Somone tell me a realll funny joke... or riddle cuz im bored >.< ;))))))))
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Let the number of bags of feed type I to be used be x and the number of bags of feed type II to be used be y, then:
We are to minimize:
C = 4x + 3y
subject to the following constraints:
From the graph of the 4 constraints above, the corner points are (0, 5), (1, 2), (4, 0).
Testing the objective function for the minimum corner point we have:
For (0, 5):
C = 4(0) + 3(5) = $15
For (1, 2):
C = 4(1) + 3(2) = 4 + 6 = $10
For (4, 0):
C = 4(4) + 3(0) = $16.
Therefore, the combination that yields the minimum cost is 1 bag of type I feed and 2 bags of type II feed.
We are to minimize:
C = 4x + 3y
subject to the following constraints:
From the graph of the 4 constraints above, the corner points are (0, 5), (1, 2), (4, 0).
Testing the objective function for the minimum corner point we have:
For (0, 5):
C = 4(0) + 3(5) = $15
For (1, 2):
C = 4(1) + 3(2) = 4 + 6 = $10
For (4, 0):
C = 4(4) + 3(0) = $16.
Therefore, the combination that yields the minimum cost is 1 bag of type I feed and 2 bags of type II feed.