Question

A factory manufactures two products, each requiring the use of three machines. The first machine can be used at most 70 hours; the second machine at most 40 hours; and the third machine at most 90 hours. The first product requires 2 hours on machine 1, 1 hour on machine 2, and 1 hour on machine 3; the second product requires 1 hour on machines 1 and 2 and 3 hours on machine 3. The profit is $40 per unit for the first product and $60 per unit for the second product. Write a LOP that will compute how many units of each product should be manufactured in order to maximize profit

Answer 1

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At Eve Y=90 point ale bolesent A10,50) As c (30.10 130.1907 Hei Ey=3 TX we (Yoylsy + (60425) F1457 * I think that manufacture poroduce x unit of Product and Ý unit of second product most Mecording to the data Fioist machine can be used 3rd machine 90 Hows. Second machine at most unhours At most do Howrs Fisist $70 per unit product of the profit is and 9. 60 per unit sketch- (x= 0, y: 301 point c can be determine Ey=2 357 सण 0135.0) by solving eqn (x=30, V=10) point 8 can be determine GI solving (X-15, y = 15) can solve | xt1y = 40 1x + 3y = 90 Obe Tective to max. OF profit 7-1407X) + 100xY) nubtect to nexource construgined are Fallow fatlysto o 1x + 1Y 240 -@) 1 x+3y = 90 In order to Find put the level of production OF of and Y marimize Her profit - 2y = -50 TY=as (x= 35, =o) There Fore profit at different equillorium TA = 140X ) + ( 60X30) = 1000 TC (40 x 30) + (60X70) = 1000 = ITP lfoo (40x95) + (6040) poofit is max i mize At point o profit = 2100 - = 2100 IT B =