Answer :
Let x units of item A and y units of item B be manufactured. Therefore, x, y 0.
As we are given,
Machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day.
According to the question, the constraints are
x + 2y 12
2x + y 12
x + y
5
He makes a profit of Rs 6.00 on item A and Rs. 4.00 on item B. Profit made by him in producing x items of A and y items of B is 6x + 4y.
Total profit Z = 6x + 4y which is to be maximized
Thus, the mathematical formulation of the given linear programming problem is
Max Z = 6x + 4y, subject to
x + 2y 12
2x + y 12
x + y
5
x, y 0
First, we will convert the inequations into equations as follows:
x + 2y = 12, 2x + y = 12, x + y = 5, x = 0 and y = 0.
The region represented by x + 2y 12
The line x + 2y = 12 meets the coordinate axes at A(12,0) and B(0,6) respectively. By joining these points, we obtain the line x + y = 12. Clearly (0, 0) satisfies the x + 2y = 12. So, the region which contains the origin represents the solution set of the inequation x + 2y 12
The region represented by 2x + y 12
The line 2x + y = 12 meets the coordinate axes at C(6,0) and D(0,12) respectively. By joining these points, we obtain the line 2x + y = 12. Clearly (0, 0) satisfies the 2x + y = 12. So, the region which contains the origin represents the solution set of the inequation 2x + y 12
The region represented by x + y
5
The line x + y
5 meets the coordinate axes at E(5,0) and F(0,4) respectively. By joining these points, we obtain the line x +
y = 5. Clearly (0, 0) satisfies the x +
y
5. So, the region which does not contain the origin represents the solution set of the inequation x +
y
5
The region represented by x 0, 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
x + 2y 12, 2x + y
12, x +
y
5, x, y
0 are as follows.
Thus the maximum profit is of Rs 40 obtained when 4 units each of item A and B are manufactured.
The corner points are D(0,6), I(4,4), C(6,0), G(5,0), and H(0,4). The values of Z at these corner points are as follows:
The maximum value of Z is 40 which is attained at I(4, 4).
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