A chemical company produces two products, X & Y. Each unit of product X requires 3 hours on operation I and 4 hours on operation II, while each unit of product Y requires 4 hours on operation I and 5 hours on operation II. Total available time for operations I & II is 20 hours and 26 hours respectively. The production of each unit of product Y also results in two units of a by-product Z at no extra cost. - Product X sells at a profit of Rs. 10/unit, while Y sells at a profit of Rs. 20/unit. By-product Z brings a unit profit of Rs. 6 if sold; in case it is not sold, the destruction cost isRs. 4/unit. Forecasts indicate that not more than 5 units of Z can be sold. Formulate the LP model to determine the quantities of X and Y to be produced, keeping Z in mind, so that the profit earned in maximum.

We have to maximize the profit that the company will earn.

Let x, y and z be the number of units of products X, Y and Z sold, while z’ be the number of units of product Z destroyed.

Maximization function: Maximize Z = 10x + 20y + 6z – 4z’

Constraints:

Max. 20 hours on operation I: 3x + 4y ≤ 20

Max. 26 hours on operation II: 4x + 5y ≤ 26

Max. 5 units of z sold: z ≤ 5

Production of Y produces 2 units of Z: y = z + z’ → y – z – z’ = 0