The South Fork Feed Company makes a feed mix from four ingredients -- oats, corn

Question

The South Fork Feed Company makes a feed mix from four ingredients -- oats, corn, soybeans and a vitamin supplement. The company has 300 pounds of oats, 400 pounds of corn, 200 pounds of soybeans, and 100 pounds of vitamin supplement available for the mix. The company has the following requirements for the mix: ?At least 30% of the mix must be soybeans. ?At least 20% of the mix must be the vitamin supplement. ?The ratio of corn to oats cannot exceed 2 to 1. ?The amount of oats cannot exceed the amount of soybeans. ?The mix must be at least 500 pounds. A pound of oats costs $0.50; a pound of corn, $1.20; a pound of soybeans, $0.60; and a pound of the vitamin supplement, $2.00. The feed company wants to know the number of pounds of each ingredient to put in the mix in order to minimize cost. a. Formulate a linear programming model for this problem. b. Solve the model by using the computer.

Explanation / Answer

v = vitamins, s = soybeans, c = corn, t = oats (since o makes a lousy variable)

v+s+c+t = 500 ; total mix requirement
v = .2 (v+s+c+t) ; 20% requirement
Combine these equations for
v = 100
but v = 100 ; ingredient limitation
so v = 100
Since v = .2 (v+s+c+t)
100 = .2 (v+s+c+t)
500 = (v+s+c+t) = 500 so
v+s+c+t = 500 and
s + c + t = 400

150 = s = 200 ; 30% requirement and ingredient limitation
t = s ; so t = 200
c = 2t

(Since there is no lower limit on corn, and it's the most expensive ingredient behind vitamins, it's pretty obvious that we need 200 soy and 200 oats to go with the 100 vitamins, but I'll try to make a more systematic approach...)

I know that you need to evaluate C = 2v + .6s + 1.2c + .5t at the vertices of the space delimited by the above constraints, but I'm not sure how to systematically find the vertices. The following is an ad hoc attempt:
1) Min s = 150, then max t = 150, c = 100
2) min s = 150, then max c = 166, t = 84
3) max s = 200, then max t = 200, c = 0
4) max s = 200, then max c = 133, t = 67
5) max t = 200 (same as 3)
6) min t = 67 (same as 4)
7) max c = 166 (same as 2)
8) min c = 0 (same as 3)

So cases 1 - 4 are your extreme points; the minimum and maximum costs can be found by evaluating the objective function C = .... at those points.

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