Maximizing Profit Understanding Y = -6x^2 + 100x - 180

This article delves into the quadratic equation y = -6x^2 + 100x - 180, exploring its significance in a real-world scenario involving the selling price of soccer balls (x) and the resulting daily profit (y). Understanding this equation allows us to determine the optimal selling price for maximizing profit. We will dissect the equation's components, analyze its graph, and ultimately, find the vertex, which represents the selling price that yields the highest daily profit. This exploration will provide a comprehensive understanding of how mathematical models, particularly quadratic equations, can be used to make informed business decisions.

Decoding the Quadratic Equation: y = -6x^2 + 100x - 180

At its core, the equation y = -6x^2 + 100x - 180 is a quadratic equation, a fundamental concept in algebra. The general form of a quadratic equation is ax^2 + bx + c, where a, b, and c are constants. In our specific equation, a = -6, b = 100, and c = -180. The negative coefficient of the x^2 term (a = -6) is crucial, as it indicates that the parabola opens downwards, meaning the graph has a maximum point, which is essential for our profit maximization problem.

The variable 'x' represents the selling price of each soccer ball. This is our independent variable, the factor we can directly control. The variable 'y' represents the daily profit derived from selling soccer balls. This is our dependent variable, as the profit is directly affected by the selling price. The equation, therefore, models the relationship between these two variables.

The coefficient '100' (b = 100) represents the linear relationship between the selling price and the profit. As the selling price increases, the profit also tends to increase, at least initially. However, this positive effect is counteracted by the quadratic term, which we will explore further.

The constant '-180' (c = -180) represents the fixed costs or initial losses, regardless of the selling price. This could include costs such as manufacturing, rent, or marketing expenses. It's important to note that even if no soccer balls are sold (x = 0), there is still a loss of $180, reflecting the initial investment.

Understanding these components provides a solid foundation for analyzing the equation's graph and identifying the selling price that maximizes profit. The next sections will delve into the graphical representation of this equation and the significance of the vertex.

Visualizing Profit: Graphing the Quadratic Equation

To truly grasp the behavior of the equation y = -6x^2 + 100x - 180, visualizing it through a graph is indispensable. The graph of a quadratic equation is a parabola, a U-shaped curve. Since the coefficient of the x^2 term is negative (a = -6), the parabola opens downwards, creating an inverted U-shape. This downward-opening shape is crucial because it signifies the existence of a maximum point, known as the vertex, which represents the highest profit achievable.

The x-axis of the graph represents the selling price of each soccer ball (x), while the y-axis represents the daily profit (y). Each point on the parabola corresponds to a specific selling price and the associated profit. By tracing the curve, we can observe how the profit changes as the selling price varies.

Initially, as the selling price increases, the profit also increases, reflected in the upward slope of the parabola's left side. This makes intuitive sense: selling soccer balls at a higher price leads to greater revenue. However, this trend doesn't continue indefinitely. As the selling price increases further, the profit eventually begins to decrease, illustrated by the downward slope of the parabola's right side. This is because at some point, the higher price may deter customers, leading to fewer sales and, consequently, lower overall profit.

The x-intercepts of the graph, the points where the parabola crosses the x-axis (y = 0), are particularly significant. These points represent the break-even points, the selling prices at which the daily profit is zero. At these prices, the revenue generated from selling soccer balls exactly covers the costs. Selling at a price below the lower x-intercept or above the higher x-intercept results in a loss.

The vertex, the highest point on the parabola, holds the key to maximizing profit. Its x-coordinate represents the optimal selling price, the price that yields the maximum daily profit. The y-coordinate of the vertex represents this maximum profit. Finding the vertex is therefore the central goal in understanding this equation and its practical implications for soccer ball sales.

Finding the Sweet Spot: Determining the Vertex and Maximum Profit

The vertex of the parabola represented by the equation y = -6x^2 + 100x - 180 is the critical point for maximizing profit. As previously established, the vertex represents the highest point on the parabola, where the y-coordinate corresponds to the maximum profit and the x-coordinate corresponds to the optimal selling price.

There are two primary methods for determining the vertex: completing the square and using the vertex formula. Let's explore both:

1. Completing the Square: This method involves rewriting the quadratic equation in vertex form, which is y = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex. This method can be more involved but provides a deeper understanding of the equation's structure.

   • First, factor out the coefficient of the x^2 term (-6) from the first two terms: y = -6(x^2 - (50/3)x) - 180.
   • Next, complete the square inside the parentheses. To do this, take half of the coefficient of the x term (-50/3), square it ((25/3)^2 = 625/9), and add and subtract it inside the parentheses: y = -6(x^2 - (50/3)x + 625/9 - 625/9) - 180.
   • Rewrite the expression inside the parentheses as a squared term: y = -6((x - 25/3)^2 - 625/9) - 180.
   • Distribute the -6 and simplify: y = -6(x - 25/3)^2 + 1250/3 - 180.
   • Combine the constants: y = -6(x - 25/3)^2 + 710/3.

   Now the equation is in vertex form. The vertex is at the point (25/3, 710/3), which is approximately (8.33, 236.67).

2. Vertex Formula: A more direct approach is to use the vertex formula. For a quadratic equation in the form y = ax^2 + bx + c, the x-coordinate of the vertex (h) is given by h = -b / 2a. Once we have the x-coordinate, we can substitute it back into the original equation to find the y-coordinate (k).

   • In our equation, a = -6 and b = 100, so h = -100 / (2 * -6) = 100 / 12 = 25/3, which is approximately 8.33.
   • Substitute x = 25/3 into the original equation to find the y-coordinate: y = -6(25/3)^2 + 100(25/3) - 180 = 710/3, which is approximately 236.67.

   Again, we find the vertex to be approximately (8.33, 236.67).

Both methods yield the same result. The x-coordinate of the vertex, approximately 8.33, represents the optimal selling price of each soccer ball to maximize profit. The y-coordinate, approximately 236.67, represents the maximum daily profit achievable at this selling price.

Practical Implications: Maximizing Soccer Ball Sales Profit

The analysis of the quadratic equation y = -6x^2 + 100x - 180 provides valuable insights for maximizing profit in soccer ball sales. The key takeaway is the identification of the optimal selling price, which we determined to be approximately $8.33 per soccer ball. At this price, the business can expect to achieve a maximum daily profit of approximately $236.67.

This finding has significant practical implications:

• Pricing Strategy: The business should set the selling price of each soccer ball close to $8.33 to maximize profitability. Deviating significantly from this price, either higher or lower, will likely result in reduced profits.
• Inventory Management: Knowing the optimal selling price and the potential profit allows for better inventory management. The business can estimate demand based on this price and adjust production or purchasing accordingly.
• Cost Analysis: The constant term in the equation, -180, represents the fixed costs. Understanding these costs is crucial for evaluating the overall profitability of the business. If fixed costs are too high, the maximum profit may be insufficient to sustain the business.
• Market Analysis: The equation assumes a specific relationship between selling price and profit. However, market conditions can change. It's essential to regularly review and adjust the equation based on real-world sales data and competitor pricing.

Beyond the specific example of soccer ball sales, this analysis demonstrates the power of mathematical modeling in business decision-making. Quadratic equations, in particular, are useful for representing scenarios where there is an optimal value, such as maximizing profit, minimizing cost, or optimizing production levels.

By understanding the underlying mathematical principles and applying them to real-world situations, businesses can make more informed decisions and improve their overall performance. The equation y = -6x^2 + 100x - 180 serves as a compelling example of how mathematics can be a powerful tool for business success.

In conclusion, the quadratic equation y = -6x^2 + 100x - 180 is more than just an abstract mathematical concept; it's a powerful tool for understanding and optimizing business outcomes. By carefully analyzing its components, graphing its curve, and determining its vertex, we can identify the optimal selling price for soccer balls that maximizes daily profit. This approach highlights the importance of mathematical modeling in making informed business decisions and driving success.