Minimizing average cost per unit

Divide to get the average cost function: $\bar{C}(x) = \frac{0.02x^2 - 8x + 2000}{x} = 0.02x - 8 + \frac{2000}{x}.$ This is the function we need to minimize for $x > 0$.

Differentiate: $\bar{C}'(x) = 0.02 - \dfrac{2000}{x^2}$. Setting $\bar{C}'(x) = 0$ gives $x^2 = \dfrac{2000}{0.02} = 100000$, so $x = \sqrt{100000} = 100\sqrt{10} \approx 316.23$ (taking the positive root since $x>0$).

Since $\bar{C}''(x) = \dfrac{4000}{x^3} > 0$ for $x>0$, this critical point is a minimum. Evaluating at $x = 100\sqrt{10}$: $$\bar{C}(100\sqrt{10}) = 0.02(100\sqrt{10}) - 8 + \frac{2000}{100\sqrt{10}} = 2\sqrt{10} - 8 + \frac{20}{\sqrt{10}} = 2\sqrt{10} - 8 + 2\sqrt{10} = 4\sqrt{10} - 8 \approx 4.65 \text{ dollars per unit}.