rd row

Multiply the 2nd row by 5/6

Add 3 times the 2nd row to the 3rd row

Multiply the 3rd row by 1/5

Add -5/3 times the 3rd row to the 2nd row

Add -2 times the 3rd row to the 1st row

Add -4/5 times the 2nd row to the 1st row

Then the RREF of A is

1

0

0

254

0

1

0

385

0

0

1

19

(e) Maximum 254 numbers of Product I, 385 numbers of Product II and 19 numbers of Product III can be carried.

	x	y	z	Total
Volume (cu.ft.)	10	8	20	38
Weight (lb.)	10	20	40	70
Value ($)	100	20	200	320