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Pet Supplies Company produces 16-ounce cans of dog food by combining meat by-products, which cost $0

Math Sep 24, 2020

Pet Supplies Company produces 16-ounce cans of dog food by combining meat by-products, which cost $0.60 per pound, and chicken by-products, which cost $0.35 per pound. Meat by-products are 55% protein and 30% fat by weight, while chicken by-products are 40% protein and 10% fat by weight. To meet customer expectations, the final product should contain at least 50% protein and between 15 and 25% fat by weight.

a) Formulate a linear programming solution to be used to determine what the composition of the dog food should be to meet the various requirements at the minimum cost.

b) Using linear programming, find the optimal solution for the objective function and the optimal value of the variables.

Expert Solution

 We know, one pound is 16 ounces. Suppose the 16-ounce can of dog food is
combined with x pounds of meat by-products and y pounds of chicken by-products.
Then the objective function is z=0.6x+0.35y, which is the cost of the can.
Now we discuss the restrictions of x and y.
1. The weight of the can is 16 ounces or 1 pound. So x+y=1
2. The total protein in the can is 55%+40%y=0.55x+0.4y>=0.5*1=0.5, since the final
product should contain at least 50% protein.
3. The total fat in the can is 30%x+10%y=0.3x+0.1y, and it should between 15% and
25% fat by weight. So 0.3x+0.1y>=0.15 and 0.3x+0.1y<=0.25
4. Of course we must have x>=0 and y>=0.
Therefore, we can formulate a linear programming problem.
Minimize the objective function z=0.6x+0.35y
with respect to the constraints
(1) x+y=1
(2) x>=0, y>=0
(3) 0.55x+0.4y>=0.5
(4) 0.3x+0.1y>=0.15
(5) 0.3x+0.1y<=0.25
(b) Now we solve the above linear programming problem.
From (1), we get x=1-y. We plug it into (3),(4),(5)
In (3), we have 0.55(1-y)+0.4y>=0.5, then y<=1/3
In (4), we have 0.3(1-y)+0.1y>=0.15, then y<=3/4
In (5), we have 0.3(1-y)+0.1y<=0.25, then y>=1/4
Then we get 1/4<=y<=1/3
So z=0.6x+0.35y=0.6(1-y)+0.35y=0.6-0.25y
Thus when y=1/3, we obtain the minimum cost z=$0.52
At this time, x=2/3 pounds, y=1/3 pounds.

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