Problem:** A company’s profit \( P \) (in thousands) is modeled by \( P = -2x^2 + 12x - 5 \), where \( x \) is the number of units sold (in hundreds). What is the maximum profit?

["Understanding Profit Maximization: How to Find the Maximum Profit from a Quadratic Model", "When managing a business, understanding how profit responds to sales volume is essential for making informed decisions. Many companies use mathematical models to predict profit, and one common form is the quadratic equation. In this article, we explore a typical profit model and determine the maximum profit achievable by optimizing the number of units sold.", "---", "### The Profit Equation: A Quadratic Model", "Consider the profit function for a company given by:", "[\nP = -2x^2 + 12x - 5\n]", "Here,\n- ( P ) represents profit in thousands of dollars,\n- ( x ) is the number of units sold in hundreds (so selling 100 units equals ( x = 1 )).", "This function is a quadratic equation in standard form ( ax^2 + bx + c ), where ( a = -2 ), ( b = 12 ), and ( c = -5 ).", "Because the coefficient of ( x^2 ) is negative (( a < 0 )), the parabola opens downward. This means the profit function has a maximum point — the vertex — where profit peaks.", "---", "### Finding the Maximum Profit Using the Vertex Formula", "The maximum value of a quadratic function occurs at the vertex. For any quadratic ( y = ax^2 + bx + c ), the ( x )-coordinate of the vertex is given by:", "[\nx = -\frac{b}{2a}\n]", "Substitute ( a = -2 ) and ( b = 12 ):", "[\nx = -\frac{12}{2(-2)} = -\frac{12}{-4} = 3\n]", "So, the optimal number of units (in hundreds) to maximize profit is ( x = 3 ), corresponding to 300 units sold.", "---", "### Calculating the Maximum Profit", "Now substitute ( x = 3 ) back into the profit equation:", "[\nP = -2(3)^2 + 12(3) - 5\n]", "[\nP = -2(9) + 36 - 5 = -18 + 36 - 5 = 13\n]", "Thus, the maximum profit is 13 thousand dollars.", "---", "### Why This Matters for Business Strategy", "Understanding the maximum profit helps businesses plan production, pricing, and marketing. Since profit drops as units sold move away from ( x = 3 ), monitoring sales closely allows swift adjustments to avoid losses. Additionally, recognizing the shape of the profit curve supports decision-making around cost control and investment.", "---", "### Conclusion", "Modeling profit with a quadratic equation enables clear insight into performance predators. By calculating the vertex of ( P = -2x^2 + 12x - 5 ), we find the maximum profit occurs at 300 units sold and amounts to $13,000. Leveraging such mathematical tools empowers smarter, data-driven business operations.", "---", "Key Takeaways:\n- Quadratic profit models peak at the vertex.\n- For ( P = -2x^2 + 12x - 5 ), maximum profit of $13,000 is achieved at ( x = 3 ).\n- Optimizing at the vertex helps maximize financial returns.", "---\nKeywords: profit maximization, quadratic profit model, business optimization, maximum profit calculation, quadratic equation business application"]