Problem 53 - Beginning and Intermediate Algebra

Question

In each problem, find the following. (a) A function \(R(x)\) that describes the total revenue received (b) The graph of the function from part ( \(a\) ) (c) The number of unsold seats that will produce the maximum revenue (d) The maximum revenue A charter flight charges a fare of \(\$ 200\) per person, plus \(\$ 4\) per person for each unsold seat on the plane. The plane holds 100 passengers. Let \(x\) represent the number of unsold seats.

Step-by-Step Solution

Answer

The functions is \( R(x) = 20000 + 400x - 4x^2 \). The number of unsold seats that will produce the maximum revenue is 50. The maximum revenue is 30000 dollars.

Step 1: Define Variables

Let the number of passengers be represented by \( (100 - x) \) where \( x \) is the number of unsold seats.

Step 2: Define the Fare Function

The fare per person can be expressed as: Fare = \( 200 + 4x \)

Step 3: Define the Revenue Function

The total revenue \(R(x)\) equals the number of passengers times the fare per person: \[ R(x) = (100 - x)(200 + 4x) \] Simplify this equation to: \[ R(x) = 20000 + 400x - 4x^2 \]

Step 4: Graph the Revenue Function

Plot the quadratic function \( R(x) = 20000 + 400x - 4x^2 \) on a graph. This is a parabola opening downwards.

Step 5: Find the Vertex of the Parabola

The vertex form of the quadratic function \( ax^2 + bx + c \) gives the maximum point. Use \( x = -\frac{b}{2a} \) to find the vertex. Here, \( a = -4 \) and \( b = 400 \): \[ x = -\frac{400}{2(-4)} = 50 \] Therefore, the number of unsold seats that will produce the maximum revenue is 50.

Step 6: Calculate the Maximum Revenue

Substitute \( x = 50 \) into \( R(x) \) to find the maximum revenue: \[ R(50) = 20000 + 400 \times 50 - 4 \times 50^2 = 30000 \] Therefore, the maximum revenue is \( 30000 \) dollars.

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function with the form: \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are coefficients and \(a eq 0\). For the exercise given, the revenue function is a quadratic function. It is described by the equation:
\(R(x) = 20000 + 400x - 4x^2\).
The characteristic feature of quadratic functions is their parabolic graph, which can open upwards or downwards depending on the sign of \(a\).
- If \(a > 0\), the parabola opens upwards, and the function has a minimum value.- If \(a < 0\), the parabola opens downwards, and the function has a maximum value.
In our case, since \(a = -4\) (which is less than zero), the parabola opens downwards, meaning the revenue function will have a maximum value at its vertex.

Maximum Revenue

To find the maximum revenue, we focus on the vertex of the revenue function's parabola. Since the parabola opens downwards (as discussed earlier), the vertex represents the highest point on the graph. The maximum revenue occurs at this vertex.
To find the vertex of a parabola given by \(ax^2 + bx + c\), we use the formula \(x = -\frac{b}{2a}\). For our function \(R(x) = 20000 + 400x - 4x^2\), we have:
- \(a = -4\)- \(b = 400\)- \(c = 20000\).
Plugging in these values, we get:
\(x = -\frac{400}{2(-4)} = 50\).
So the number of unsold seats that will produce the maximum revenue is \(50\). We can then substitute \(x = 50\) back into \(R(x)\) to find the maximum revenue:
\(R(50) = 20000 + 400 \times 50 - 4 \times 50^2 = 30000\).
Therefore, the maximum revenue is \(30000\) dollars.

Parabola Vertex

The vertex of a parabola is the critical point where it changes direction. For a downward-opening parabola, this is the maximum point, while for an upward-opening parabola, it is the minimum point. For the quadratic function \(R(x) = 20000 + 400x - 4x^2\), the vertex is crucial in finding the maximum revenue.
To determine the vertex, use the formula \(x = -\frac{b}{2a}\) where \(a\) and \(b\) are coefficients from the quadratic equation \(ax^2 + bx + c\). For our function, \(a = -4\) and \(b = 400\) resulting in:
\(x = -\frac{400}{2(-4)} = 50\).
This tells us that the number of unsold seats for maximum revenue is \(50\). Substituting \(x = 50\) into \(R(x)\) provides the maximum revenue value of \(30000\) dollars. In essence, the vertex calculation shows both the optimal number of unsold seats and the maximum revenue achievable.

Algebraic Problem-Solving

Algebraic problem-solving involves breaking down a problem into manageable steps using algebraic principles. In the exercise at hand, we begin by defining our variables and expressions.
1. **Define Variables:** Let \(x\) represent the number of unsold seats. The number of passengers, therefore, is \(100 - x\). 2. **Define Fare:** The fare can be expressed as \(200 + 4x\). 3. **Revenue Function:** The revenue is the number of passengers multiplied by the fare: \(R(x) = (100 - x)(200 + 4x)\). Simplifying, we get the quadratic function: \(R(x) = 20000 + 400x - 4x^2\). 4. **Graphing the Function:** Plotting \(R(x)\) shows a parabola opening downward. 5. **Identify Vertex:** Using \(x = -\frac{b}{2a}\), we find \(x = 50\). Substituting back into \(R(x)\) gives the maximum revenue.
Each step utilizes fundamental algebraic techniques to derive meaningful insights and solutions. Understanding and following these steps is key to solving similar algebraic problems efficiently.