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Asked by Shashikant Last Modified
Let x is no. of rackets and y is no. of bats. We need to maximise the profit by producing more rackets than bats since profit on racket is more than that on bat.
But the constraint is on both m/c time and craftsman time, i.e 1.5x + 3y <= 42 and 3x +y <= 24, By solving this inequality, we see that maxm. no. of rackets and bats can be made as 4 and 12 respectively.
Hence the maxm. profit = 4 X 20 + 12 X 10 = Rs. 200
read lessVishaka Bhutia
Home Tutor
(i) Let the number of rackets and the number of bats to be made be x and y respectively.
The machine time is not available for more than 42 hours.
The craftsman’s time is not available for more than 24 hours.
The factory is to work at full capacity. Therefore,
1.5x + 3y = 42
3x + y = 24
On solving these equations, we obtain
x = 4 and y = 12
Thus, 4 rackets and 12 bats must be made.
(i) The given information can be complied in a table as follows.
Tennis Racket | Cricket Bat | Availability | |
Machine Time (h) | 1.5 | 3 | 42 |
Craftsman’s Time (h) | 3 | 1 | 24 |
∴ 1.5x + 3y ≤ 42
3x + y ≤ 24
x, y ≥ 0
The profit on a racket is Rs 20 and on a bat is Rs 10.
The mathematical formulation of the given problem is
Maximize … (1)
subject to the constraints,
1.5x + 3y ≤ 42 … (2)
3x + y ≤ 24 … (3)
x, y ≥ 0 … (4)
The feasible region determined by the system of constraints is as follows.
The corner points are A (8, 0), B (4, 12), C (0, 14), and O (0, 0).
The values of Z at these corner points are as follows.
Corner point | Z = 20x + 10y | |
A(8, 0) | 160 | |
B(4, 12) | 200 | → Maximum |
C(0, 14) | 140 | |
O(0, 0) | 0 |
Thus, the maximum profit of the factory when it works to its full capacity is Rs 200.
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