Answer :
The animal feed producer makes two types of grain, A and B. Each unit of grain A contains 2 grams of fat, 1 gram of protein, and 80 calories. Each unit of grain B contains 3 grams of fat, 3 grams of protein, and 60 calories.
Suppose that the producer wants each unit of the final product to yield at least 18 grams of fat, at least 12 grams of protein, and at least 480 calories.
If each unit of A costs 10 cents and each unit of B costs 12 cents, how many units of each type of grain should the producer use to minimize the cost?
First, let x be the number of units of grain A and y be the number of units of grain B, which are used to minimize the cost of the feed.
Let the function C(x, y) denote the cost of producing x units of grain A and y units of grain B.C(x,y) = 10x + 12y
where each unit of A costs 10 cents, and each unit of B costs 12 cents. The producer wants each unit of the final product to yield at least 18 grams of fat, at least 12 grams of protein, and at least 480 calories. Each unit of grain A contains 2 grams of fat, 1 gram of protein, and 80 calories; therefore, x units of grain A contain 2x grams of fat, x grams of protein, and 80x calories.
Similarly, y units of grain B contain 3y grams of fat, 3y grams of protein, and 60y calories.
Therefore, the following inequalities must be satisfied:2x + 3y >= 181x + 3y >= 12 80x + 60y >= 480 We use the graphing technique to solve this problem by finding the feasible region and using a corner point method. From the above inequalities, we plot the following equations on a graph and find the feasible region.
2x + 3y = 18,1x + 3y = 12,80x + 60y = 480
This is a plot of the feasible region. Now we need to find the corner points of the feasible region and evaluate C(x, y) at each point.(0, 4), (4.5, 1.5), (6, 0), (0, 12), and (9, 0) are the corner points of the feasible region.
We use these points to compute the minimum cost.
C(0,4) = 10(0) + 12(4)
= 48,C(4.5,1.5)
= 10(4.5) + 12(1.5)
= 57,C(6,0)
= 10(6) + 12(0)
= 60,C(0,12)
= 10(0) + 12(12)
= 144,C(9,0) = 10(9) + 12(0) = 90
Therefore, the minimum cost is 48 cents, which is obtained when 0 units of grain A and 4 units of grain B are used. The producer should use 0 units of grain A and 4 units of grain B to minimize the cost of producing the feed.
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