Answer :

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


x and y ≥ 0


The tabular form of the given data in the question is



The profit on a package of screws A is Rs 7 and on package of screws B is Rs 10.


the constraints are


4x+ 6y ≤ 240


And 6x + 3y ≤ 240 and x, y ≥ 0


And total profit is Z = 7x+ 10y


The mathematical formulation of the data is


Maximizing Z = 7x+ 10y


Subject to constraints


4x+ 6y ≤ 240


And 6x + 3y ≤ 240 and x, y ≥ 0


The feasible reason is determined by the system of constraints is:



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


The value of Z at these points is



The maximum value of Z is at point B(30, 20)


Therefore factory should produce 30 packages of screw A and 20 packages of screw B to get the maximum profit of Rs 410.


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