A factory manufactures two types of screws, *A* and *B*, each type requiring the use of two machines - an automatic and a hand-operated. It takes 4 minute on the automatic and 6 minutes on the hand-operated machines to manufacture a package of screws '*A*', while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a package of screws '*B*'. Each machine is available for at most 4 hours on any day. The manufacturer can sell a package of screws '*A*' at a profit of 70 P and screws '*B*' at a profit of Rs 1. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.

#### Solution

Let the factory manufacture *x* screws of type *A* and *y* screws of type *B* on each day. Therefore,

*x *≥ 0 and *y *≥ 0

The given information can be compiled in a table as follows.

Screw A |
Screw B |
Availability | |

Automatic Machine (min) | 4 | 6 | 4 × 60 = 240 |

Hand Operated Machine (min) | 6 | 3 | 4 × 60 = 240 |

\[4x + 6y \geq 240\]

The manufacturer can sell a package of screws '*A*' at a profit of Rs 0.7 and screws '*B*' at a profit of Re 1.

Total profit, Z = 0.7*x* + 1*y*

The mathematical formulation of the given problem is

Maximize Z = 0.7*x* + 1*y** *

subject to the constraints,

*x*,* y* ≥ 0

First we will convert inequations into equations as follows:

4*x* + 6*y* = 240, 6*x* + 3*y* = 240, *x* = 0 and *y* = 0

Region represented by 4*x* + 6*y* ≤ 240:

The line 4*x* + 6*y* = 240 meets the coordinate axes at *A*_{1}(60, 0) and *B*_{1}(0, 40) respectively. By joining these points we obtain the line 4*x* + 6*y* = 240. Clearly (0,0) satisfies the* *4*x* + 6*y* = 240. So,the region which contains the origin represents the solution set of the inequation 4*x* + 6*y* ≤ 240.

Region represented by 6*x* + 3*y* ≤ 240:

The line 6*x* + 3*y* = 240 meets the coordinate axes at* **C*_{1}(40, 0) and *D*_{1}(0, 80) respectively. By joining these points we obtain the line 6*x* + 3*y* = 240. Clearly (0,0) satisfies the inequation 6*x* + 3*y* ≤ 240. So,the region which contains the origin represents the solution set of the inequation 6*x* + 3*y* ≤ 240.

Region represented by *x* ≥ 0 and* y* ≥ 0:

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations *x* ≥ 0, and *y *≥ 0.

The feasible region determined by the system of constraints 4*x* + 6*y* ≤ 240, 6*x* + 3*y* ≤ 240, *x* ≥ 0, and *y* ≥ 0 are as follows.

The corner points are* **C*_{1}(40, 0), *E*_{1}(30, 20) and *B*_{1}(0, 40).

The values of Z at these corner points are as follows.

Corner point | Z = 7x + 10y |

C_{1}(40, 0) |
280 |

E_{1}(30, 20) |
410 |

B_{1}(0, 40) |
400 |

The maximum value of Z is 410 at (30, 20).

Thus, the factory should produce 30 packages of screws *A* and 20 packages of screws *B* to get the maximum profit of Rs 410.