There are 3 foods with the following Protein/Fat/Carbs (grams/ounce): 2,3,4; 3,3,1; 3,3,2. We want a meal using the 3 foods that results in 24g of protein, 27g of fat, 20g of carbs. How much of each food is needed?

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There are 3 foods with the following Protein/Fat/Carbs (grams/ounce): 2,3,4; 3,3,1; 3,3,2. We want a meal using the 3 foods that results in 24g of protein, 27g of fat, 20g of carbs. How much of each food is needed?

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Answer 1 · 2016-10-27T00:38:02

$(A, B, C) = (3, 4, 2)$ $(A,B,C)=(3,4,2)$ ounces of food

Explanation:

I'm going to use the Food letters - A, B, C - to be the number of ounces of that food that should be used.

I'm going to put the chart here (numbers in grams/ounce):

$Food 0 Protein 0 Fat 0 Carbs$ $"Food"color(white)(0)"Protein"color(white)(0)"Fat"color(white)(0)"Carbs"$

$00 A 0000020000030004$ $color(white)(00)"A"color(white)(00000)"2"color(white)(00000)"3"color(white)(000)"4"$

$00 B 0000030000030001$ $color(white)(00)"B"color(white)(00000)"3"color(white)(00000)"3"color(white)(000)"1"$

$00 C 0000030000030002$ $color(white)(00)"C"color(white)(00000)"3"color(white)(00000)"3"color(white)(000)"2"$

And we need 24g of Protein, 27g of Fat, 20g Carbs, so:

(P)rotein: $2 A + 3 B + 3 C = 24$ $2A+3B+3C=24$
(F)at: $0003 A + 3 B + 3 C = 27$ $color(white)(000)3A+3B+3C=27$
(C)arbs: $04 A + 1 B + 2 C = 20$ $color(white)(0)4A+1B+2C=20$

I'm first going to work on expressing the B values in terms of A and C. I'll do that by subtracting $3 \times C$ $3xxC$ and first P and then F:

3C: $000012 A + 3 B + 6 C = 60$ $color(white)(0000)12A+3B+6C=60$
P: $0000002 A + 3 B + 3 C = 24$ $color(white)(000000)2A+3B+3C=24$

$3 C - P$ $3C-P$ : $10 A + 000003 C = 36$ $10A+color(white)(00000)3C=36$

3C: $000012 A + 3 B + 6 C = 60$ $color(white)(0000)12A+3B+6C=60$
F: $0000003 A + 3 B + 3 C = 27$ $color(white)(000000)3A+3B+3C=27$

$3 C - F$ $3C-F$ : $9 A + 0000003 C = 33$ $9A+color(white)(000000)3C=33$

And I can now subtract $3 C - F$ $3C-F$ from $3 C - P$ $3C-P$ to get rid of C and solve for A:

$3 C - P$ $3C-P$ : $10 A + 000003 C = 36$ $10A+color(white)(00000)3C=36$
$3 C - F$ $3C-F$ : $9 A + 0000003 C = 33$ $9A+color(white)(000000)3C=33$

$(3 C - P) - (3 C - F)$ $(3C-P)-(3C-F)$ : $A = 3$ $A=3$

And now we can substitute this back in to one of our equations (I'll do both to double check our answer this far - I'll use colours to help highlight what we're doing):

$3 C - P$ $color(blue)(3C-P$ : $10 A + 3 C = 36$ $10A+3C=36$

$3 C - P$ $3C-P$ : $10 (3) + 3 C = 36$ $10(3)+3C=36$

$3 C - P$ $3C-P$ : $30 + 3 C = 36$ $30+3C=36$

$3 C - P$ $3C-P$ : $3 C = 6$ $3C=6$

$3 C - P$ $color(blue)(3C-P$ : $C = 2$ $color(red)(C=2$

$3 C - F$ $color(green)(3C-F$ : $9 A + 3 C = 33$ $9A+3C=33$

$3 C - F$ $3C-F$ : $9 (3) + 3 C = 33$ $9(3)+3C=33$

$3 C - F$ $3C-F$ : $27 + 3 C = 33$ $27+3C=33$

$3 C - F$ $3C-F$ : $3 C = 6$ $3C=6$

$3 C - F$ $color(green)(3C-F$ : $C = 2$ $color(red)(C=2$

Ok - C checks out as equaling 2. Now let's substitute into one of the originals (and I'll do all 3 to show it works in all the original equations):

(P)rotein: $2 A + 3 B + 3 C = 24$ $2A+3B+3C=24$

(P)rotein: $2 (3) + 3 B + 3 (2) = 24$ $2(3)+3B+3(2)=24$

(P)rotein: $6 + 3 B + 6 = 24$ $6+3B+6=24$

(P)rotein: $3 B = 12$ $3B=12$

(P)rotein: $B = 4$ $B=4$

(F)at: $3 A + 3 B + 3 C = 27$ $3A+3B+3C=27$

(F)at: $3 (3) + 3 B + 3 (2) = 27$ $3(3)+3B+3(2)=27$

(F)at: $9 + 3 B + 6 = 27$ $9+3B+6=27$

(F)at: $3 B = 12$ $3B=12$

(F)at: $B = 4$ $B=4$

(C)arbs: $4 A + 1 B + 2 C = 20$ $4A+1B+2C=20$

(C)arbs: $4 (3) + 1 B + 2 (2) = 20$ $4(3)+1B+2(2)=20$

(C)arbs: $12 + 1 B + 4 = 20$ $12+1B+4=20$

(C)arbs: $B = 4$ $B=4$

Everything checks out!

$(A, B, C) = (3, 4, 2)$ $(A,B,C)=(3,4,2)$ ounces of food

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There are 3 foods with the following Protein/Fat/Carbs (grams/ounce): 2,3,4; 3,3,1; 3,3,2. We want a meal using the 3 foods that results in 24g of protein, 27g of fat, 20g of carbs. How much of each food is needed?

1 Answer

Explanation:

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