Chapter 4: Applications of Derivatives - Section 4.5 - Applied Optimization - Exercises 4.5 - Page 226: 56

$x=10$

Step 1. Given $c(x)=x^3-20x^2+20000x$, the average cost can be found as $A_c=\frac{c(x)}{x}=x^2-20x+20000$ Step 2. To minimize the average cost, let its derivative be zero to get $A'_c=2x-20=0$ and $x=10$, which gives $A_c=(10)^2-20(10)+20000=19900$ Step 3. Check $A''_c=2\gt0$ and we know it is a minimum.