Question

The Skating Rink has a small rink available for skating parties. Customers can choose

between a two-hour party and a three-hour party. The staff and facilities can handle 5
parties in 12 hour period. If the revenue from a two-hour party is $50.00 and the revenue
for a three hour party is $80.00, how many of each type of party would maximize the revenue?

Answer 1

To maximize revenue, the center should offer 0 two-hour parties and 5 three-hour parties.

Step-by-step explanation:

To maximize revenue, we need to determine the number of two-hour and three-hour parties that should be offered. Let's suppose the number of two-hour parties is represented by 'x', and the number of three-hour parties is represented by 'y'.

Given that the staff and facilities can handle 5 parties in a 12-hour period, we have the following constraint: x + y = 5.

To maximize revenue, we need to find the values of x and y that maximize the function R = 50x + 80y, where R represents the total revenue. This is a linear programming problem, and can be solved using graphical or algebraic methods.

By substituting y = 5 - x into the revenue function, we get R = 50x + 80(5 - x).

Simplifying, R = 50x + 400 - 80x = -30x + 400.

To maximize revenue, we want to find the maximum value of R. In this case, the revenue will be maximized at the highest point on the graph of the function, which will occur at the vertex of the parabola.

The vertex of the parabola can be found by using the formula x = -b/2a, where a = -30 and b = 0. Plugging in these values, we find x = -0/2(-30) = 0.

Substituting this value back into the equation, we find R = -30(0) + 400 = 400.

Therefore, to maximize revenue, the center should offer 0 two-hour parties and 5 three-hour parties.