A theater puts on a play and it's overhead is $750. It will charge adult's $5 per ticket and students $3 per ticket. The theater holds 200 people. What is the maximum profit? (How many adult tickets will be sold and how many student tickets will be sold?)
The maximum profit would be realized if only adult tickets were sold.
(200 * 5) - 750 = ?
To find the maximum profit, we need to determine the number of adult and student tickets that will be sold.
Let's assume the theater sells "a" adult tickets and "s" student tickets.
Given that the theater holds 200 people, we can form an equation: a + s = 200 (equation 1).
The overhead cost of the theater is $750. This amount needs to be subtracted from the total revenue to calculate the profit.
The revenue generated from adult tickets is 5a, and the revenue generated from student tickets is 3s. Therefore, the total revenue is 5a + 3s.
To maximize profit, we need to maximize the revenue while still covering the overhead cost. So, the profit equation becomes: 5a + 3s - 750.
Now we can solve this problem using a linear programming approach:
Step 1: Solve equation 1 for 'a' in terms of 's':
a = 200 - s.
Step 2: Substitute the value of 'a' in the profit equation:
Profit = 5(200 - s) + 3s - 750.
Step 3: Simplify and express the equation as a function of 's':
Profit = 1000 - 5s + 3s - 750
Profit = -2s + 250.
Step 4: Find the maximum value of 's' by finding the vertex of the function:
The s-coordinate of the vertex can be found by taking the derivative of the profit equation and setting it to zero:
d(Profit)/ds = -2 = 0.
-2s = -2.
s = 1.
Step 5: Substitute the found value of 's' back into equation 1 to find 'a':
a = 200 - s = 200 - 1 = 199.
Therefore, the maximum profit is achieved when the theater sells 199 adult tickets and 1 student ticket.