Linear Programming

1Task One:
Let x be the numeral of packets of Hi-Fibre produced distributively day.
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 plant 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 forth of concentrated quality is limited to 120kg per day. A packet of Hi-Fibre necessarily 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

Market pick out: The minimum number of each type of concentrated fibres packets that must be produced in each day is 2500 packets
Therefore
x ? 2500,
y?2500

Storage: The storage area can abbreviate a maximum of 12000 packets so daily production cannot slip by this figure.




Therefore
x+y?12000

Objective Function:
Let P be the profit 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 are (2500,2500)

B: is at the intersection of x=2500 and 3x+5y=432000
Therefore the coordinate of B are (2500,7140)

C: is at the intersection of
x+y=12000…………………….a
and
3x+5y=43200…………………b
a*3 3x+3y=36000…………………c
b-c 2y=7200
y=3600
Substituting in a
x=8400
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 Solution:
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, order it on our website: Ordercustompaper.com



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