A manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry and then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for production of A and B and the number of man-hours available for the firm is as follows :

Gadgets |
Foundry |
Machine shop |

A | 10 | 5 |

B | 6 | 4 |

Time available (hour) | 60 | 35 |

Profit on the sale of A is ₹ 30 and B is ₹ 20 per units. Formulate the LPP to have maximum profit.

#### Solution

Let the number of gadgets A produced by the firm be x and the number of gadgets B produced by the firm be y.

The profit on the sale of A is ₹ 30 per unit and on the sale of B is ₹ 20 per unit.

∴ total profit is z = 30x + 20y

This is a linear function that is to be maximized. Hence it is the objective function. The constraints are as per the following table:

Gadgets |
Foundry |
Machine shop |
Total available Time (in hour) |

A | 10 | 5 | 60 |

B | 6 | 4 | 35 |

From the table total man-hours of labour required for x units of gadget A and y units of gadget B in foundry is (10x + 6y) hours and total man-hours of labour required in machine shop is (5x + 4y) hours.

Since the maximum time available in foundry and machine shops are 60 hours and 35 hours respectively. Therefore, the constraints are 10x + 6y ≤ 60, 5x + 4y ≤ 35.

Since, x and y cannot be negative, we have x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as:

Maximize z = 30x + 20y, subject to

10x + 6y ≤ 60,

5x + 4y ≤ 35,

x ≥ 0, y ≥ 0