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*A SMALL MANUFACTURER HAS EMPLOYED 5 SKILLED N 10 SEMI SKILLED MEN AND MAKES AN ARTICLE IN 2 QUALITIES , A DELUXE MODEL AND AN ORDINARY MODEL. THE MAKING OF A DELUX MODEL REQUIRES 2 HR OF WORK BY A SKILLED LABOUR N 2H OF WORK OF SEMI SKLLED MAN. THE ORDINARY MODL REQUIRES 1H BY SKILLED N 3 H BY SEMI SKILLED . ACCORDING TO UNION RULE NO MAN MAY WORK FOR MORE THAN 8H PER DAY THE MANUFACTURER GAINS RS15ON DELUX N RS 10 ON ORDINARY .. IN ORDER TO MAXIMISE PROFIT HOW MANY OF EACH TYPE SHOULD BE MADE*

Let the company produce x deluxe model and y ordinary model article

Hence, the total time spent by 5 skilled men = 2x+y and it should be less than 8hours/person

Hence, 2x+y $\le $40 (1)

Also, the total time spent by 10 semi-skilled men = 2x+3y and it should be less than 8hours/person

Hence, 2x+3y $\le $80 (2)

Also, $\ge $x0 (3) and y$\ge $0 (4)

Profit = 15x+10y

We need to maximise profit given the constraints. So, for that draw all the 4 lines on the graph and check the common area, you will form a hexagon OABC with A(0,80/3); B(10,20); C(20,0); O(0,0)

Since the feasible region is a bound region, we can check the profit function at all the vertices to find the point of maxima

At point A: 15(0)+10(80/3) = 800/3

At point B: 15(10)+10(20) = 350

At point C: 15(20)+10(0) = 300

At point 0: 15(0)+10(0) = 0

So, the maxima lies at point C with x =10and y = 20 and the maximum profit = Rs 350.

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