Question 1139813: An animal breeder can buy four types of food for Vietnamese pot-bellied pigs. Each case of Brand A contains 25 units of fiber, 30 units of protein, and 30 units of fat. Each case of Brand B contains 100 units of fiber, 80 units of protein, and 70 units of fat. Each case of Brand C contains 275 units of fiber, 210 units of protein, and 190 units of fat. Each case of Brand D contains 100 units of fiber, 80 units of protein, and 60 units of fat. How many cases of each brand should the breeder mix together to obtain a food that provides 3950 units of fiber, 3060 units of protein, and 2740 units of fat?
Let x represent the number of cases of Brand A, y represent the number of cases of Brand B, z represent the number of cases of Brand C, and w represent be the number of cases of Brand D. There are four ways in which the breeder can mix brands to obtain a food that provides 3950 units of fiber, 3060 units of protein, and 2740 units of fat.
If w=0, the solution is
If w=1, the solution is
If w=2, the solution is
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
There are constraint equations for fiber, protein, and fat.
fiber: 
protein: 
fat: 
That's a system of three equation in four unknowns; it will have (if the problem is well formulated) a family of solutions in non-negative integers.
Solving the system algebraically would be very tedious. A solution using matrices on a graphing calculator produces the following set of equations relating w, x, y, and z:
x-w = 0
y+4w = 12
z-w = 10
Use those equations to express x, y, and z in terms of w; then substitute the values shown for w to find the different ways the breeder can get the right amounts of fiber, protein, and fat.
Note: The problem as you state it says there are four ways to get the right amounts, but it seems to be asking for only the solutions where w is 0, 1, or 2. There is in fact a fourth value of w that provides a solution.
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