Problem 62 - Thomas Calculus

Question

The 800-room Mega Motel chain is filled to capacity when the room charge is \(\$ 50\) per night. For each \(\$ 10\) increase in room charge, 40 fewer rooms are filled each night. What charge per room will result in the maximum revenue per night?

Step-by-Step Solution

Answer

The maximum revenue per night is achieved at a room charge of \( \$125 \).

Step 1: Define the Variables

Let the current room charge be \( x \). Since the initial condition charges \( \\(50 \), and for each \( \\)10 \) increase, 40 fewer rooms are filled, let the number of such increments be \( n \). Thus the room charge becomes \( 50 + 10n \), and the number of rooms filled is \( 800 - 40n \).

Step 2: Establish the Revenue Function

The revenue \( R \) is calculated as the product of the number of rooms filled and the charge per room. Thus, the revenue function is \( R(n) = (50 + 10n)(800 - 40n) \).

Step 3: Expand the Revenue Function

Expand the function: \[R(n) = (50 + 10n)(800 - 40n) = 50 \times 800 - 50 \times 40n + 10n \times 800 - 10n \times 40n\] \[= 40000 - 2000n + 8000n - 400n^2 = 40000 + 6000n - 400n^2 \]

Step 4: Rewrite the Revenue Function in Standard Form

The quadratic function can be rewritten as: \( R(n) = -400n^2 + 6000n + 40000 \). This is in the standard form \( ax^2 + bx + c \), where \( a = -400 \), \( b = 6000 \), and \( c = 40000 \).

Step 5: Find the Vertex of the Parabola

Since the revenue function is quadratic, its maximum value occurs at the vertex. Use the vertex formula \( n = -\frac{b}{2a} \): \[ n = -\frac{6000}{2 \times (-400)} = \frac{6000}{800} = 7.5 \].

Step 6: Calculate the Optimal Room Charge

Substitute \( n = 7.5 \) back into the equation for the room charge: \( 50 + 10 \times 7.5 = 50 + 75 = 125 \). Thus, the room charge that maximizes revenue is \( \$125 \).

Key Concepts

Revenue Maximization

Revenue maximization is all about finding the point where a business, or in this case a motel, earns the most money. The idea is to tweak pricing or other variables to reach this peak. In the context of the Mega Motel chain, understand that every small change in room pricing directly impacts the number of rooms filled.
Here's what happens:

• At a lower price, more rooms get filled, but at a lower revenue per room.
• By increasing the price, fewer rooms will be filled, but the revenue per room increases.

To find the "sweet spot" or the maximum revenue, you have to balance these forces. This is why we developed a revenue function using the variables of room price and number of rooms occupied.
The revenue function we derived was:\[ R(n) = (50 + 10n)(800 - 40n) \]This equation helps to visualize how revenue changes with price adjustments and informs us exactly where that maximum revenue point occurs.

Quadratic Functions

Quadratic functions are mathematical tools that help us understand the relationships between variables that include squared terms. These functions can model a variety of real-world situations, like projectile motion or in our case, revenue changes.
A quadratic function is often written as:\[ ax^2 + bx + c = y \]In our Mega Motel example, the revenue function simplified into this standard quadratic equation form:\[ R(n) = -400n^2 + 6000n + 40000 \]Notice:

• The function has a squared term, "\(-400n^2\)," making it a quadratic.
• The coefficients, in this case, help in shaping the parabola of the graph.
• The "\(a\)" coefficient is negative, which means the parabola opens downward, indicating a maximum point.

This is crucial because it lets us calculate exactly where the maximum revenue occurs, which is at the vertex of the parabola.

Vertex of a Parabola

The vertex of a parabola can tell us a lot about a quadratic function. For profit or revenue-related problems, the vertex can often indicate the optimal point, like achieving maximum revenue.
To find the vertex in a quadratic equation of the form:\[ ax^2 + bx + c \]we use the vertex formula:\[ n = -\frac{b}{2a} \]In the context of the Mega Motel example, substituting the values gives:\[ n = -\frac{6000}{2 \times (-400)} = 7.5 \]This tells us that 7.5 increments of \( \\(10 \) each from the original price is the optimal increase for maximum revenue.
The vertex also translates into actionable decision-making because from this calculation, we determine the optimal room charge to be \( \\)125 \). This ensures maximum profitability given the constraints of room filling and pricing.