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Question
A rubber company is engaged in producing three types of tyres A, B and C. Each type requires processing in two plants, Plant I and Plant II. The capacities of the two plants, in number of tyres per day, are as follows:
| Plant | A | B | C |
| I | 50 | 100 | 100 |
| II | 60 | 60 | 200 |
The monthly demand for tyre A, B and C is 2500, 3000 and 7000 respectively. If plant I costs Rs 2500 per day, and plant II costs Rs 3500 per day to operate, how many days should each be run per month to minimize cost while meeting the demand? Formulate the problem as LPP.
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Solution
Let plant I be run for x days and plant II be run for y days
Then,
| Tyres | Plant I (x) | Plant II (y) | Demand |
| A | 50 | 60 | 2500 |
| B | 100 | 60 | 3000 |
| C | 100 | 200 | 7000 |
Minimum demand for Tyres A,B and C is 2500, 3000 and 7000 respectively.The demand can be more than the minimum demand.
Therefore,the inequations will be
\[50x + 60y \geq 2500\]
\[100x + 60y \geq 3000\]
\[100x + 200y \geq 7000\]
Also, the objective function is Z = 2500x + 3500y
Hence, the required LPP is as follows:
Minimise Z = 2500x + 3500y
subject to
\[50x + 60y \geq 2500\]
\[100x + 60y \geq 3000\]
\[100x + 200y \geq 7000\]
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