the standard weight of a special purpose brick is 5kg and it must contain two basic ingredients B1 and B2.B1 costs rs.5/kg and B2 costs rs.8/kg . strength consideration dictate that the brick should conatin not more than 4kg of B1 and minimum cost of brick satisfying the above condition . formulate this situation as an LPP and solve it.
This is a total calculation question and here is the help regarding equation formation and try to solve it for your own practice.
LP Formulation:
Let the amount of B1 be x (in kg)
Let the amount of B2 be y (in kg)
Minimize 5x + 8y
Subject To:
x + y = 5 (since the weight of the brick must be 5 kg)
x ≤ 4 (since there must be no more than 4 kg of B1)
y ≥ 2 (since there must be a min. of 2 kg of B2)
x ≥ 0
The problem now asks you to find the minimum cost of the brick graphically. So now you'll have to graph each constraint listed above to find the corner points of the feasible region, and then determine which corner point offers you the optimal solution.
the minimum cost is: 31
With 3 kg of B1, and 2 kg of B2.
LP Formulation:
Let the amount of B1 be x (in kg)
Let the amount of B2 be y (in kg)
Minimize 5x + 8y
Subject To:
x + y = 5 (since the weight of the brick must be 5 kg)
x ≤ 4 (since there must be no more than 4 kg of B1)
y ≥ 2 (since there must be a min. of 2 kg of B2)
x ≥ 0
The problem now asks you to find the minimum cost of the brick graphically. So now you'll have to graph each constraint listed above to find the corner points of the feasible region, and then determine which corner point offers you the optimal solution.
the minimum cost is: 31
With 3 kg of B1, and 2 kg of B2.