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Question
A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30, respectively. The company makes a profit of Rs 80 on each piece of type A and Rs 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?
Solution
Let x pieces of type A and y pieces of type B be manufactured per week.
Let Z denote the total profit. Then,
Z = 80x + 120y
Since each piece of A requires 9 labour hours of fabricating and each piece of B requires 12 labour hours of fabricating, the total labour hours required for fabrication are 9x + 12y. This must be less than or equal to the total hours available for fabrication. Hence,
9x + 12y ≤180
i.e. 3x + 4y ≤60 ... (i)
Also, each piece of A requires 1 labour hour for finishing and each piece of B requires 3 labour hours for finishing. Hence, the total labour hours required for finishing are 1x + 3y. But the maximum number of hours available for finishing is 30.
∴ 1x + 3y ≤30 ... (ii)
Since, the number of pieces cannot be negative, therefore,
x≥0, y≥0
Hence, the linear programming problem for the given problem is as follows:
Maximise Z = 80x + 120y, subject to the constraints
3x + 4y ≤60
x + 3y ≤30
and x≥0, y≥0
To solve this LPP graphically, we have to first convert the inequalities into equations and draw the corresponding lines.
The feasible region of the LPP is shaded in the figure. The coordinates of the corner points of the feasible region are A (0, 10), B (12, 6), C (20, 0).
The values of the objective function at these points are given in the following table:
| Point (x, y) | Value of objective function Z = 80x + 120y |
| A (0, 10) | Z = 80 × 0 + 120 × 10 = 1200 |
| B (12, 6) | Z = 80 × 12 + 120 × 6 = 1680 |
| C (20, 0) | Z = 80 × 20 + 120 × 0 = 1600 |
Clearly, Z is maximum at (12, 6). The maximum value of Z is 1680.
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