Problem 84 - Algebra and Trigonometry

Question

Long-Distance cost A telephone company offers two long-distance plans: Plan A: \(\quad \$ 25\) per month and 5\(€\) per minute Plan B: \(\$ 5\) per month and 12\(€\) per minute For how many minutes of long-distance calls would Plan B be financially advantageous?

Step-by-Step Solution

Answer

Plan B becomes advantageous at 3 or more minutes.

Step 1: Define the Cost for Each Plan

Let's establish equations for both plans. For Plan A, the cost function can be defined as follows: the monthly cost plus the cost per minute. Hence the cost for Plan A is \( C_A = 25 + 5x \), where \( x \) is the number of minutes. Similarly, for Plan B, the cost will be \( C_B = 5 + 12x \).

Step 2: Set Up an Inequality

We need to find out when Plan B becomes cheaper than Plan A. This can be done by setting up the inequality \( C_B < C_A \). Substituting the cost functions into the inequality gives us \( 5 + 12x < 25 + 5x \).

Step 3: Solve the Inequality

Solve the inequality \(5 + 12x < 25 + 5x \). First, subtract \(5x\) from both sides to get \( 7x < 20 \). Then, divide both sides by 7 to solve for \( x \), which gives \( x < \frac{20}{7} \approx 2.86 \).

Step 4: Determine the Minimum Whole Number

Since the number of minutes cannot be a fraction, round \(2.86\) up to the nearest whole number. Therefore, Plan B becomes advantageous at or above \(3\) minutes.

Key Concepts

Understanding Linear Inequality

Linear inequalities are vital in comparing costs, such as finding the number of minutes when a telephone plan becomes cheaper. A linear inequality looks like a linear equation (like a straight line) but uses inequality symbols such as less than (<) or greater than (>) instead of an equal sign.
For the given problem, we need to establish when Plan B is cheaper than Plan A. This situation is described by setting up the inequality using the cost functions designed in the solution.
The inequality we solve here is:

• Plan A cost: \( C_A = 25 + 5x \)
• Plan B cost: \( C_B = 5 + 12x \)

When \( C_B < C_A \), Plan B is cheaper, leading to the inequality \( 5 + 12x < 25 + 5x \). This inequality is solved by manipulating terms, much like you would in an equation, moving all terms involving \( x \) to one side and constant terms to the other. Finally dividing both sides by the coefficient of \( x \) simplifies to \( x < \frac{20}{7} \). Solving inequalities can rapidly determine conditions for decision-making, like which phone plan to choose.

The Role of Function Modeling

Function modeling is about creating equations that represent real-world scenarios. In the telephone plan comparison, each cost structure is modeled by a linear function:

• Plan A: \( C_A = 25 + 5x \)
• Plan B: \( C_B = 5 + 12x \)

Each function translates the fixed monthly charge and the additional cost per minute into a mathematical language, allowing us to easily compare them using algebra.
Function modeling helps in simplifying complex problems by using equations to depict crucial elements. Here, it allowed us to succinctly represent each plan's cost efficiently and find the solution to which plan is more economical based on usage. This model highlights the importance of translating verbal descriptions into mathematical ones to facilitate problem-solving through analysis and calculation.

Mastering Rounding Numbers

Rounding numbers is pivotal when dealing with decimals to simplify results, especially when they require whole number outcomes, like in this example where the number of minutes cannot be a fraction.
After solving the inequality \( x < \frac{20}{7} \approx 2.86 \), the number is rounded. Since you can't talk for a fractional part of a minute on the phone, we need to consider the nearest whole number.
Typically, we round up in these contexts, reflecting where a plan becomes advantageous at the first full minute beyond the calculated value. Thus, \( x \approx 2.86 \) rounds up to \( 3 \), meaning after 3 minutes, Plan B starts to be financially beneficial.
Rounding is practical and essential in real-life applications, ensuring solutions are adaptable to practical usage, which often doesn't accommodate partial values.