The $75 is the price that maximizes revenue, generating $22,500 more than the initial price.
Maximizing Revenue for State Park Annual Pass using Calculus
Step 1: Define the Variables and Function
P: Price of the annual pass ($).
X: Number of annual passes purchased.
R(P): Revenue generated at price P.
We are given that:
Initial price, P0 = $100
Initial number of passes purchased, X0 = 750
Increase in purchases per $5 decrease in price, ΔX/ΔP = 15
The revenue function can be modeled as:
R(P) = P * X
Step 2: Express X as a function of P (Demand Function)
We know that X increases as P decreases. This can be represented by a linear relationship with a negative slope:
X = a - bP
where:
a: Represents the maximum number of people willing to buy a pass (intercept).
b: Represents the sensitivity of purchase to price change (slope).
Using the given information:
X0 = a - bP0 = 750 (Initial condition)
ΔX/ΔP = -b = -15 (Sensitivity)
Solving for a and b:
a = X0 + bP0 = 750 + 15(100) = 2250
b = -15
Therefore, the demand function is:
X = 2250 - 15P
Step 3: Maximize Revenue (R(P))
To maximize revenue, we need to find the price P that maximizes the product of P and X (R(P)).
Substituting the demand function into the revenue function:
R(P) = P * (2250 - 15P)
Expanding and simplifying:
R(P) = 2250P - 15P^2
This is a quadratic function with a maximum at the vertex. To find the vertex, we take the derivative of R(P) and set it equal to 0:
R'(P) = 2250 - 30P = 0
Solving for P:
P = 75
Therefore, the price that maximizes revenue is $75.
Step 4: Verification
Let's check if our answer is reasonable:
The initial price was $100, and revenue was P0X0 = $100750 = $75,000.
At the optimal price of $75, the number of passes purchased increases to X = 2250 - 15(75) = 1350.
Revenue at the optimal price is R(75) = 75*1350 = $97,500.
This confirms that $75 is the price that maximizes revenue, generating $22,500 more than the initial price.