A farmer grows three crops, corn oats, and soybeans, which he mixes together to feed his cows and pigs. At least 40% of the feed mix for the cows must be corn. The feed mix for the pigs must contain at least twice as much soybeans as corn. He has harvested 1000 bushels of corns, 500 bushel of oats, and 1000 bushels of soybeans. He needs 1000 bushels of each feed mix for his livestock. The unused corn, oats and soybeans can sold for $4, $3.50, and $3,25 a bushel, respectively (thus, these amounts also represent the cost of the crops used to feed the livestock.)
How many bushels of each crop should be used in each feed mix in order to produce sufficient food for the livestock at minimal cost?
its a BIG M METHOD problem
i just need the SUBJECT TO CONSTRAINTS and the MAXIMIZE or MINIMIZE.. please
i cant get it totally how to solve it.. let me try little by little... for cows 40% of total feed must be corn... he need 1000 of each bushel makes 3000 in total.. so corn use is 40% of total x = 1200 bushels of corn.
ok... thanks for the effort. :)
let me know if i go some where wrong... now it says for pigs oats are use twice as much as corn. corn was used 1200 so oats would be 2400
sorry not oats .. was soyabeans
yeah..
seems like some information is missing from here
no, they are not.
still there?
yes still working on it...
ok does he use 3000 on his whole live stock i mean (cow +pigs) or just one of it?
ok. :)
just one of it
whole livestock, rather
means 3000 on cow + pigs?
yeah
i search this on net: min 4(corn_1+corn_2) + 3.5 (oats_1+oats_2) + 3.25 (soybean_1+ soybean_2), but i don't understand
lol i dont get this either..
but you know about big m method?
yes i kinda did it days back..
oh, ok
is it correct? maximize -P= -4x1+3.50x2+3.25x3?
how come u have got this.. ?
well,its my answer
hmm not even getting any near :( sorry
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