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A state park charges $100 for an annual pass. At this rate 750 people purchase passes every year. For each $5 decrease in price 15 more people purchase a pass. What price should the park charge in order to maximize revenue? Use calculus and show your work.

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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.

User Mark Jansen
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