A breed of cattle needs at least 10 proteins and 8 fats units per day. Feed type I provides 6 protein and 2 fat units at $4 per

Let x represent the amount of feed type I and y the amount of feed type II to be utilized. Our goal is to minimize the following:

C = 4x + 3y

while adhering to these constraints:



Analyzing the graph of these four constraints yields corner points at (0, 5), (1, 2), and (4, 0).

Testing the objective function at each corner point to find the minimum:

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.

Thus, the combination that results in the lowest cost is one bag of feed type I and two bags of feed type II.