Suppose the average costs of a mining operation depend on the number of machines used, and average costs, in dollars, are given by C(x) = 2500x + 722,500 / x , x > 0,

Question

Suppose the average costs of a mining operation depend on the number of machines used, and average costs, in dollars, are given by C(x) = 2500x + 722,500/ x, x > 0

What is the minimum average cost? 

Answer 1

C '(x) = 2500 - 722500/x² = 0; x² = 722500/2500; x² = 289; x = 17;

C''(x) = 1445000/x³; C''(17) = 1445000/17³ > 0. So, at x = 17 evarage cost is minimum,

so min C(x) = C(17) = 2500·17 + 722500/17 = $85000

Answer 2

Minimum average Cost is where the average cost and marginal cost curves intersect, but with this cost function, they never intersect. But they do get very close.

AC = C/x, MC = C'

set C' = C/x and solve for x

C' = 2500 +-722,500/x^2

C/x = 2500 + 722,500/x^2

only way to make them equal is when x= infinity. The larger x, the closer they get.

Or another way to look at this is focus on the C/x function. Minimum C/x is 2500 when x = infinity

You never reach minimum Average Cost, you just approach 2500 as a limit as x approaches infinity