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Question
A factory owner purchases two types of machines, A and B, for his factory. The requirements and limitations for the machines are as follows:
| Area occupied by the machine |
Labour force for each machine |
Daily output in units |
|
| Machine A Machine B |
1000 sq. m 1200 sq. m |
12 men 8 men |
60 40 |
He has an area of 7600 sq. m available and 72 skilled men who can operate the machines.
How many machines of each type should he buy to maximize the daily output?
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Solution
Let x machines of type A and y machines of type B were purchased.
Number of machines cannot be negative.
Therefore,
\[x, y \geq 0\]
We are given,
| Area occupied by the machine |
Labour force for each machine |
Daily output in units |
|
| Machine A Machine B |
1000 sq. m 1200 sq. m |
12 men 8 men |
60 40 |
The area of 7600 sq m is available and there are 72 skilled men available to operate the machines.
Therefore, the constraints are
\[1000x + 1200y \leq 7600\]
\[\text{ and } 12x + 8y \leq 72\]
Total daily output = Z = \[60x + 40y\]\
which is to be maximised.
Thus, the mathematical formulation of the given linear programming problem is
Max Z = \[60x + 40y\]
subject to
\[1000x + 1200y \leq 7600\]
\[12x + 8y \leq 72\]
First we will convert inequations into equations as follows :
1000x + 1200y = 7600, 12x + 8y = 72, x = 0 and y = 0
Region represented by 1000x + 1200y ≤ 7600:
The line 1000x + 1200y = 7600 meets the coordinate axes at \[A_1 \left( \frac{38}{5}, 0 \right)\] and \[B_1 \left( 0, \frac{19}{3} \right)\] respectively. By joining these points we obtain the line
1000x + 1200y = 7600. Clearly (0,0) satisfies the 1000x + 1200y = 7600. So, the region which contains the origin represents the solution set of the inequation 1000x + 1200y ≤ 7600.
Region represented by 12x + 8y ≤ 72:
The line 12x + 8y = 72 meets the coordinate axes at C1(6, 0) and D1(0, 9) respectively. By joining these points we obtain the line 12x + 8y = 72 .Clearly (0,0) satisfies the inequation 12x + 8y ≤ 72. So,the region which contains the origin represents the solution set of the inequation 12x + 8y ≤ 72.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 1000x + 1200y ≤ 7600, 12x+ 8y ≤ 72, x ≥ 0, and y ≥ 0 are as follows.
The corner points are O(0, 0)
| Corner point | Z= 60x + 40y |
| O | 0 |
| B1 | 253.3 |
| E1 | 360 |
| C1 | 360 |
The maximum value of Z is 360 which is attained at E1(4, 3) and C1(6, 0).
Thus, the maximum output is Rs 360 obtained when 4 units of type A and 3 units of type B or 6 units of type A are manufactured.
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