1. **State the problem:** We are given the cost function $$c(x) = x^3 - 6x^2 + 15x$$ where $x$ represents thousands of units produced. We want to find if there is a production level $x$ that minimizes the average cost, and if so, find that $x$. 2. **Recall the formula for average cost:** The average cost function $A(x)$ is given by dividing the total cost by the number of units produced: $$A(x) = \frac{c(x)}{x}$$ for $x > 0$. 3. **Write the average cost function:** $$A(x) = \frac{x^3 - 6x^2 + 15x}{x}$$ Simplify by canceling $x$: $$A(x) = \frac{\cancel{x}(x^2 - 6x + 15)}{\cancel{x}} = x^2 - 6x + 15$$ 4. **Find critical points of $A(x)$:** To minimize $A(x)$, find where its derivative is zero. Calculate the derivative: $$A'(x) = \frac{d}{dx}(x^2 - 6x + 15) = 2x - 6$$ 5. **Set derivative equal to zero and solve:** $$2x - 6 = 0$$ $$2x = 6$$ $$x = 3$$ 6. **Check if this critical point is a minimum:** Calculate the second derivative: $$A''(x) = \frac{d}{dx}(2x - 6) = 2$$ Since $$A''(3) = 2 > 0$$, the function is concave up at $x=3$, so $x=3$ is a minimum point. 7. **Interpret the result:** The production level that minimizes average cost is $x=3$, which means 3000 units. **Final answer:** The production level that minimizes average cost is $$\boxed{3}$$ (thousands of units).