The owners of a theme park know that an average of 50,000 people visit the park each day:
They are presently charging $15.00 for an admission ticket. Each time in the past that they have raised the admission price, an average of 2,500 fewer people have come to the park for each $1.00 increase in ticket price. What ticket price will maximize the revenue from ticket sales?


To solve the problem of determining the ticket price that will maximize the revenue from ticket sales, we can use the following steps:

### Step 1: Define variables

- Let $ p $ represent the ticket price.
- Let $ N(p) $ represent the number of people who visit the park daily when the ticket price is $ p $.

We know the following:
- At a price of $15, the number of visitors is 50,000.
- For every $1 increase in the ticket price, 2,500 fewer people visit the park.

### Step 2: Express the number of visitors as a function of the ticket price

At a base price of $15:
- $ N(15) = 50,000 $.

For every $1 increase in the price, 2,500 fewer people visit. Therefore, the number of visitors as a function of price $ p $ can be written as:
\[
N(p) = 50,000 - 2,500(p - 15)
\]

### Step 3: Define the revenue function

Revenue $ R(p) $ is the product of the number of visitors and the ticket price:
\[
R(p) = N(p) \times p = \left(50,000 - 2,500(p - 15)\right) \times p
\]

### Step 4: Simplify the revenue function

Expand the equation:
\[
R(p) = \left(50,000 - 2,500p + 37,500\right) \times p
\]
\[
R(p) = (87,500 - 2,500p) \times p
\]
\[
R(p) = 87,500p - 2,500p^2
\]

### Step 5: Maximize the revenue function

To maximize the revenue, take the derivative of $ R(p) $ with respect to $ p $ and set it equal to zero:
\[
\frac{dR}{dp} = 87,500 - 5,000p
\]
Set the derivative equal to zero to find the critical point:
\[
87,500 - 5,000p = 0
\]
\[
5,000p = 87,500
\]
\[
p = \frac{87,500}{5,000} = 17.50
\]

### Step 6: Verify it's a maximum

To confirm this is a maximum, check the second derivative:
\[
\frac{d^2R}{dp^2} = -5,000
\]
Since the second derivative is negative, the function has a maximum at $ p = 17.50 $.

### Conclusion

The ticket price that maximizes the revenue is **$17.50**.

---

Would you like more details on any of the steps or have further questions?

Here are 5 related questions for practice:
1. What would be the revenue if the ticket price is $17.50?
2. How many visitors would come to the park if the price is set to $18.00?
3. What is the relationship between ticket price and the number of visitors?
4. How can you find the price that minimizes the number of visitors?
5. What would the revenue be if the ticket price were reduced to $14.00?

**Tip**: To maximize a function, always check the second derivative—it confirms whether you have a maximum or a minimum point.

The owners of a theme park know that an average of 50,000 people visit the park each day. They are presently charging $15.00 for an admission ticket. Each time in the past that they have raised the admission price, an average of 2,500 fewer people have come to the park for each $1.00 increase in ticket price. What ticket price will maximize the revenue from ticket sales?

Discover how to maximize theme park revenue using quadratic functions and derivatives. This example explores ticket price changes and their impact on visitor numbers, leading to the optimal price for maximum revenue. Suitable for students in Grades 10-12, covering key algebra and calculus concepts.

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