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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