Answer:
x + 2y ≤ 100 and x + 3y ≤ 400
Maximum profit = 6x + 5y.
Explanation:
Let there be x number of small dishes and y number of large dishes to maximize the profit.
So, total profit is P = 6x + 5y .......... (1)
Now, the small dish uses 1 cup of sauce and 1 cup of cheese and the large dish uses 2 cups of sauce and 3 cups of cheese.
So, as per given conditions,
x + 2y ≤ 100 ........ (1) and
x + 3y ≤ 400 .......... (2)
Therefore, those are the constraints for the problem. (Answer)