Problem 53 - Calculus Early Transcendentals

Question

Suppose a tour guide has a bus that holds a maximum of 100 people. Assume his profit (in dollars) for taking \(n\) people on a city tour is \(P(n)=n(50-0.5 n)-100 .\) (Although \(P\) is defined only for positive integers, treat it as a continuous function.) a. How many people should the guide take on a tour to maximize the profit? b. Suppose the bus holds a maximum of 45 people. How many people should be taken on a tour to maximize the profit?

Step-by-Step Solution

Answer

In conclusion, the tour guide should take 50 people on the tour to maximize profit when the bus has a maximum capacity of 100 people, and should take 45 people when the bus has a maximum capacity of 45 people. This is due to the nature of the profit function and the given constraints on the capacity of the bus.

Step 1: Part a: Optimizing profit with a maximum capacity of 100 people

First, we have to find the derivative of the profit function \(P(n)\) with respect to \(n\). We are given the profit function \(P(n) = n(50 - 0.5n) - 100\). Now, let's find its derivative:\(P'(n) = \frac{dP(n)}{dn} = \frac{d}{dn} (n(50-0.5n)-100)\).By using the product rule, we see that \(P'(n) = (50 - 0.5n) - 0.5n = 50 - n\). Now, let's find the critical points by setting \(P'(n) = 0\): \[0 = 50 - n\] Thus, \(n = 50\) is our critical point. Since we are dealing with a quadratic function that opens downward, we know that this maximum profit will occur at this critical point n=50. So the guide should take 50 people on the tour to maximize profit when the bus can hold up to 100 people.

Step 2: Part b: Optimizing profit with a maximum capacity of 45 people

Since the maximum capacity of the bus is now 45 people, we need to check whether the optimal number of people we found in part a (50 people) is still valid or not. If the optimal number of people is still within the allowed capacity, we don't need to change the previous solution. However, if the optimal number is outside of the allowed capacity, we need to adjust our solution accordingly. Since the optimal number of people found in part a (50 people) is more than the allowed maximum capacity of 45 people, we need to consider the maximum capacity as our optimal number of people. In other words, the guide should now take 45 people on a tour to maximize the profit when the bus has a maximum capacity of 45 people.

Key Concepts

Derivative

In calculus, the derivative of a function is a crucial concept used to determine the rate at which a quantity changes.
When we have a profit function like in our exercise, taking the derivative helps us find the rate of change of profit concerning the number of people, \(n\).

To find a derivative, particularly in the given problem, we use rules such as the product rule, which applies when multiplying two functions, like in \(P(n) = n(50-0.5n) - 100\).
The derivative, \(P'(n) = 50 - n\), tells us how changes in \(n\) affect profit. When \(P'(n)\) is zero, we have a point where profit changes direction, called a critical point.

Critical Points

Critical points are values of \(n\) where the derivative of our function is zero or undefined.
These points are very important because they help identify where a function reaches its maximum or minimum values.

In our problem, setting \(P'(n) = 0\) gives \(n = 50\), which is a critical point. This indicates that at 50 people, the profit does not increase with the addition of more attendees.

• If the function is quadratic and opens downward, like in this case, the critical point represents a maximum profit point.
• Checking whether the point is a maximum or minimum can be done through the second derivative test, but knowing the nature of the function can also be sufficient.

This is why at 50 attendees, the guide achieves maximum profit, provided the bus capacity allows it.

Quadratic Function

Quadratic functions are polynomial functions of degree two, with the general form \(ax^2 + bx + c\), where \(a, b,\) and \(c\) are constants.
The graph of a quadratic function is a parabola, and when addressing profit, it's significant because it helps visualize the maximum or minimum point.

In our given problem, the profit function \(P(n) = n(50 - 0.5n) - 100\) is a quadratic function that can be expanded into \(-0.5n^2 + 50n - 100\).

• The \(-0.5\) coefficient of \(n^2\) tells us the parabola opens downward, meaning it has a maximum point.
• Finding where the parabola peaks or dips helps determine the optimum number of people to achieve maximum profit. In our case, this occurs at the critical point \(n=50\).

Understanding the nature of quadratic functions is vital to making informed decisions in optimization problems.

Maximum Profit

Maximizing profit is the primary goal, especially in business applications.
To find the maximum profit, we use calculus concepts like derivatives and critical points.

In the context of the tour guide problem, determining the number of people that maximizes profit requires adjusting based on constraints like bus capacity. In part (a) of the exercise, provided there is no constraint, maximum profit occurs at the critical point \(n=50\).

• In part (b), we imposed a constraint reducing the capacity to 45 people. Hence, though the critical point suggested 50, the guide must consider the bus's capacity and take only 45 participants for maximum profit.
• This scenario illustrates how external constraints affect profit maximization strategies.

Conceptualizing and calculating maximum profit through optimization are essential skills in understanding and solving real-world problems.