Problem 87 - Applied Calculus

Question

GENERAL: Newsletters A newsletter has a maximum audience of 100 subscribers. The publisher estimates that she will lose 1 reader for each dollar she charges. Therefore, if she charges \(x\) dollars, her readership will be \((100-x)\) a. Multiply this readership by \(x\) (the price) to find her total revenue. Multiply out the resulting quadratic function. b. What price should she charge to maximize her revenue? [Hint: Find the value of \(x\) that maximizes this quadratic function.]

Step-by-Step Solution

Answer

The publisher should charge $50 to maximize revenue, achieving a total revenue of $2500.

Step 1: Understanding Revenue

Revenue is calculated by multiplying the number of subscribers by the price per subscription. Here, the readership is given as \((100-x)\) and the price is \(x\). Thus, the total revenue \(R\) can be expressed as a function of \(x\): \[ R(x) = x(100 - x) \]

Step 2: Expanding the Quadratic Function

Now, let's expand the quadratic function \(R(x) = x(100 - x)\). Use the distributive property to multiply out the expression:\[ R(x) = 100x - x^2 \] This is the quadratic function in standard form \(-x^2 + 100x\).

Step 3: Finding the Vertex of the Parabola

The function \(-x^2 + 100x\) is a downward opening parabola. The maximum point occurs at its vertex. To find the vertex, we use the formula for the x-coordinate of the vertex of a parabola \(ax^2 + bx + c\), which is given by:\[ x = -\frac{b}{2a} \] In our function, \(a = -1\) and \(b = 100\). Thus,\[ x = -\frac{100}{2(-1)} = 50 \] This means that the price that maximizes revenue is \(x = 50\).

Step 4: Calculating Maximum Revenue

To find the maximum revenue, substitute \(x = 50\) back into the revenue function:\[ R(50) = 50(100 - 50) = 50 \times 50 = 2500 \] This calculation shows that the maximum revenue is \(2500\) dollars when the price charged is \(50\) dollars.

Key Concepts

Revenue Maximization through Quadratic Functions

To maximize revenue, we must understand the relationship between price and readership. When applying a quadratic function like \( R(x) = -x^2 + 100x \), revenue is modeled with respect to pricing strategies. Quadratic functions are instrumental due to their parabolic nature, allowing us to determine the peak - the vertex - which signifies maximum values.
Maximizing revenue involves setting the price point so that total earnings reach their highest. In this exercise, the revenue is expressed as \( x(100-x) \), where \( x \) is the price. Expanding it to \( -x^2 + 100x \) shows us the structure of the problem: the price that maximizes revenue falls at the vertex of the parabola.
This function indicates a trade-off between price and number of subscribers. By solving it, you pinpoint optimal pricing strategies to harness maximum profits.

Understanding the Vertex of a Parabola

The vertex of a parabola holds essential information about the maximum or minimum value of a quadratic function. In a revenue context, this maximum identifies the highest possible revenue at a specific price.
The quadratic equation in the example is \( -x^2 + 100x \). The formula \( x = -\frac{b}{2a} \) provides the x-coordinate of the vertex. Here, \( a = -1 \) and \( b = 100 \), leading to \( x = 50 \). This x-value corresponds to the price maximizing revenue.
Understanding this aspect of quadratics empowers you to calculate ideal scenarios effortlessly, ensuring the business captures the height of its monetary potential without trial and error. Identifying the vertex translates into realizing the specific conditions under which the business operates optimally.

Characteristics of a Downward Opening Parabola

In a quadratic function like \( -x^2 + 100x \), the negative coefficient of \( x^2 \) indicates a downward opening parabola. This means the parabola arches downward, making its vertex the highest point on the curve, rather than lowest.
Such parabolas are typical in maximizing problems because they showcase a clear peak - the maximum value we are often seeking in optimization problems like revenue maximization. A downward opening parabola gives a visual representation where the curve rises to a point, and then declines.
Understanding this shape helps illustrate why the vertex signifies the maximum. Here, the top of the parabola directly correlates to the optimal pricing point for maximizing revenues. Thus, the nature of a downward opening parabola serves essential roles in solving real-world problems, such as optimizing profits or other maximum-seeking scenarios.