A company's profit is modeled by \(P(x) = -2x^2 + 80x - 150\), where \(x\) is the number of units sold. What is the maximum profit?
["Title: Maximizing Company Profit with Quadratic Model (P(x) = -2x^2 + 80x - 150)", "Understanding how profit is influenced by units sold is crucial for businesses seeking growth and sustainability. One powerful way to analyze profit is by modeling it mathematically—specifically using a quadratic function. In this article, we explore how the profit function (P(x) = -2x^2 + 80x - 150) helps determine the maximum profit achievable, and we calculate the exact value and the number of units that yield this optimal result.", "---", "### Understanding the Profit Function", "The given profit model is:", "[\nP(x) = -2x^2 + 80x - 150\n]", "This is a quadratic function with a negative leading coefficient, meaning its graph is a downward-opening parabola. The maximum profit occurs at the vertex of this parabola. Since profit drops as units sold go beyond this point, identifying the vertex is key to strategic decision-making.", "---", "### Finding the Maximum Profit", "For any quadratic function in the form (P(x) = ax^2 + bx + c), the vertex occurs at:", "[\nx = -\frac{b}{2a}\n]", "Here, (a = -2), (b = 80), so:", "[\nx = -\frac{80}{2 \ imes (-2)} = -\frac{80}{-4} = 20\n]", "Thus, the optimal number of units to sell for maximum profit is 20 units.", "---", "### Calculating Maximum Profit", "Substitute (x = 20) into the profit function:", "[\nP(20) = -2(20)^2 + 80(20) - 150\n]", "Calculate step-by-step:", "[\nP(20) = -2(400) + 1600 - 150 = -800 + 1600 - 150 = 650\n]", "So, the maximum profit is $650 when 20 units are sold.", "---", "### Why This Matters for Business Strategy", "Modeling profit as a quadratic function enables businesses to:", "- Identify the ideal production and sales level for maximum returns\n- Avoid overselling, which lowers profit due to marginal cost increases\n- Plan inventory, pricing, and marketing efforts around peak profitability\n- Make data-driven decisions about expanding capacity or adjusting pricing", "---", "### Conclusion", "The profit function (P(x) = -2x^2 + 80x - 150) reaches its maximum when 20 units are sold, yielding a peak profit of $650. By modeling profit with this quadratic equation, companies gain valuable insight into optimal performance levels and can align operations accordingly. Maximizing profit isn’t just about selling more—it’s about selling smart, at precisely the right volume.", "---", "Keywords: Profit maximization, quadratic profit model, maximize profit function, business optimization, (P(x) = -2x^2 + 80x - 150), maximum profit calculation, business strategy, quadratic applications in business", "For more insights on leveraging mathematical models to boost business performance, follow us on [Your Business Insights Blog] and explore more advanced analytical tools."]
