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in Linear Programming by (27.7k points)
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A small manufacture has employed 5 skilled men and 10 semiskilled men and makes an article in two qualities, a deluxe model and an ordinary model. The making of a deluxe model requires 2hours work by a skilled man and 2hours work by a semiskilled man. The ordinary model requires 1 hour by a skilled man and 3 hours by a semiskilled man. By union rules, no man can work more than 8 hours per day. The manufacture gains ₹15 on the deluxe model and ₹10 on the ordinary model. How many of each type should be made in order to maximize his total daily profit? Also, find the maximum daily profit.

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Best answer

Let x and y be number of deluxe article manufactured and ordinary article manufactured. 

∴According to the question, 

2x + y ≤ 40, 2x + 3y ≤ 80, x ≥ 0, y ≥ 0

Maximize Z = 15x + 10y 

The feasible region determined by 2x + y ≤ 40, 2x + 3y ≤ 80, x ≥ 0, y ≥ 0 is given by

The corner points of feasible region are A(0,0) , B(0,80/3) , C(10,20),D(20,0). 

The value of Z at corner points are

Corner Point Z = 15x + 10y
A(0, 0) 0
B(0, 80/3) 266.67
C(10, 20) 350 Maximum
D(20, 0) 300

The maximum value of Z is 350 at point (10,20). 

Hence, the manufacturer should produce 10 types of deluxe article and 20 types of ordinary article to make maximum profit of Rs.350.

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