Problem 126 - College Algebra Essentials

Question

Use the strategy for solving word problems, modeling the verbal conditions of the problem with a linear inequality. A company manufactures and sells blank audio cassette tapes. The weekly fixed cost is \(\$ 10,000\) and it costs \(\$ 0.40\) to produce each tape. The selling price is \(\$ 2.00\) per tape. How many tapes must be produced and sold each week for the company to generate a profit?

Step-by-Step Solution

Answer

To generate a profit, the company needs to produce and sell more than 6250 tapes weekly.

Step 1: Identify Costs and Revenues

First, the cost components need to be identified. The company has a weekly fixed cost of \$10,000 and a production cost of \$0.40 per tape. The company sells each tape for \$2.00, which represents the company's revenue per tape.

Step 2: Formulate the Cost and Revenue Equations

The total weekly cost, \( C \), can be modeled as \( C = 10000 + 0.40T \), where \( T \) is the number of tapes produced and sold each week. Similarly, the weekly revenue, \( R \), can be modeled as \( R = 2.00T \).

Step 3: Set Up the Profit Inequality

The company generates a profit when the revenue exceeds the costs, or mathematically, when \( R > C \). Substituting the cost and revenue equations results in the inequality \( 2.00T > 10000 + 0.40T \).

Step 4: Solve the Inequality

Solving this inequality will yield the minimum number of tapes that need to be sold to ensure a profit. Subtracting \( 0.40T \) from both sides gives \( 1.60T > 10000 \). Dividing by 1.60 yields \( T > 6250 \). So, to generate a profit, the company needs to produce and sell more than 6250 tapes weekly.

Step 5: Validate and Understand the Result

This means that if the company sells exactly 6250 tapes, they will break even. The company will start making a profit if they sell anything above this number. Thus, the solution is reasonable given that costs are linear and constant per tape, and all tapes produced are sold at a constant price.

Key Concepts

Word Problems in Linear Inequalities

Word problems are a common way to apply math to real-life scenarios, helping us solve practical challenges. In this case, we look at a manufacturing company trying to determine the number of tapes they need to sell in order to make a profit. Generally, word problems like these involve:

• Identifying the known values and variables: fixed costs, variable costs per unit, selling price per unit.
• Understanding the relationship between these values: forming equations and inequalities to model the problem.
• Using logical reasoning to interpret queries like how many units must be sold for profit.

Linear inequalities are especially useful in these scenarios because they allow us to set a condition, such as the revenue needing to exceed costs, leading us to the solution through systematic steps.
Always start by clearly defining your variables and translating the given situation into an inequality or equation.

Cost-Revenue Analysis

Cost-revenue analysis is key to determining profitability in business scenarios. In the problem, the fixed costs are established at \(\\)10,000\( weekly, with a variable cost of \)\\(0.40\) per tape.
The next step is to define the revenue, which depends on how many tapes are sold, each priced at \(\\)2.00$. These two components are expressed in equations:

• Total Cost, \( C = 10,000 + 0.40T \), where \( T \) is the number of tapes.
• Total Revenue, \( R = 2.00T \).

In cost-revenue analysis, the goal is to understand how many units (tapes, in this case) need to be sold for revenue to outweigh costs, leading to profit. Breaking down these components clearly helps form a solid foundation for solving the inequality and grasping the broader business implications.

Profit Calculation from Linear Inequalities

Profit is what remains when costs are subtracted from revenue. Using linear inequalities, we can express the condition for profit as \( R > C \). For the company in our problem, substituting known cost and revenue equations gives:\[ 2.00T > 10,000 + 0.40T \]
To solve this inequality:

• Simplify the inequality by subtracting \( 0.40T \) from both sides.
• This results in \( 1.60T > 10,000 \).
• Divide both sides by 1.60 to isolate \( T \).

This gives \( T > 6250 \), meaning more than 6250 tapes must be sold weekly for a profit. The calculation highlights how businesses use these mathematical models to decide on production and sales strategies.