Problem 19 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The owner of a luxury motor yacht that sails among the 4000 Greek islands charges $$\$ 600 /$$ person/ day if exactly 20 people sign up for the cruise. However, if more than 20 people sign up (up to the maximum capacity of 90 ) for the cruise, then each fare is reduced by $$\$ 4$$ for each additional passenger. Assuming at least 20 people sign up for the cruise, determine how many passengers will result in the maximum revenue for the owner of the yacht. What is the maximum revenue? What would be the fare/passenger in this case?

Step-by-Step Solution

Answer

In this scenario, the maximum revenue of \(28,900 is achieved when there are 85 passengers, and the fare per passenger at this point is \(340.

Step 1: Write the function for revenue

Let's use x as the number of passengers. The fare per person is given by 600 - 4(x-20) when x > 20, and it is 600 for x = 20. Thus, the revenue function, R(x), in terms of passengers x can be written as: R(x) = x * (600 - 4(x-20))

Step 2: Simplify the revenue function

Let's simplify the function R(x): R(x) = x * (600 - 4x + 80) R(x) = x * (680 - 4x) R(x) = -4x^2 + 680x

Step 3: Find the critical points of the function

Now, let's find the critical points of the function R(x) by differentiating it with respect to x: R'(x) = -8x + 680 Setting R'(x) = 0, we get: -8x + 680 = 0 8x = 680 x = 85

Step 4: Test the maximum point in the revenue function

Now we will test the only critical point, x = 85, to see if it results in a maximum value for the revenue function using the second derivative. R''(x) = -8 (since the second derivative is a constant) Since R''(85) < 0, we can conclude that x = 85 is a maximum point in the revenue function.

Step 5: Calculate the maximum revenue and the fare per passenger

Now that we know the number of passengers that will result in the maximum revenue (x = 85), let's find the maximum revenue and the fare per passenger in this case. The maximum revenue is: R(85) = -4*(85^2) + 680*85 = 28900 The fare per passenger, in this case, is: Fare = 600 - 4(85-20) = 600 - 4*65 = 340 So, when there are 85 passengers, the maximum revenue for the yacht owner is \(28,900, and the fare per passenger is \)340.

Key Concepts

Revenue Function

In mathematics, particularly in calculus, the revenue function is an essential tool for businesses aiming to maximize their income. A revenue function is a way to describe the total income a business might receive based on different levels of sales or customers. For the yacht example, we use the number of passengers (denoted as \(x\)) to express how the revenue changes. If 20 passengers sign up, the fare is set at \(\\(600\) per person. However, for each additional passenger, beyond 20, the fare lowers by \(\\)4\). The revenue function becomes:

• \(R(x) = x \cdot (600 - 4(x-20))\)

The term \((600 - 4(x-20))\) adjusts the fare based on the number of passengers, driving both the fare and ultimately the revenue.

Critical Points

Critical points in calculus refer to the values of \(x\) where the first derivative of a function is zero or undefined. These points are crucial as they help us find maximum or minimum values of a function, such as the maximum revenue. To identify critical points for the revenue function \(R(x)\), we first take its derivative, which gives us the rate of change of revenue with respect to passengers:

• \(R'(x) = -8x + 680\)

Setting \(R'(x) = 0\) provides:

• \(-8x + 680 = 0\)
• \(8x = 680\)
• \(x = 85\)

At \(x = 85\), the function stops increasing and might be at a peak, which is crucial for determining maximum revenue.

Second Derivative Test

The second derivative test helps us confirm whether a critical point is a maximum, minimum, or a saddle point. For the yacht problem, once we have identified \(x = 85\) as a critical point from the first derivative \(R'(x)\), we apply the second derivative to establish the nature of this point:

• \(R''(x) = -8\)

Since \(R''(85) = -8\), which is less than zero, this indicates that the function is concave down at \(x = 85\). This confirms \(x = 85\) as a maximum point, guaranteeing the highest revenue at this number of passengers.

Fare Calculation

Calculating the fare is crucial for understanding how changes in passenger numbers impact revenue. Initially, fares start at \(\\(600\) per person for exactly 20 passengers. As more passengers join, beyond 20, each fare decreases by \(\\)4\) per additional passenger. For example, for 85 passengers, the fare per person is calculated as:

• Fare = \(600 - 4(85-20)\)
• Fare = \(600 - 4 \times 65\)
• Fare = \(340\)

Hence, while the boat accommodates more passengers, reducing fares could still result in maximum revenue, which is \(28,900\) in this case when 85 passengers each pay \(\$340\).