8:00 AM Class Group 4
Meat and Potatoes Group Project
Group 4: Chris Cardenas, Chase Chemero, Joseph Ciarleglio, Jennifer Rosa, Thanh Tran
Math 1070Q
Linear Programming Project
December 1, 2008
Group Project Meat and Potatoes
We first assign the number of portions ordered of potatoes and meat 'X' and 'Y' respectively. One portion of potatoes (1X) contains 3 units of carbohydrates, 4 units of vitamins, 1 unit of protein and costs 75 cents. One portion of meat (1Y) contains 1 unit of carbohydrates, 3 units of vitamins, 3 units of protein and costs 2 dollars. Because we know there can only be positive numbers of portions ordered we can infer non-negativity clauses (X ≥0 and Y ≥0). The minimum daily requirements of a balanced diet are 8 units of carbohydrates, 19 units of vitamins and 7 units of protein. These give us our constraints (or conditions).
3X+Y ≥8 ----This constraint that defines the minimum amount of carbohydrates we need daily. One portion of potatoes delivers 3 units of carbohydrates and one portion of meat delivers only one unit of carbohydrates. We need 8 total units of carbohydrates altogether so 3X+Y ≥ 8 (3 times the number of portions ordered of potatoes plus 1 times the number of portions ordered of meat is greater or equal to 8 total units of carbohydrates).
4X+3Y ≥19 -------This constraint defines the minimum amount of vitamins we need daily. One portion of potatoes delivers 4 units of vitamins and one portion of meat delivers 3 units of vitamins. We need 19 total units of vitamins altogether so 4X+3Y ≥19 (4 times the number of portions ordered of potatoes plus 3 times the number of portions ordered of meat is greater or equal to 19 total units of vitamins).
X+3Y ≥7 ------This constraint defines the minimum amount of protein we need daily. One portion of potatoes delivers 1 unit of protein and one unit of meat delivers 3 units of protein. We need 7 units of protein altogether so X+3Y ≥7 (1 times the number of portions ordered of potatoes plus 3 times the number of portions ordered of meat is greater than or equal to 7 total units of protein).
Because we know the costs of both X and Y we can derive our objective function: .75X+2Y=M
Our objective function is crucial to our model because it is what we seek to minimize. M is the cost when ordering X portions of potatoes and Y portions of meat. By minimizing M and still meeting minimum daily requirements we have answered this problem.