SOLUTION: While stranded on an island, the crew of a sailboat has access to only three sources of food, as shown in the table below. One of the crew members designs a daily diet to supply ea

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Question 174531This question is from textbook Algebra 2
: While stranded on an island, the crew of a sailboat has access to only three sources of food, as shown in the table below. One of the crew members designs a daily diet to supply each person with 120g of fat, 220g of carbohydrates, and 80g of protein.
_______________ A______B_______C
Fat____________ 10 g____ 4g_____ 12 g
Carbohydrates___11 g___77 g_____0g
Protein_________4 g_____1 g_____16 g
A. Write a system of three equations in three variables to find the number of portions of each food each person must have to meet the daily diet.
B. Use an augmented matrix to solve the system of equations from part (a).
C. Suppose food C runs out. How would this change the number of portions of food required each day?
 This question is from textbook Algebra 2

Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
120g of fat, 220g of carbohydrates, and 80g of protein.
_______________ A______B_______C
Fat____________ 10 g____ 4g_____ 12 g
Carbohydrates___11 g___77 g_____0g
Protein_________4 g_____1 g_____16 g
:
:
A)10A+4B+12C=120
11A+77B =220
4A+ 1B+16C=80
:
B)using cramers rule
:
find determinant of main matrix%28matrix%283%2C3%2C10%2C4%2C12%2C11%2C77%2C0%2C4%2C1%2C16%29%29using column 3---> 12%28matrix%282%2C2%2C11%2C77%2C4%2C1%29%29-0%28matrix%282%2C2%2C10%2C4%2C4%2C1%29%29+16%28matrix%282%2C2%2C10%2C4%2C11%2C77%29%29---->12(-297)+16(726)=8052
:
A=det%28matrix%283%2C3%2C120%2C4%2C12%2C220%2C77%2C0%2C80%2C1%2C16%29%29/8052---->so using the 3rd column--->det = 12%28matrix%282%2C2%2C220%2C77%2C80%2C1%29%29-0+16%28matrix%282%2C2%2C120%2C4%2C220%2C77%29%29--->12(-5940)+16(8360)=62480
:highlight%28A=62480%2F8052=7.76%29
:
:
B=det%28matrix%283%2C3%2C10%2C120%2C12%2C11%2C220%2C0%2C4%2C80%2C16%29%29/8052---->so using the 3rd column--->det = 12%28matrix%282%2C2%2C11%2C220%2C4%2C80%29%29-0+16%28matrix%282%2C2%2C10%2C120%2C11%2C220%29%29---->0+16(880)=14080
:highlight%28B=14080%2F8052=1.75%29
:
C=det%28matrix%283%2C3%2C10%2C4%2C120%2C11%2C77%2C220%2C4%2C1%2C80%29%29/8052---->using column 1--->det= 10%28matrix%282%2C2%2C77%2C220%2C1%2C80%29%29-11%28matrix%282%2C2%2C4%2C120%2C1%2C80%29%29+4%28matrix%282%2C2%2C4%2C120%2C77%2C220%29%29--->10(5940)-11(200)+4(-8360)--->59400-2200-33440=
23760
:highlight%28C=23760%2F8052=4.37%29
:
:
C)if C was no longer available we would have the equation
:
10A+4B+0C=120
11A+77B+0C=220
4A+ B+0C=80
:
we cannot use Cramers rule because there would be no unique solution as the determinant you would be dividing by would be zero.
:according to this graph there is no unique solution to this scenario.
It appears you could reach 2 of the conditions in any order but not all 3 at once.