A companyâs profit \( P \) is given by \( P(x) = -5x^2 + 150x - 1000 \), where \( x \) is the number of units sold. Find the number of units sold to maximize profit.
["Maximizing Profit: Finding the Optimal Number of Units Sold for ( P(x) = -5x^2 + 150x - 1000 )", "In any business, maximizing profit is a key objective, especially when pricing and sales volume directly affect financial outcomes. For a company whose profit function is defined as ( P(x) = -5x^2 + 150x - 1000 ), where ( x ) represents the number of units sold, identifying the optimal sales volume is crucial for peak performance.", "### Understanding the Profit Function", "The given profit function is a quadratic equation in the standard form:", "[\nP(x) = ax^2 + bx + c\n]", "Here, ( a = -5 ), ( b = 150 ), and ( c = -1000 ). Because the coefficient ( a ) is negative, the parabola opens downward, meaning it has a maximum point—representing the maximum achievable profit.", "### How to Find the Maximum Profit", "The vertex of a parabola given by ( ax^2 + bx + c ) occurs at:", "[\nx = -\frac{b}{2a}\n]", "This formula gives the value of ( x ) that maximizes (or minimizes) the quadratic function, depending on the sign of ( a ). Since ( a = -5 < 0 ), the vertex corresponds to the maximum profit.", "### Substituting into the Formula", "Plug ( a = -5 ) and ( b = 150 ) into the vertex formula:", "[\nx = -\frac{150}{2(-5)} = -\frac{150}{-10} = 15\n]", "### Interpretation", "The number of units the company must sell to maximize profit is 15 units.", "At this sales volume:", "- The profit is maximized.\n- The downward curvature of the profit function ensures no higher profit is achievable at any other production level.\n- Selling fewer than 15 units yields diminishing returns, while selling more than 15 units results in reduced profit due to the negative quadratic coefficient.", "### Verifying the Maximum", "For completeness, calculate ( P(15) ):", "[\nP(15) = -5(15)^2 + 150(15) - 1000 = -5(225) + 2250 - 1000 = -1125 + 2250 - 1000 = 125\n]", "Thus, the maximum profit is $125 when 15 units are sold.", "### Conclusion", "For a company with profit modeled by ( P(x) = -5x^2 + 150x - 1000 ), the optimal number of units to sell to maximize profit is 15 units. Business leaders can use this insight to align production, pricing, and marketing strategies with the profit-maximizing sales volume, ensuring efficient use of resources and maximized financial gains.", "---", "Keywords: profit maximization, quadratic profit function, maximize profit, optimal sales volume, business strategy, operations management, P(x), revenue optimization", "Meta Description: Discover how to maximize profit using the quadratic function ( P(x) = -5x^2 + 150x - 1000 ). Find the optimal 15 units sold for maximum earnings."]
