A company’s profit \( P(x) \) in thousands of dollars is modeled by the function \( P(x) = -2x^2 + 40x - 150 \), where \( x \) is the number of units sold in hundreds. What is the maximum profit and how many units must be sold to achieve it?

["Maximizing Profit: The Optimal Sales Level and Maximum Profit Using a Quadratic Model", "Understanding how profit is affected by sales volume is crucial for any business aiming to grow sustainably. For a company whose profit ( P(x) ) is modeled by the quadratic function ( P(x) = -2x^2 + 40x - 150 ), where ( x ) represents the number of units sold in hundreds, identifying the maximum profit and the corresponding sales volume can guide strategic decisions. This article explains how to determine the maximum profit and the optimal number of units to sell using mathematical modeling.", "### What the Function Represents", "The profit function ( P(x) = -2x^2 + 40x - 150 ) is a downward-opening parabola because the coefficient of ( x^2 ) is negative (( -2 )). This means the function has a single peak — the vertex — which corresponds to the maximum profit. The variable ( x ) is defined as thousands of units sold, so dividing ( x ) by 100 gives the actual number of units.", "### Finding the Maximum Profit", "The maximum value of a quadratic function ( ax^2 + bx + c ) occurs at ( x = -\frac{b}{2a} ). For our function:", "- ( a = -2 )\n- ( b = 40 )\n- ( c = -150 )", "Compute the vertex:", "[\nx = -\frac{40}{2 \ imes (-2)} = -\frac{40}{-4} = 10\n]", "So, the maximum profit occurs when 10 hundred units are sold, or 1,000 units.", "### Calculating the Maximum Profit", "Substitute ( x = 10 ) into the profit function:", "[\nP(10) = -2(10)^2 + 40(10) - 150 = -2(100) + 400 - 150 = -200 + 400 - 150 = 50\n]", "Since ( P(x) ) is measured in thousands of dollars, the maximum profit is $50,000.", "### Why This Matters", "Selling 1,000 units at this volume maximizes the company’s earnings, identified through calculus-based optimization embedded in the quadratic model. Recognizing this peak helps in forecasting revenue, planning production, and strategically setting sales targets to avoid underperformance or wasted resources.", "### Summary", "- Maximum Profit: $50,000\n- Units to Sell for Maximum Profit: 1,000 units (i.e., ( x = 10 ))", "Leveraging profit modeling tools like ( P(x) = -2x^2 + 40x - 150 ) empowers businesses to make data-driven decisions, optimize operations, and enhance profitability with precision.", "---", "Category: Business Mathematics | Profit Maximization | Quadratic Functions | Sales Strategy | Small Business Finance\nKeywords: profit function, maximize profit, quadratic profit model, optimal sales units, business growth, revenue optimization"]