Linear Programming

1Task One: Let x be the numeral of packets of Hi-Fibre produced individually twenty-four hour period. Let y be the frame of packets of Hi-Fibre Plus produced all(prenominal) day. linear Constraints: Time: There is one oven and one package vegetation that operates for 12 hours a day. The actual number of hours inevitable is 3x+5y seconds. thusly 3x+5y?12*60*60 => 3x+5y?43200 Material: The come frontwards of concentrated quality is limited to 120kg per day. A packet of Hi-Fibre inescapably 5g of concentrated fibre and a packet of Hi-Fibre Plus necessarily 10g of concentrated fibre. thence the actual amount of concentrated fibre used each day is : 5x+10y grams 120kg=120000g Therefore 5x+10y?120 000 securities industry pick out: The minimum number of each image of concentrated fibres packets that must be produced in each day is 2500 packets Therefore x ? 2500, y?2500 Storage: The storage d omain can abbreviate a maximum of 12000 packets so fooling production cannot slip by this figure.

Therefore x+y?12000 Objective Function: Let P be the wage in pence each day from the trade of Hi-Fibre and Hi-Fibre Plus produced in each day; where P is in pence Therefore P=12x+15y Graph: Vertices: A: the coordinate ar (2500,2500) B: is at the intersection of x=2500 and 3x+5y=432000 Therefore the coordinate of B be (2500,7140) C: is at the intersection of x+y=12000.a and 3x+5y=43200b a*3 3x+3y=36000c b-c 2y=7200 y=3600 Substituting in a x=84 00 Therefore the coordinate of C are (8400,! 3600) D: is at the intersection of y=2500 and x+y=12000 Therefore the coordinate of D are (9500,2500) Optimal effect: P(A)=12p*2500 + 15p*2500 =67500p P(B)=12p*2500+15p*7140=137100p P(C)=12p*8400+15p*3600=154800p P(D)=12p*9500+15p*2500=151500p Therefore the maximum profit each day is 154800p, the optimal number of Hi-Fibre and Hi-Fibre Plus produced per day is 8400 packets and 3600 packets. Task Two: a) If the optimal number of...If you want to get a full essay, put it on our website: OrderCustomPaper.com

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