A rubber company is engaged in producing three types of tyres A, B and C.

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