1. Zenith Co. produces three different models of TV for its customer. Each model
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Question
1. Zenith Co. produces three different models of TV for its customer. Each model I, II, and III earns a profit of $15, $5, and $30 respectively. Raw materials needed for each model and availability of raw materials are given in the following table.
Raw Materials
Pounds of raw material for each TV
Maximum availability (lbs)
Model I
Model II
Model III
Glass
3
2
3
150
Plastic
5
1
6
275
Metal
4
2
5
235
Using the Solver method, find the optimal solution.
Raw Materials
Pounds of raw material for each TV
Maximum availability (lbs)
Model I
Model II
Model III
Glass
3
2
3
150
Plastic
5
1
6
275
Metal
4
2
5
235
Explanation / Answer
The problem in mathematical notation is
p=profit
x=# of model I
y=# of model II
z=# of model III
Maximize p = 15x + 5y + 30z subject to
3x + 2y + 3z <= 150
5x + 1y + 6z <= 275
4x + 2y + 5z <= 235
x >= 0
y >= 0
z >= 0
Using simplex algorithm
Table #1
x y z s1 s2 s3 s4 s5 s6 p
3 2 3 1 0 0 0 0 0 0 150
5 1 6 0 1 0 0 0 0 0 275
4 2 5 0 0 1 0 0 0 0 235
1 0 0 0 0 0 -1 0 0 0 0
0 1 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-15 -5 -30 0 0 0 0 0 0 1 0
Table #2
x y z s1 s2 s3 s4 s5 s6 p
3 2 3 1 0 0 0 0 0 0 150
5 1 6 0 1 0 0 0 0 0 275
4 2 5 0 0 1 0 0 0 0 235
-1 0 0 0 0 0 1 0 0 0 0
0 1 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-15 -5 -30 0 0 0 0 0 0 1 0
Table #3
x y z s1 s2 s3 s4 s5 s6 p
3 2 3 1 0 0 0 0 0 0 150
5 1 6 0 1 0 0 0 0 0 275
4 2 5 0 0 1 0 0 0 0 235
-1 0 0 0 0 0 1 0 0 0 0
0 -1 0 0 0 0 0 1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-15 -5 -30 0 0 0 0 0 0 1 0
Table #4
x y z s1 s2 s3 s4 s5 s6 p
3 2 3 1 0 0 0 0 0 0 150
5 1 6 0 1 0 0 0 0 0 275
4 2 5 0 0 1 0 0 0 0 235
-1 0 0 0 0 0 1 0 0 0 0
0 -1 0 0 0 0 0 1 0 0 0
0 0 -1 0 0 0 0 0 1 0 0
-15 -5 -30 0 0 0 0 0 0 1 0
Table #5
x y z s1 s2 s3 s4 s5 s6 p
0.5 1.5 0 1 -0.5 0 0 0 0 0 12.5
0.833333 0.166667 1 0 0.166667 0 0 0 0 0 45.8333
-0.166667 1.16667 0 0 -0.833333 1 0 0 0 0 5.83333
-1 0 0 0 0 0 1 0 0 0 0
0 -1 0 0 0 0 0 1 0 0 0
0.833333 0.166667 0 0 0.166667 0 0 0 1 0 45.8333
10 0 0 0 5 0 0 0 0 1 1375
Optimal Solution is : x = 0, y = 0, z = 45.8333 and p = 1375
Since z is the number of setsof model III take it to be 45. Then profit is p = 1350.
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