What Does P(20) = 150 Mean In The Context Of The Problem?

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In the world of mathematics, functions are powerful tools that help us model real-world situations. One such application is in understanding profit and loss scenarios in business. The function provided, p(x) = 8x - 10, is a linear function that models the profit (in dollars) from selling x books. Understanding how to interpret these functions is essential for making informed decisions. In this article, we will delve deep into this function, focusing particularly on the meaning of p(20) = 150, and explore its broader implications.

Let's start by dissecting the function p(x) = 8x - 10. In this equation:

  • p(x) represents the profit in dollars.
  • x signifies the number of books sold.
  • The number 8 is the profit per book. Each book sold contributes $8 to the total revenue.
  • The number -10 represents a fixed cost or initial investment, such as the cost of printing or marketing the books. This cost is incurred regardless of the number of books sold.

This function tells us that for every book sold, the profit increases by $8, but we need to account for an initial cost of $10. Therefore, to start making a profit, more than a few books need to be sold.

Before we focus on p(20) = 150, let's briefly interpret p(10) = 70. This tells us that when 10 books are sold, the profit is $70. We can verify this by substituting x = 10 into the function:

  • p(10) = 8(10) - 10 = 80 - 10 = 70

This calculation confirms that selling 10 books results in a profit of $70. This helps set the stage for understanding what happens when we sell even more books.

Now, let's turn our attention to the core question: What does p(20) = 150 mean in the context of the problem? In simple terms, p(20) = 150 means that when 20 books are sold, the profit is $150. This is a specific point on the graph of the function p(x), where the x-coordinate is 20 (books sold) and the y-coordinate is 150 (profit in dollars). To further illustrate, we can calculate this:

  • p(20) = 8(20) - 10 = 160 - 10 = 150

Thus, our calculation confirms that selling 20 books yields a profit of $150. This value takes into account both the revenue generated from selling the books and the initial costs incurred.

Breaking Down the Profit

To understand the $150 profit better, we can break it down:

  • Revenue from selling 20 books: 20 books * $8/book = $160
  • Initial cost: $10
  • Profit: Revenue - Initial Cost = $160 - $10 = $150

This breakdown makes it clear how the profit is derived from the revenue and the initial costs. It showcases the importance of selling enough books to cover the initial investment and then generate a profit. Understanding this breakdown is crucial for making strategic decisions about pricing, production, and marketing.

Understanding the function p(x) = 8x - 10 and the meaning of p(20) = 150 has practical implications for anyone in the business of selling books, or any product for that matter. Here are a few key takeaways:

  1. Profitability Threshold: The function helps determine the number of books that need to be sold to break even (i.e., when profit is zero). By setting p(x) = 0, we can solve for x:

    • 0 = 8x - 10
    • 8x = 10
    • x = 10/8 = 1.25

    Since we can't sell a fraction of a book, we need to sell at least 2 books to start making a profit. This is a crucial piece of information for setting sales targets.

  2. Scaling Profit: The linear nature of the function means that for each additional book sold, the profit increases by a constant amount ($8 in this case). This makes it easy to predict profit at different sales volumes. For instance, if we want to make a profit of $500, we can solve for x:

    • 500 = 8x - 10
    • 510 = 8x
    • x = 510/8 = 63.75

    So, we need to sell approximately 64 books to make a profit of $500. This allows for planning and setting realistic financial goals.

  3. Cost Management: The function also highlights the impact of fixed costs on profitability. If the initial cost were higher, we would need to sell more books to break even and achieve a desired profit level. Therefore, managing costs is just as important as increasing sales.

  4. Pricing Strategy: The profit per book ($8) is a key factor in determining the overall profitability. If the selling price per book is increased, the profit per book would also increase, leading to higher overall profits for the same number of books sold. This helps in making informed decisions about pricing strategies.

To further illustrate the utility of this function, let's consider a few hypothetical scenarios:

Scenario 1: Increased Initial Costs

Suppose the initial costs increase from $10 to $50 due to additional marketing expenses. The new profit function would be p(x) = 8x - 50. Now, let's calculate the profit from selling 20 books:

  • p(20) = 8(20) - 50 = 160 - 50 = 110

In this scenario, selling 20 books yields a profit of $110, which is less than the previous $150. This demonstrates how increased costs can impact profitability. To break even with these higher costs, we need to sell more books:

  • 0 = 8x - 50
  • 8x = 50
  • x = 50/8 = 6.25

We now need to sell at least 7 books to start making a profit, compared to the previous 2 books.

Scenario 2: Decreased Profit per Book

Suppose the profit per book decreases from $8 to $6 due to a discount offered to customers. The new profit function becomes p(x) = 6x - 10. Now, let's calculate the profit from selling 20 books:

  • p(20) = 6(20) - 10 = 120 - 10 = 110

With a lower profit per book, selling 20 books results in a profit of $110, again less than the original $150. This highlights the importance of the profit margin on each item sold. The break-even point in this scenario is:

  • 0 = 6x - 10
  • 6x = 10
  • x = 10/6 = 1.67

We still need to sell 2 books to break even, but the overall profitability is reduced.

Scenario 3: Combining Increased Costs and Decreased Profit

If we combine both scenarios, with initial costs increasing to $50 and profit per book decreasing to $6, the profit function is p(x) = 6x - 50. The profit from selling 20 books is:

  • p(20) = 6(20) - 50 = 120 - 50 = 70

Selling 20 books now results in a profit of just $70, significantly lower than the original $150. The break-even point is:

  • 0 = 6x - 50
  • 6x = 50
  • x = 50/6 = 8.33

We need to sell at least 9 books to break even in this combined scenario. This comprehensive view illustrates the complex interplay between costs, pricing, and sales volume in determining profitability.

To further enhance our understanding, let's visualize the function p(x) = 8x - 10. The graph of this function is a straight line with a slope of 8 and a y-intercept of -10. The slope indicates the rate of change of profit with respect to the number of books sold. In this case, for every additional book sold, the profit increases by $8. The y-intercept represents the initial cost or loss when no books are sold.

On the graph:

  • The point (10, 70) represents p(10) = 70, meaning that when 10 books are sold, the profit is $70.
  • The point (20, 150) represents p(20) = 150, meaning that when 20 books are sold, the profit is $150.
  • The x-intercept (where the line crosses the x-axis) represents the break-even point, which we calculated to be approximately 1.25 books. Since we can’t sell a fraction of a book, we need to sell 2 books to start making a profit.

A visual representation can often make it easier to grasp the overall trend and relationship between variables, reinforcing the mathematical understanding of the function.

In conclusion, understanding the function p(x) = 8x - 10 and interpreting values like p(20) = 150 provides valuable insights into profit calculations. p(20) = 150 specifically means that by selling 20 books, the profit earned is $150, considering both the revenue generated and the initial costs incurred. This understanding has practical implications for businesses, aiding in setting sales targets, managing costs, and making informed pricing decisions. By analyzing and interpreting mathematical functions in real-world contexts, we can make better, data-driven decisions. The ability to translate mathematical expressions into practical understanding is a crucial skill in many areas, from business and finance to science and engineering. Mastering this skill enables a more nuanced and strategic approach to problem-solving and decision-making.