Answer:
For a point (3, 1), we have;
Michael will buy 3 sandwich and 1 hot meals and such that the number of meals = 1 + 3 = 4 lunches
The cost of the 4 lunches is 2×3 + 3×1 = $9
Step-by-step explanation:
The given parameters are;
2·x + 3·y ≤ 15
x + y ≥ 3
Where:
x = The number of sandwiches Micheal will buy
y = The number of hot lunches that will be bought by Micheal
We plot the equations to obtain the possible range of values are;
x ≥ 3 - y
2×(3 - y) + 3×y ≤ 15
when x = 0, we have
2×0 + 3×y ≤ 15
y ≤ 5
Similarly we have;
2×x + 3×(3 - x) ≤ 15
when y = 0, we have
2×x + 3×0 ≤ 15
2×x ≤ 15
x ≤ 10
Therefore, we have the possible solution as follows;
Either x = and y =
x = 1 and y = 2
x = 2 and y = 1
x = 3 and y = 3
x = 4 and y = 1
x = 5 and y = 1
x = 6 and y = 1
Therefore, where the solution is the point (3, 1), we have;
Michael will buy 3 sandwich and 1 hot meals and such that the number of meals = 1 + 3 = 4 lunches
The cost of the 4 lunches is 2×3 + 3×1 = $9.