from my homework
a small manufacturing plant makes three types of inflatabl boats: one-person, two-person, and 4 person models. each boa requires the services of three departments as listed in the tabble. the cutting, assemply, and packaging depepartments have available a masimum of 380,330, and 120 labor-hours per week respectively.
table
collumns are one person, 2 person, 4 person respectivly
rows
cutting 0.5hours 1hour 1.5hour
assembly 0.6hours 0.9hour 1.2hour
packaging 0.2hour 0.3hour 0.5hour
(use gauss-jordan elimination when possible and show all work and reasoning)
x=number of boats for one-person
y=number of boats for two-person
z=number of boats for four-person
We have the next system of equations.
0.5x+y+1.5z=380
0.6x+0.9y+1.2z=330
0.2x+0.3y+0.5z=120
we can solve this systema of equations by Gauss-Jordan elimination method.
0.5 1 1.5 380
0.6 0.9 1.2 330
0.2 0.3 0.5 120
0.5 1 1.5 380
0 -0.15 -0.3 -63 (0.5R₂-0.6R₁)
0 -0.05 -0.05 -16 (0.5R₃-0.2R₁)
0.5 1 1.5 380
0 -0.15 -0.3 -63
0 0 0.0075 0.75 (0.15R₃-0.05R₂)
0.5 1 1.5 380
0 -0.15 -0.3 -63
0 0 1 100 (R₃/0.0075)
0.5 1 0 230 (R₁-1.5R₃)
0 -0.15 0 -33 (R₂+0.3R₃)
0 0 1 100
0.5 1 0 230
0 1 0 220 (R₂/-0.15)
0 0 1 100
1 2 0 460 (2R₁)
0 1 0 220
0 0 1 100
1 0 0 20 (R₁-2R₂)
0 1 0 220
0 0 1 100
Answer A)
20 boats for one-person
220 boats for two-person
100 boats for four-person.
B)We have 120 hour more of work.
C) We have 100 boats manofactured less. We only manofacture 220 boats for two-person and 20 boats for one-person
Solutions
To solve the problem lets use the variables "x,y, and z"
x=number of boats for one-person
y=number of boats for two-person
z=number of boats for four-person
⇒ System of equations
0.5x+y+1.5z=380
0.6x+0.9y+1.2z=330
0.2x+0.3y+0.5z=120
To solve these equations we have to use the Gauss-Jordan elimination method.
\(\left[\begin{array}{ccc}0.5&1&1.5\\0.6&0.9&1.2\\0.2&0.3&0.5\end{array}\right]\)
\(\left[\begin{array}{ccc}0.5&1&1.5\\0&-0.15&-0.3\\0&-0.05&-16\end{array}\right]\)
(0.5R₂-0.6R₁)
(0.5R₃-0.2R₁)
\(\left[\begin{array}{ccc}0.5&1&1.5\\0&-0.15&-0.3\\0&0&0.0075\end{array}\right]\)
(0.15R₃-0.05R₂)
\(\left[\begin{array}{ccc}0.5&1&1.5\\0&-0.15&-0.3\\0&0&1\end{array}\right]\)
(R₃/0.0075)
\(\left[\begin{array}{ccc}0.5&1&0\\0&0.15&0\\0&0&1\end{array}\right]\)
(R₁-1.5R₃)
(R₂+0.3R₃)
\(\left[\begin{array}{ccc}0.5&1&0\\0&1&0\\0&0&1\end{array}\right]\)
(R₂/-0.15)
\(\left[\begin{array}{ccc}1&2&0\\0&1&0\\0&0&1\end{array}\right]\)
(2R₁)
\(\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\)
(R₁-2R₂)
These numbers have to be added to each one
1 - 380, 330, 120
2 - 380, -63, -16
3 - 380, -63, 0.75
4- 380, -63, 100
5 - 230, -33, 100
6 - 230, 220, 100
7 - 460, 220, 100
8 - 20, 220, 100
Answer A)
20 boats for one-person
220 boats for two-person
100 boats for four-person.
B)We have 120 hour more of work.
C) We have 100 boats manufactured less. We only manufacture 220 boats for two-person and 20 boats for one-person