Let's go through the problem step by step: ### **Given:** The profit function is: \[ P(x) = 0.8 \sqrt{x^3 - 6x^2 + 4 - 1.6\ln(x)} - 125 \] The company has already done \( x = 10 \) moves. We are to **estimate the marginal profit** (i.e., \(\frac{dP}{dx}\)) when \( x = 10 \). --- ## **Step 1: Find the derivative \( P'(x) \)** Let’s denote: \[ f(x) = x^3 - 6x^2 + 4 - 1.6\ln(x) \] So, \[ P(x) = 0.8 \sqrt{f(x)} - 125 \] The derivative is: \[ P'(x) = 0.8 \cdot \frac{1}{2\sqrt{f(x)}} \cdot f'(x) \] \[ = \frac{0.8}{2} \cdot \frac{f'(x)}{\sqrt{f(x)}} \] \[ = 0.4 \cdot \frac{f'(x)}{\sqrt{f(x)}} \] --- ## **Step 2: Find \( f'(x) \)** Recall: \[ f(x) = x^3 - 6x^2 + 4 - 1.6\ln(x) \] So, \[ f'(x) = 3x^2 - 12x - \frac{1.6}{x} \] --- ## **Step 3: Evaluate at \( x = 10 \)** ### **Calculate \( f(10) \):** \[ f(10) = (10)^3 - 6(10)^2 + 4 - 1.6\ln(10) \] \[ = 1000 - 600 + 4 - 1.6 \times 2.3026 \] \[ = 404 - 3.68416 \] \[ = 400.31584 \] ### **Calculate \( f'(10) \):** \[ f'(10) = 3(10)^2 - 12(10) - \frac{1.6}{10} \] \[ = 3(100) - 120 - 0.16 \] \[ = 300 - 120 - 0.16 \] \[ = 179.84 \] --- ## **Step 4: Substitute into \( P'(x) \) at \( x = 10 \):** \[ P'(10) = 0.4 \cdot \frac{179.84}{\sqrt{400.31584}} \] Calculate \( \sqrt{400.31584} \): \[ \sqrt{400.31584} \approx 20.0079 \] So, \[ P'(10) = 0.4 \cdot \frac{179.84}{20.0079} \] \[ = 0.4 \cdot 8.990 \] \[ = 3.596 \] --- ## **Final Answer:** \[ \boxed{3.60} \] So, the marginal profit when the company makes one more move after 10 moves is **approximately \$3.60**.

The marginal profit at a given point is the derivative of the profit function with respect to the number of moves, representing the instantaneous rate of change of profit. It is estimated by evaluating the derivative \( P'(x) \) at the specific value \( x = 10 \).