Minimizing average cost per unit

Differentiate $A(x) = 0.001x^2 - 0.3x + 40 + 1000x^{-1}$ to get $A'(x) = 0.002x - 0.3 - 1000x^{-2}$. Setting $A'(x) = 0$ and multiplying through by $x^2$ gives $0.002x^3 - 0.3x^2 - 1000 = 0$. Solving this equation numerically (or by inspection) for $x>0$ yields $x = 200$.

To confirm this is a minimum, note $A''(x) = 0.002 + 2000x^{-3}$, which is positive for all $x>0$, so $x=200$ is indeed a minimum. Substituting $x=200$ into $A(x)$: $A(200) = 0.001(200)^2 - 0.3(200) + 40 + \frac{1000}{200} = 40 - 60 + 40 + 5 = 25$. So the minimum average cost is $25$ dollars per unit, occurring at a production level of $200$ units.