A company is making two products A and B. The cost of producing one unit of products A and B are Rs 60 and Rs 80 respectively. As per the agreement, the company has to supply at least 200 units of product B to its regular customers. One unit of product A requires one machine hour whereas product B has machine hours available abundantly within the company. Total machine hours available for product A are 400 hours. One unit of each product A and B requires one labour hour each and total of 500 labour hours are available. The company wants to minimize the cost of production by satisfying the given requirements. Formulate the problem as a LPP.

Matching exercise

Let production of product A be x units and of be B y units.
Given,
Profit on 1 unit of product A = Rs.60
Profit on 1 unit of product B = Rs.80
So, profit on x units of A and y units of B is 60x and 80y respectively.
Let z = total profit,
So, we have
z = 60x + 80y
Given, a minimum supply of product B is 200
So, y ≥ 200 (First constraint)
Given that, production of one unit of product A requires 1 hour of machine hours, so x units of product A require x hours but total machine time available for product A is 400 hours
So, x ≤ 400 (Second constraint)

Given, each unit of product A and B requires one hour of labour hour, so x units of product A require x hours and y units of product B require y hours of labour hours, but total labour hours available is 500 so
x + y ≤ 500 (Third constraint)
Hence, mathematical formulation of LPP is,
Find x and y which
Minimize z = 60x + 80y
Subject to constraints,
y ≥ 200
x ≤ 400
x + y ≤ 500
and also, as production cannot be less than zero, so x, y ≥ 0