. A manufacturer produces three products, A,B,C. each unit of products A,B,C requires the following amounts of time in hours in each of the indicated departments.
Product
Department A B C
Machining 0.2 0.3 0.5
Assembly 0.4 0.2 0.4
Painting 0.3 0.5 0.2
Suppose that the total time available per week in each department is 105 hrs, 116 hrs, and 83 hrs respectively. Determine the number of units of each product to be produced In a week to use up all the available time.(use the Gauss - Jordan method )
To solve this problem using the Gauss-Jordan method, we need to set up a system of linear equations based on the given information.
Let's assume that the number of units of product A, B, and C produced in a week is x, y, and z, respectively.
Based on the time required in each department, we can set up the following equations:
0.2x + 0.3y + 0.5z = 105 -- (1) (Machining department)
0.4x + 0.2y + 0.4z = 116 -- (2) (Assembly department)
0.3x + 0.5y + 0.2z = 83 -- (3) (Painting department)
Now, let's put these equations in matrix form:
| 0.2 0.3 0.5 | | x | | 105 |
| 0.4 0.2 0.4 | * | y | = | 116 |
| 0.3 0.5 0.2 | | z | | 83 |
To solve this system of equations using Gauss-Jordan method, we can perform the following row operations:
1. Multiply Row 1 by 10:
| 2 3 5 | | x | | 1050 |
| 4 2 4 | * | y | = | 1160 |
| 3 5 2 | | z | | 830 |
2. Multiply Row 2 by 5:
| 2 3 5 | | x | | 1050 |
| 20 10 20 | * | y | = | 5800 |
| 3 5 2 | | z | | 830 |
3. Multiply Row 3 by 2:
| 2 3 5 | | x | | 1050 |
| 20 10 20 | * | y | = | 5800 |
| 6 10 4 | | z | | 1660 |
4. Subtract 2 times Row 1 from Row 2:
| 2 3 5 | | x | | 1050 |
| 0 4 10 | * | y | = | 3700 |
| 6 10 4 | | z | | 1660 |
5. Subtract 3 times Row 1 from Row 3:
| 2 3 5 | | x | | 1050 |
| 0 4 10 | * | y | = | 3700 |
| 0 1 -11 | | z | | -350 |
6. Multiply Row 2 by 1/4:
| 2 3 5 | | x | | 1050 |
| 0 1 2.5| * | y | = | 925 |
| 0 1 -11 | | z | | -350 |
7. Subtract Row 2 from Row 3:
| 2 3 5 | | x | | 1050 |
| 0 1 2.5| * | y | = | 925 |
| 0 0 -13.5| | z | | -1275|
8. Multiply Row 3 by -1/13.5:
| 2 3 5 | | x | | 1050 |
| 0 1 2.5| * | y | = | 925 |
| 0 0 1 | | z | | 94.44|
9. Subtract 2.5 times Row 3 from Row 2:
| 2 3 5 | | x | | 1050 |
| 0 1 0 | * | y | = | 704.16|
| 0 0 1 | | z | | 94.44 |
10. Subtract 5 times Row 3 from Row 1:
| 2 3 0 | | x | | 1040.56|
| 0 1 0 | * | y | = | 704.16 |
| 0 0 1 | | z | | 94.44 |
11. Subtract 3 times Row 2 from Row 1:
| 2 0 0 | | x | | 980.56 |
| 0 1 0 | * | y | = | 704.16 |
| 0 0 1 | | z | | 94.44 |
So the solution to the system of equations is:
x = 490.28
y = 704.16
z = 94.44
Therefore, the manufacturer needs to produce approximately 490 units of product A, 704 units of product B, and 94 units of product C in a week in order to use up all the available time.