Final answer:
The store will maximize its revenue by selling a total of 80 cans of tennis balls. This is found by calculating the derivative of the revenue function and determining the number of additional cans beyond 20 that maximizes the function, which is 60 cans plus the initial 20 cans.
Step-by-step explanation:
The question involves a revenue maximization problem in which a sporting goods store sells tennis balls with a tiered pricing structure. The price is $2.40 per can for up to 20 cans. For each can above 20, the price per can decreases by $0.02, up to a limit of 60 cans. To calculate the revenue for cans above 20, let's denote the number of cans above 20 as x. The price per can for these would be $(2.40 - 0.02x), and the total revenue generated from these additional cans would be x times the adjusted price. The store's total revenue R from selling x additional cans is:
R = (20 * $2.40) + x * ($2.40 - 0.02x)
To maximize revenue, we need to find the value of x that makes the derivative of the revenue function equal to zero. The derivative of the revenue function with respect to x is:
R' = $(2.40 - 0.04x)
Setting this equal to zero gives us the number of cans that maximizes revenue.
x = 60 cans
However, we cannot forget that x represents cans over 20. So, to find the total cans that maximize revenue, we add the initial 20 cans to x:
Total cans for maximum revenue = 20 + x = 20 + 60 = 80 cans
Therefore, the store will maximize its revenue by selling 80 cans of tennis balls in one transaction.