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

Gadget |
Foundry |
Machine-shop |

A | 10 | 5 |

B | 6 | 4 |

Firm's capacity per week | 1000 | 600 |

The profit on the sale of *A* is Rs 30 per unit as compared with Rs 20 per unit of *B*. The problem is to determine the weekly production of gadgets *A* and *B*, so that the total profit is maximized. Formulate this problem as a LPP.

#### Solution

Let *x* and *y* number of gadgets *A* and *B* respectively being produced in order to maximize the profit.

Since, each unit of gadget* A* takes 10 hours to be produced by machine *A* and 6 hours to be produced by machine *B* and each unit of gadget* B* takes 5 hours to be produced by machine *A* and 4 hours to be produced by machine *B*.

Therefore, the total time taken by the Foundry to produce *x* units of gadget A and *y *units of gadget B is

Hence, 10

*x*

*+*

*6*

*y*

*≤ 1000*

*.*

This is our first constraint.

The total time taken by the machine-shop to produce

*x*units of gadget

*A*and

*y*units of gadget

*B*is 5

*x*+ 4

*y*. This must be less than or equal to the total hours available.

Hence, 5

*x*+ 4

*y*

*≤ 600*

This is our second constraint.

Since

*x*and

*y*are non negative integers

*,*therefore

*A*is Rs 30 per unit as compared with Rs 20 per unit of

*B*. Therefore, profit gained on

*x*and

*y*number of gadgets A and

*B*is Rs 30

*x*and Rs 20

*y*respectively.

Let Z denotes the total cost

Therefore, Z= Rs (30

*x*

*+ 20*

*y*)

Hence, the above LPP can be stated mathematically as follows:

Maximize Z = 30

*x*

*+ 20*

*y*

subject to

10

*x*

*+ 6*

*y*

*≤ 1000*

*,*

5

*x*+ 4

*y*

*≤ 600*

*x, y ≥*0