Problem 51 - Algebra for College Students

Question

A cable TV company has 1000 subscribers, each of whom pays \(\$ 15\) per month. On the basis of a survey, the company believes that for each decrease of \(\$ 0.25\) in the monthly rate, it could obtain 20 additional subscribers. At what rate will the maximum revenue be obtained, and how many subscribers will there be at that rate? 1100 subscribers at \(\$ 13.75\) per month

Step-by-Step Solution

Answer

Maximum revenue is obtained with 1100 subscribers at $13.75 a month.

Step 1: Define Revenue Function

The revenue obtained by the company is the product of the number of subscribers and the monthly rate. Let \( x \) be the number of 0.25 dollar decreases. Then, the number of subscribers will be \( 1000 + 20x \) and the rate will be \( 15 - 0.25x \). The revenue function is given by \( R(x) = (1000 + 20x)(15 - 0.25x) \).

Step 2: Simplify Revenue Function

Expand the revenue function: \[ R(x) = (1000 + 20x)(15 - 0.25x) \]\[ R(x) = 1000(15 - 0.25x) + 20x(15 - 0.25x) \]\[ R(x) = 15000 - 250x + 300x - 5x^2 \]\[ R(x) = 15000 + 50x - 5x^2 \].

Step 3: Determine Maximum Revenue

The function \( R(x) = 15000 + 50x - 5x^2 \) is a quadratic function opening downwards (since the coefficient of \( x^2 \) is negative). The maximum revenue occurs at the vertex of the parabola. Use the vertex formula \( x = -\frac{b}{2a} \) where \( a = -5 \) and \( b = 50 \).

Step 4: Calculate Optimal x-value

Substitute the values into the vertex formula:\[ x = -\frac{50}{2(-5)} = -\frac{50}{-10} = 5 \].The optimal number of decreases is 5.

Step 5: Calculate Maximum Rate and Subscribers

Substitute \( x = 5 \) back into the expressions for the rate and the number of subscribers:- Monthly rate: \( 15 - 0.25 \times 5 = 15 - 1.25 = 13.75 \).- Number of subscribers: \( 1000 + 20 \times 5 = 1000 + 100 = 1100 \).

Key Concepts

Quadratic Functions

Quadratic functions are a fundamental aspect of algebra and appear often in various real-world scenarios. They are polynomials of degree two, typically expressed in the standard form as \( y = ax^2 + bx + c \). The most distinctive feature of quadratic functions is their graph, which is a parabola. Key characteristics of parabolas include:

• They have a highest or lowest point known as the vertex depending on the sign of \( a \).
• If \( a > 0 \), the parabola opens upwards, and if \( a < 0 \), it opens downwards.
• The axis of symmetry is a vertical line that passes through the vertex.

Understanding quadratic functions is crucial for solving problems related to maximizing or minimizing values, such as in the original exercise with the revenue function.

Vertex Formula

The vertex of a quadratic function is a critical point where the maximum or minimum value of the function occurs. To find the vertex, we use the vertex formula: \[ x = -\frac{b}{2a} \]This formula derives from the standard form \( y = ax^2 + bx + c \) of the quadratic equation. It helps pinpoint exactly where the function changes direction in its parabola form.The vertex provides:

• The value of \( x \) for which the function obtains its highest or lowest point.
• Insight into how changes in \( b \) and \( a \) affect the location and nature of the parabola.

In the context of revenue optimization, as seen in the problem, identifying the vertex helps in determining the optimal monthly rate and number of subscribers for maximum revenue.

Revenue Function

A revenue function is a specific application of algebra that involves computing the total income generated from sales or subscriptions. It is typically expressed as the product of the quantity sold and the price per unit.In the exercise, the revenue function was derived as:\[ R(x) = (1000 + 20x)(15 - 0.25x) \]Where:

• The first term \( 1000 + 20x \) represents the number of subscribers.
• The second term \( 15 - 0.25x \) denotes the monthly rate per subscriber.

Upon expanding and simplifying, this revenue function can be written in the form of a quadratic equation. Analyzing this equation allows companies to make strategic decisions regarding pricing and subscription strategies to optimize total revenue.

Educational Algebra Problems

Educational algebra problems often simulate real-life scenarios to help students apply mathematical concepts and bolster problem-solving skills. These problems improve critical thinking and familiarize students with algebraic translations of everyday situations. When tackling such problems:

• Identify what the variables represent in the context of the problem.
• Translate the situation into algebraic expressions and equations.
• Use suitable algebraic techniques, such as solving quadratic equations, to find solutions.
• Interpret the solutions in context, ensuring they make sense in the real-world scenario described.

Solutions to these problems often illuminate broader concepts, like how changing one element of a business plan can significantly impact outcomes, aided by practical application of quadratic functions and other algebraic tools.