Problem 54 - Calculus Early Transcendentals

Question

Suppose you own a tour bus and you book groups of 20 to 70 people for a day tour. The cost per person is \$30 minus \(\$ 0.25\) for every ticket sold. If gas and other miscellaneous costs are \(\$ 200,\) how many tickets should you sell to maximize your profit? Treat the number of tickets as a nonnegative real number.

Step-by-Step Solution

Answer

Answer: 60 tickets.

Step 1: Setting up the profit equation

To set up the equation representing the profit, we need to consider the revenue (number of tickets sold times the price per ticket) and the costs (fixed cost and the per-person cost): Profit = Revenue - Total Cost P = R - TC Revenue is given by the price per ticket multiplied by the number of tickets sold. As the cost per person is \(30-0.25n\), the revenue equation will be: R = n(30-0.25n) Total Cost consists of the fixed cost (gas and miscellaneous costs = $200), and there is no variable cost associated with selling tickets: TC = 200 Now let's plug these values into the profit equation: P = n(30-0.25n) - 200

Step 2: Differentiating the profit equation

Next, we need to find the critical points of the profit function by taking the derivative with respect to the number of tickets, n: dP/dn = d(n(30-0.25n) - 200)/dn Using the product rule and constant rule: dP/dn = (30-0.25n) - n(0.25) = 30 - 0.50n

Step 3: Setting the derivative equal to 0 and solving for n

Set the derivative equal to 0 to find the critical points: 0 = 30 - 0.50n Now, solve for n: 0.50n = 30 n = 60

Step 4: Interpret the result

Since the number of tickets has only one critical point, it must be a maximum. Therefore, to maximize your profit, you should sell 60 tickets. Note that this result makes sense in the given context, as it falls within the allowed range of 20 to 70 people per tour.

Key Concepts

Profit maximization

When we talk about profit maximization in calculus, we are focusing on determining the optimal number of units or items to sell in order to achieve the highest possible profit. In this context, profit is defined as the difference between the revenue (money received from sales) and total costs (expenses to produce or purchase the items).

For the tour bus company, maximizing profit involves determining the number of tickets to sell in a way that earnings from sales exceed the costs as much as possible. This is done by establishing a profit equation, incorporating both revenue and costs into a single expression. By finding out how profit changes with respect to ticket sales, we can identify the sales level that yields the highest profit, which is our ultimate goal.

Derivative application

Derivatives are important tools in calculus that help us understand how a function changes. In optimization problems, like profit maximization, derivatives help us find where the slope of our profit function changes direction, or in other words, where it reaches a maximum or minimum point. For the tour bus example, we take the derivative of the profit function with respect to the number of tickets.

This derivative, often represented as \(dP/dn\), allows us to examine how small changes in the number of tickets sold affect the profit. To find the maximum profit, we need to determine where this derivative equals zero, as this indicates a flattening of the slope at a peak point. By solving the equation \(dP/dn = 0\), we find the critical points, which are potential candidates for maximum profit.

Critical points

Critical points are specific values that make the derivative of a function equal to zero or undefined. They represent situations where the function could have a local maximum, minimum, or a saddle point.

In the tour bus example, the critical point was found by solving \(0 = 30 - 0.50n\), resulting in \(n = 60\). This means, at 60 tickets, the slope of the profit function is zero, indicating a potential maximum profit point.

• To confirm whether this point is actually a maximum, and not a minimum or inflection point, we verify additional criteria, such as ensuring no extreme points lie outside the considered range (20 to 70 people).
• With only one such critical point in this context and given it's within the acceptable range, it offers us confidence that it is indeed a maximum.

Revenue function

The revenue function is a mathematical representation of earnings based on sales level or quantity sold. It multiplies the number of units (in this case, tickets) by the price per unit. For the tour bus situation, it's expressed as \(R = n(30 - 0.25n)\).

This function tells us how total revenue changes when more tickets are sold. Here, the expression \(30 - 0.25n\) means the ticket price starts at \(30, reducing by \)0.25 for each additional ticket due to discounts offered for bulk sales.

• The revenue function combined with the fixed costs allows us to create the profit equation by subtracting total costs from revenue.
• Understanding how the revenue changes with ticket sales is crucial, as it highlights not only potential earnings but also the impact of pricing strategies on profit.