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Question

A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.

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Solution

Consider the manufacturer produce

Tabulate the given data as,

Screw A | Screw B | Availability | |

Automatic machine (min) | 4 | 6 | |

Hand Operated machine (min) | 6 | 3 | |

The required constraints are,

The objective function (profit) which needs to maximize is,

The line

x | 0 | 60 |

y | 40 | 0 |

Also, when

This is true, so the graph have the shaded region towards the origin.

The line

x | 0 | 40 |

y | 80 | 0 |

Also, when

This is true, so the graph have the shaded region towards the origin.

By the substitution method, the intersection points of the lines

Plot the points of all the constraint lines,

It can be observed that the corner points are

Substitute these points in the given objective function to find the maximum value of Z.

Corner points | |

| 280 |

| 410 (Maximum) |

| 400 |

The maximum value of Z is 410 at the point

Thus, 30 packages of screw A and 20 packages of screw B should be produced each day to get the maximum profit of 410.

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