A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws ‘A’ while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a packet of screws ‘B’. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws ‘A’ at a profit of 70 paise and screws ‘B’ at a profit of

Question

A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws ‘A’ while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a packet of screws ‘B’. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws ‘A’ at a profit of 70 paise and screws ‘B’ at a profit of < 1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit? Formulate the above LPP and solve it graphically and find the maximum profit.

Accepted Answer

From the given information,Let x= the number of packets of screw ‘A’ manufactured in a dayy= the number of packets of screw ‘B’ manufactured in a day.Therefore, ≥0,≥0 ItemNumberMachine AMachine BProfitScrew Ax4 minutes6 minutes₹0.7Screw B y6 minutes3 minutes₹1Max. time available 4 hrs= 240 min.4 hrs= 240 min.  Then, the constraints are :4x+6y≤240 or 2x+3y≤1206x+3y≤240 or 2x+y≤80and total profit,Z=0.7x+ySo our LPP will beMax.Z=0.7x+ySubject to the constraints :2x+3y≤1202x+y≤80and x,y≥0Now, 2x+3y≤120 and 2x+y≤80 x060y400x040y800Now, let’s plte the points on the graph, from that we will get the feasible region OABC as shown (Shaded) below.Corner pointsValue of Z=0.7 x+yC(0,40)0.7(0)+40=40B(30,20)0.7(30)+20=41 maximumA(40,0)0.7(40)+0=28O(0,0)0.7(0)+0=0Notice that the maximum value of Z is 41 at B(30, 20)Therefore, the maximum profit is Rs. 41 which obtained when 30 packets of screw A and 20 packets of screw B produced.Which are required answers.

Corner points	Value of Z=0.7 x+y
$C (0, 40)$	$0 .7 (0) + 40 = 40$
$B (30, 20)$	$0 .7 (30) + 20 = 41 maximum$
$A (40, 0)$	$0 .7 (40) + 0 = 28$
$O (0, 0)$	$0 .7 (0) + 0 = 0$

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Item	Number	Machine A	Machine B	Profit
Screw A	x	4 minutes	6 minutes	$₹ 0.7$
Screw B	y	6 minutes	3 minutes	$₹ 1$
Max. time available		4 hrs= 240 min.	4 hrs= 240 min.