1. **Problem Statement:** Suppose a company produces and sells $x$ units of a product. The cost to produce $x$ units is given by the cost function $$C(x) = 500 + 20x + 0.05x^2,$$ where 500 is the fixed cost, 20x is the variable cost per unit, and $0.05x^2$ represents increasing costs due to overtime and inefficiencies. The revenue from selling $x$ units is given by the revenue function $$R(x) = 50x - 0.1x^2,$$ where the price decreases as more units are sold due to market saturation. Find the number of units $x$ that maximizes the company's profit and calculate the maximum profit. 2. **Formula and Rules:** Profit is defined as revenue minus cost: $$P(x) = R(x) - C(x).$$ To find the maximum profit, we need to find the critical points by taking the derivative of $P(x)$ with respect to $x$ and setting it equal to zero. Then, use the second derivative test to confirm if it is a maximum. 3. **Calculate the profit function:** $$P(x) = (50x - 0.1x^2) - (500 + 20x + 0.05x^2) = 50x - 0.1x^2 - 500 - 20x - 0.05x^2 = (50x - 20x) - (0.1x^2 + 0.05x^2) - 500 = 30x - 0.15x^2 - 500.$$ 4. **Find the derivative of the profit function:** $$P'(x) = \frac{d}{dx}(30x - 0.15x^2 - 500) = 30 - 0.3x.$$ 5. **Set the derivative equal to zero to find critical points:** $$30 - 0.3x = 0 \implies 0.3x = 30 \implies x = \frac{30}{0.3} = 100.$$ 6. **Second derivative test:** $$P''(x) = \frac{d}{dx}(30 - 0.3x) = -0.3.$$ Since $P''(100) = -0.3 < 0,$ the profit function is concave down at $x=100,$ indicating a maximum. 7. **Calculate the maximum profit:** $$P(100) = 30(100) - 0.15(100)^2 - 500 = 3000 - 0.15(10000) - 500 = 3000 - 1500 - 500 = 1000.$$ **Answer:** The company maximizes profit by producing and selling 100 units, resulting in a maximum profit of 1000.