9:30 AM Class Group 3
Timothy Cawley, Julian Chan, Stephen Miske, Pedro Muniz, Jonathan Trudeau
Our task for this project is to achieve the daily requirements of a
balanced diet while spending the least amount of money possible and it
must be solved algebraically by using a linear equation.
Foods:
Each portion of potatoes consists of 3 units of carbohydrates, 4 units
of Vitamins, 1 unit of protein and it costs 75 cents.
Each portion of meat consists of 1 unit of carbohydrates, 3 units of
vitamins, 4 units of protein and it costs 2 dollars.
The daily requirements for a balanced diet consists of 8 units of
carbohydrates, 19 units of vitamins, and 7 units of protein.
Linear equation:
Variables:
X= units of carbohydrates
Y= units of vitamins
Z= units of vitamins
3= the total amount of money needed to purchase the potatoes.
2= total amount of money needed to purchase the single portion of meat.
5= total amount of money spent to achieve the daily requirements.
12x +16y + 4z=3
1x + 3y + 3z=2
13x + 19y + 7z = 5
To arrive at our final equation we took the portions of 3 potatoes and
then plugged them into our linear equation for out first line, which came
to a total amount of 3 dollars. We took the values of one portion of meat
needed to achieve our daily values and put those into our linear equation,
which came out to a total amount of 2 dollars. We then took the total
amount of the two portions and plugged them into the last line which gave
us the total amount for the daily requirements and we had spent a total
amount of 5 dollars.