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15. A factory manufacturers 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 screw A, while it takes 6 minutes on automatic and 3 minutes on 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 ?. 7 and screws B at a profit of ?. 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce a day in order to maximize his profit? Determine the maximum profit.

0 2 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

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    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    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

    The profit on a package of screws A is Rs 7 and on the package of screws B is Rs 10. Therefore, the constraints are

    4x + 6y ≤ 240

    6x + 3y ≤ 240

    Total profit, Z=7x +10y

    The mathematical formulation of the given problem is

    Maximize Z=7x +10y …(1)

    subject to the constraints,

    4x + 6y ≤ 240      ....(2)

    6x + 3y ≤ 240     .....(3)

    x, y ≥ 0 … (4)

    The feasible region determined by the system of constraints is

    The corner points are A (40, 0), B (30, 20), and C (0, 40).

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

    The maximum value of Z is 410 at

    ...more

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