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Question 1123220: A western shop wishes to purchase 300 felt and 200 straw cowboy hats. Bids have been received from the wholesalers. Beta Hatters has agreed to supply not more than 200 hats, Kappa Hatters not more than 250, Delta Hatters nor more than 150. The owner of the shop has estimated that his profit per hat sold for Beta Hatters would be P30/felt and P40/straw, from Kappa Hatters P38/felt and P35/straw, and from Delta Hatters P40/felt and P36/straw. Set up a linear programming to maximize the owner’s profit.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i solved this as follows:
a = number of felt hats from beta
b = number of straw hats from beta
c = number of felt hats from kappa
d = number of straw hats from kappa
e = number of felt hats from delta
f = number of straw hats from delta
your objective function becomes:
profit = 30a + 40b + 38c + 35d + 40e + 36f
30a + 40b is the profit from beta (felt + straw)
38c + 35d is the profit from kappa (felt + straw)
40e + 36f is the profit from delta (felt + straw)
your constraints are:
a + c + e = 200 because 200 felt hats are required in total.
b + d + f = 300 because 300 straw hats are required in total.
a + b <= 200 because that's the most hats beta can supply.
c + d <= 250 because that's the most hats kappa can supply.
e + f <= 150 because that's the most hats delta can supply.
a,b,c,d,e,f >= 0 because number of hats must be greater than or equal to 0.
i put these requirements into the following linear programming softwarethat uses the simplex method.
http://www.zweigmedia.com/RealWorld/simplex.html
it told me the optimum solution was:
p = 19400; a = 0, b = 200, c = 50, d = 100, e = 150, f = 0
that's profit = 19400 when:
a = 0 (felt hats by beta)
b = 200 (straw hats by beta)
c = 50 (felt hats by kappa)
d = 100 (straw hats by kappa)
e = 150 (felt hats by delta)
f = 0 (straw hats by delta)
total straw = 300 as required.
total felt = 200 as required.
here's a picture of the setup for the software used.
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