
Yefim S. answered 03/22/21
Math Tutor with Experience
Average cost Cav(x) = C(x)/x = 0.002x + 100 + 8000/x;
Cav(x)' = 0.002 - 8000/x2 = 0; x2 = 8000/0.002 = 4000000; x = 2000
To prove that this is minimum let find Cav(x)'' = 16000/x3; Cav'''(2000) = 0.000002 > 0.
By second derivative test we have that production level x = 2000 gives minimum average cost