Question: Muir Manufacturing produces two.......... Edit
Let x be the number of rolls of grade X carpet and y be the number of rolls of grade Y carpet.
Now Z be the function of x & y.
Thus the objective function is,
Since each roll of grade X carpet and of grade Y carpet uses 50 units & 40 units of synthetic fiber respectively.
50x+40y ≤ 3000 ------ (1)
Since each roll of grade X carpet and of grade Y carpet requires 25 hours & 28 hours of production time respectively.
25x+28y ≥ 1800 ------ (2)
Since each roll of grade X carpet and of grade Y carpet needs 20 units & 15 units of foam backing respectively.
20x+15y ≤ 1500 ------ (3)
Therefore the required constraint equations are,
50x+40y ≤ 3000
25x+28y ≥ 1800
20x+15y ≤ 1500
And also x ≥ 0 & y ≥ 0
We will solve the given problem by graphical method.
Here, maximum Z = 200x+160y ------ (4)
Now we write the constraint equations as equations,
This can be written as,
(x/60) + (y/75) = 1 ------ (5)
(x/72) + (y/64.286) = 1 ------ (6)
(x/75) + (y/100) = 1 ------ (7)
Now using the straight lines (5) , (6) & (7) we draw the following graph,
o ' ' ' ' ' ' ' ' ' ' ' X
Now the region bounded by the straight lines (5), (6) & the Y axis named as ABCA is the feasible region.
Now solving x=0, & the straight line (5) we get the point A (0, 75).
Similarly solving x=0 & the straight line (6) we get the point B (0, 450/7).
And solving (5) & (6) we get the point C (78, 75/2).
Since this is a maximizing problem we will take that values of x & y for which the objective function gives the maximum value.
Now putting the value of x & y from A, B & C in (4) we get the values of Z are 12000, (72000/7) ≈ 10285.7143 & 21600 respectively.
Therefore the feasible solutions of this problem are,
x =78 & y = 75/2.
Therefore the optimal solution for total profit is,
maximum Z = 21600
The optimal solution for the decision variables is,
X = 78 & y = 75/2
And the optimal solution for the total profit is,
i.e, the maximum value of the objective function is,
maximum Z = 21600.
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