A lunch stand makes $.75 profit on each chef's salad and $1.20 profit on each Caesar salad. On a typical weekday, it sells between 40 and 60 chef's salads and between 35 and 50 Caesar salads. The total number sold has never exceeded 100 salads. How many of each type should be prepared in order to maximize profit?
Let's call the number of chef's salads sold x and the number of Caesar salads sold y.
The profit from selling x chef's salads is 0.75x.
The profit from selling y Caesar salads is 1.20y.
The total profit is the sum of these two profits, so the equation for the profit is 0.75x + 1.20y.
We are told that the number of chef's salads sold is between 40 and 60, so 40 ≤ x ≤ 60.
We are also told that the number of Caesar salads sold is between 35 and 50, so 35 ≤ y ≤ 50.
The total number of salads sold is x + y, and it cannot exceed 100, so x + y ≤ 100.
To maximize profit, we want to find the maximum value of 0.75x + 1.20y subject to the constraints 40 ≤ x ≤ 60, 35 ≤ y ≤ 50, and x + y ≤ 100.
We can use linear programming to solve this problem. However, since the constraints are relatively small, we can use a brute force method to find the solution.
Let's start by trying the lowest and highest values for x and y:
x = 40, y = 35: 0.75(40) + 1.20(35) = 30 + 42 = 72
x = 40, y = 50: 0.75(40) + 1.20(50) = 30 + 60 = 90
x = 60, y = 35: 0.75(60) + 1.20(35) = 45 + 42 = 87
x = 60, y = 50: 0.75(60) + 1.20(50) = 45 + 60 = 105
Since x + y cannot exceed 100, we can eliminate the last solution.
So, the maximum profit is obtained when x = 40 and y = 50, which means 40 chef's salads and 50 Caesar salads should be prepared to maximize profit.