Answer :

The given data can be shown in a table as follows:



Now, let the required weekly production of gadgets A and B be x and y respectively


As it is given that profit on each gadget A is Rs. 30 and that on B is Rs. 20, so profit on x and y number of gadgets A and B respectively are 30x and 20y.


If z = total profit then we have,


z = 30x + 20y


Further, it is also given that the production of A and B requires 10 hours per week and 6 hours per week in the foundry. Also, the maximum capacity of the foundry is given as 1000 hours.


Now, x units of A and y units of B will require 10x + 6y hours


So, we have


10x + 6y ≤ 1000


This is our first constraint.


Given, production of one unit gadget A requires 5x hours per week and y units of gadget B requires 4y hours per week, but the maximum capacity of the machine shop is 600 hours per week.


So, 5x + 4y ≤ 600


This is our second constraint.


Hence, the mathematical formulation of LPP is:


Find x and y which will maximize z = 30x + 20y


Subject to constraints,


10x + 6y ≤ 1000


5x + 4y ≤ 600


and also, as production cannot be less than zero, so x, y ≥ 0


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