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.