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Question 174542This question is from textbook Algebra 2
: Hi. As of now I've been for almost two hours trying to solve the same problem, and I just can't. I don't know if it's correct, or not, the thing is that i have pages after pages of work, and I just don't seem to arrive to reasonable answer. could you please help?
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 proteins.
_______________ 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.
Ok, so this one I know it has to be:
10a+4b+12c=120
11a+77b+0c=220
4a+1b+16c=80
B. Use an augmented matrix to solve the system of equations from part (a). Round each answer to the nearest tenth.
Here I just got lost. I watched some videos and did everything they did, and somehow i got a=428/27, b=0, and c=-540/17. But then i tried to do everything by solve in the graphing calculator and the answers were completely different
C. Suppose food C runs out. How would this change the number of portions of food required each day?
Please, help.
This question is from textbook Algebra 2
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! I have a feeling no one is helping you on this because we are all stumped by it too!! I'm retired, and particularly bored today, so I thought I would give it a looksee! I would say first of all, for you to go with your calculator solution!!
I haven't worked with matrices for a VERY long time, but, if your equations are correct and I think they are correct, you need to convert from this system of equations to one in which you have eliminated coefficients to leave 0s and 1s. You may have to try a few times and I must admit that I didn't get the calculator answers the first time. This is what I did. First, reduce the equations to smaller numbers:
10a + 4b + 12c = 120 (Divide by 2)
11a + 77b + 0c = 220 (Divide by 11)
4a + 1b + 16c = 80
5a + 2b + 6c = 60
a + 7b + 0c = 20 (Mult by -4 and add to eq #3)
4a + 1b + 16c = 80
5a + 2b + 6c = 60
a + 7b + 0c = 20 (Mult by -5 and add to eq #1)
0a - 27b + 16c = 0
5a + 2b + 6c = 60
0a - 33b + 6c = -40
0a - 27b + 16c = 0
Next, eliminate the b coefficients by multiplying Eq #2 times -27 and Eq #3 times 33 and add together. (The first two equations remain the same!)
5a + 2b + 6c = 60
0a - 33b + 6c = -40
0a + 0b + 366c = 1080
Next, divide the Eq #3 by 366:
5a + 2b + 6c = 60
0a - 33b + 6c = -40
0a + 0b + 1c = 1080/366
So, c= 1080/366 = 2.9508 approximately, which is the calculator value that I got. I'm sure you can finish this now. By the way, my other calculator values were a=7.759 and b= 1.7486 approximately.
Good luck!!
R^2
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