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, 90 units of protein, and 80 units of fat.
Each case of Brand C contains 275 units of fiber, 240 units of protein, and 230 units of fat
Each case of Brand D contains 200 units of fiber, 180 units of protein, and 140 units of fat.
How many cases of each brand should the breeder mix together to obtain a food that provides 3700 units of fiber, 3270 units of protein, and 3040 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 3700 units of fiber, 3270 units of protein, and 3040 units of fat.
This is a mixture problem and a hard one. You need to find the percentage of each nutrient for each brand.
BRAND A -- represented as x
Find percentage of each nutrient:
FIBER: 25/(25+ 30 + 30) = 25/85
Protein: 30/(25+ 30 + 30) = 30/85
FAT: 30/(25+ 30 + 30) = 30/85
BRAND B -- represented as y
Find percentage of each nutrient:
FIBER: 100/(275 + 240 + 230) = 100/270
Protein: 90/(275 + 240 + 230) = 90/270
FAT: 80/(275 + 240 + 230) = 80/270
BRAND C -- represented as z
Find percentage of each nutrient:
FIBER: 275/(275 + 240 + 230) = 275/745
Protein: 240/(275 + 240 + 230) = 240/745
FAT: 230/(275 + 240 + 230) = 230/745
BRAND D -- represented as w
Find percentage of each nutrient:
FIBER: 200/(200 + 180 + 140) = 200/520
Protein: 180/(200 + 180 + 140) = 180/520
FAT: 140/(200 + 180 + 140) = 140/520
SET UP YOUR EQUATIONSFIBER:
(25/85)x + (100/270)y + (275/745)z + (200/520)w = 3700
PROTEIN:
(30/85)x + (90/270)y + (240/745)z + (180/520)w = 3270
FAT:
(30/85)x + (80/270)y + (230/745)z + (140/520)w = 3040
Total:
x + y + z + w = (3700 + 3270 + 3040)
Now SOLVE, you should get more than 1 solution though!
\(w=r_1\)
\(x=\frac{17r_1}{104}\)
\(y=\frac{-(135r_1-210600)}{52}\)
\(z=\frac{(149r_1+619840)}{104}\)
A similar question with slightly different numbers solution is shown below
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