Let's break down the problem step by step: ### Given: The profit function for the moving company is: \[ P(x) = 0.8x^3 - 6x^2 + 4 - 1.6 \ln(x) - 125 \] The company has already done \( x = 10 \) moves. We are asked: **Estimate the marginal profit when they make one more move.** --- ### Step 1: Find the Marginal Profit Function The **marginal profit** is the derivative of the profit function with respect to \( x \): \[ P'(x) = \frac{d}{dx}\left[0.8x^3 - 6x^2 + 4 - 1.6\ln(x) - 125\right] \] Differentiate term by term: - \( \frac{d}{dx}(0.8x^3) = 0.8 \cdot 3x^2 = 2.4x^2 \) - \( \frac{d}{dx}(-6x^2) = -6 \cdot 2x = -12x \) - \( \frac{d}{dx}(4) = 0 \) - \( \frac{d}{dx}(-1.6\ln(x)) = -1.6 \cdot \frac{1}{x} = -\frac{1.6}{x} \) - \( \frac{d}{dx}(-125) = 0 \) So, \[ P'(x) = 2.4x^2 - 12x - \frac{1.6}{x} \] --- ### Step 2: Plug \( x = 10 \) into the Marginal Profit Function \[ P'(10) = 2.4(10^2) - 12(10) - \frac{1.6}{10} \] Calculate each term: - \( 2.4 \times 100 = 240 \) - \( 12 \times 10 = 120 \) - \( \frac{1.6}{10} = 0.16 \) Plug these in: \[ P'(10) = 240 - 120 - 0.16 = 120 - 0.16 = 119.84 \] --- ### Step 3: Interpret the Result - The marginal profit when the company has done 10 moves is **approximately \$119.84**. - This means, **when the company makes the 11th move, the profit increases by about \$119.84**. --- ## **Final Answer** \[ \boxed{ \text{The marginal profit when they make one more move after 10 moves is approximately \$119.84.} } \]