If w=0 the solution is: If w=1 the solution is: If w=2 the solution is: An anima
ID: 1719346 • Letter: I
Question
If w=0 the solution is:
If w=1 the solution is:
If w=2 the solution is:
An animal breeder can buy four of food for Vietnamese pot-bellied pigs. Each case of Brand A contains 35 units of fiber, 40 units of protein, and 40 units of fat. Each case of Brand B contains 140 units of fiber, 130 units of protein, and 120 units of fat. Each case of Brand C contains 245 units of fiber, 220 units of protein, and210 units of fat. Each case of Brand D contains 420 units of fiber, 390 units of protein, and 350 units of fat. How many cases of each brand should the breeder mix together to obtain a food that provides 4305 units of fiber, 3930 units of protein, and 3690 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 4305 units of fiber, 3930 units of protein, and 3690 units of fat. If w=0, the solution isExplanation / Answer
The given data tabulated as follows:
35x+140y+245 z +420 w=4305 or x+4y+7z + 12w = 123
4x+13y+22z + 39 w = 393
4x+12y+21z+35w =369
Case I : w=0
x+4y+7z = 123
4x+13y+22z = 393
4x+12y+21z =369
Solving we get
{ x = 0, y = 15, z = 9 }.
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Case II : w = 1
x+4y+7z + 12 = 123
4x+13y+22z + 39 = 393
4x+12y+21z+35 =369
Solving we have
{ x = 1, y = 10, z = 10 }.
Case iii : w = 2
x+4y+7z + 12(2)= 123
4x+13y+22z + 39 (2) = 393
4x+12y+21z+35(2) =369
{ x = 2, y = 5, z = 11 }
Brand A B C D Total Fibre 35 140 245 420 4305 Protein 40 130 220 390 3930 Fat 40 120 210 350 3690