Average Cost and Integrals

A company has found that the total cost of manufacturing x items is C(x) = 20e^(x/6) . Given that the average cost is defined as C (with a bar) (x) = C(x)/x, determine the value of x that minimizes C (with a bar) (x).

This one's confusing =/

I will denote C(x) (with a bar) as A(x).

Since A(x) = C(x)/x, we have:
A(x) = C(x)/x = 20e^(x/6)/x.

By re-writing A(x) as 20x^(-1)*e^(x/6) and applying the Product Rule gives:
A'(x) = 20(-1)x^(-2)*e^(x/6) + (10/3)x^(-1)*e^(x/6)
= -20e^(x/6)/x^2 + 10e^(x/6)/(3x)
= [10x*e^(x/6) - 60e^(x/6)]/(3x^2), by combining fractions
= 10e^(x/6)(1 - 6x)/(3x^2), by factoring.

Setting A'(x) = 0 gives:
10e^(x/6)(1 - 6x)/(3x^2) = 0 ==> x = 1/6.

Therefore, x = 1/6 will minimize A(x).

I hope this helps!