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.

Asked by Sakshi | 1 year ago | 99

Now, let x and y be the required days of plant I and II to minimize cost.

Given:

Plant I costs per day = Rs 2500

So, cost to run plant x requires per month = Rs 2500x

Plant II costs per day = Rs 3500

So, cost to run plant y requires per month = Rs 3500y

Let total cost per month be ‘Z’

So, Z = 2500x + 3500y

First Constraint:

Given:

Production of type A from plant I requires = 50

Production of type A from plant II requires = 60

So,

x unit of production of type A from plant I requires = 50x

y unit of production of type A from plant II requires = 60y

Total demand of type A per month = 2500

i.e., 50x + 60y ≥ 2500

Second Constraint:

Given:

Production of type B from plant I requires = 100

Production of type B from plant II requires = 60

So,

x unit of production of type B from plant I requires = 100x

y unit of production of type B from plant II requires = 60y

Total demand of type B per month = 3000

i.e., 100x + 60y ≥ 3000

Third Constraint:

Given:

Production of type C from plant I requires = 100

Production of type C from plant II requires = 200

So,

x unit of production of type C from plant I requires = 100x

y unit of production of type C from plant II requires = 200y

Total demand of type C per month = 7000

i.e., 100x + 200y ≥ 7000

Hence, the required mathematical formulation of linear programming is:

Minimize Z = 2500x + 3500y

Subject to

50x + 60y ≥ 2500

100x + 60y ≥ 3000

100x + 200y ≥ 7000

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