Question

Tour Service You operate a tour service that offers the following rates: \(\cdot \$ 200\) per person if 50 people (the minimum number to book the tour) go on the tour. \(\cdot\) For each additional person, up to a maximum of 80 people total, the rate per person is reduced by \(\$ 2 .\) It costs \(\$ 6000\) (a fixed cost) plus \(\$ 32\) per person to conduct the tour. How many people does it take to maximize your profit?

Short answer

The number of people that maximize the profit can be found by solving the derivative of the profit function for zero and checking that the solution is within the range 50 to 80.

Step-by-step solution

Formulate the revenue and cost functions

First, based on the problem, the revenue from the tour is based on the number of people who go on the tour. When the number of people \(x\) is greater than or equal to 50 and less than or equal to 80, each person pays \(\$200 - 2(x - 50)\). So, the revenue \(R(x)\) is given by: \(R(x) = (200 - 2(x - 50)) \times x \n\nSimilarly the cost of conducting a tour is given to be \(\$6000 + 32x\). So, the cost \(C(x)\) is given by: \(C(x) = 6000 + 32x\)

Formulate the profit function

The profit from the tour service is the revenue minus the cost, so the profit function \(P(x)\) is given by: \(P(x) = R(x) - C(x)\) Substituting from step 1, we have the profit function as: \(P(x) = ((200 - 2(x - 50)) \times x) - (6000 + 32x)\)

Find the maximum profit

The maximum or minimum of a function is found at the point where the derivative is zero. Hence, to find the number of people that maximizes the profit, we need to calculate the derivative \(P'(x)\) of the profit function and find the value of \(x\) that makes this derivative zero. Note: The value of \(x\) we are interested in is between 50 and 80 inclusive, as we can only have between 50 and 80 people.

Determine the number of people

Solve the equation \(P'(x) = 0\) for \(x\) between 50 and 80 (inclusive). The solution of this equation gives the number of people that maximize your profit.

Key concepts

Revenue and Cost Functions

Understanding revenue and cost functions is crucial in economics and business, as they form the backbone of profit analysis. The revenue function provides a relationship between the number of goods or services sold and the total income generated from these sales. In our tour service example, the revenue function, denoted as R(x), changes based on the number of participants, x. The initial rate is \(200 per person, which decreases by \)2 for each additional person beyond the base number of 50.

The cost function, indicated as C(x), helps us understand the total expenses involved in offering a service or product. In this case, it includes a fixed cost of \(6000, which does not change regardless of the number of tourists, and a variable cost of \)32 per person.To further clarify, as the number of participants changes, the revenue per person adjusts and so does the total revenue. At the same time, the total cost increases linearly with each additional tourist. These functions give us an insight into how the pricing strategy and costs affect the overall profitability of the tour service.

Profit Optimization

The goal of any business is to maximize profit, which is the difference between revenue and costs. Profit optimization requires us to determine the price and production level that yields the highest profit. In calculus, this involves finding the maximum value of a profit function. The profit function, P(x), is constructed by subtracting the cost function from the revenue function: P(x) = R(x) - C(x).

In the provided exercise, the profit function is already formulated. To optimize profit, we need to evaluate this function within the given range of participants (50 to 80). The goal is to find the point where the profit is at its peak, which is where additional sales neither increase nor decrease the profit. This critical point, where the slope of the profit graph is zero, can be discovered by utilizing calculus, specifically by finding where the derivative of the profit function equals zero.

Derivatives in Economics

The concept of derivatives is a powerful tool in economics, particularly when it comes to analyzing functions representing revenue, cost, and profit. A derivative essentially represents the rate at which a function is changing at any given point, which in economic terms, can indicate the rate of change in cost, revenue, or profit with respect to the number of goods or services produced or sold.In our tour service instance, taking the derivative of the profit function, P'(x), and setting it to zero will identify the level of production (or number of tourists, in this context) that maximizes profit. Derivatives help businesses determine the most profitable operational point. To illustrate, when the derivative is positive, the function is increasing, indicating that adding more tourists would initially raise profit. When the derivative becomes negative, it suggests adding more tourists actually starts to reduce profit. Where the derivative shifts from positive to negative indicates a maximum profit point – the sweet spot for the number of tourists the business should aim to attract to the tour.