Question

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?

Answer 1

`(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"color(white)(0)"Protein"color(white)(0)"Fat"color(white)(0)"Carbs"`

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

`color(white)(00)"B"color(white)(00000)"3"color(white)(00000)"3"color(white)(000)"1"`

`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: `2A+3B+3C=24`
(F)at: `color(white)(000)3A+3B+3C=27`
(C)arbs: `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 `3xxC` and first P and then F:

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

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

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

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

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

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

`(3C-P)-(3C-F)`: `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):

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

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

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

`3C-P`: `3C=6`

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

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

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

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

`3C-F`: `3C=6`

`color(green)(3C-F`: `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: `2A+3B+3C=24`

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

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

(P)rotein: `3B=12`

(P)rotein: `B=4`

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

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

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

(F)at: `3B=12`

(F)at: `B=4`

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

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

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

(C)arbs: `B=4`

Everything checks out!

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