Before we can minimize the average cost, we first need to construct the average cost function. This requires us to divide the total cost function by the quantity, x.

¯c(x)=c(x)x=3x3−12x2+9000xx=3x2−12x+9000c¯(x)=c(x)x=3x3−12x2+9000xx=3x2−12x+9000

Let's now minimize this function using Calculus. The first step in the procedure to minimize any function is to differentiate the function we want to work with. The derivative of the average cost function is called the marginal average cost. The power rule is the only rule that we need to apply to find this derivative.

¯c′(x)=6x−12c¯′(x)=6x−12

Let's now find the critical point of this function by setting the derivative equal to zero.

6x−12=06x=12x=26x−12=06x=12x=2

Next, we need to classify this critical point, and the second derivative test is the quickest technique that we can use to do that.

¯c′′(x)=6c¯″(x)=6

When the second derivative is positive at a critical point, it means that the critical point is the location of a local minimum. Thus, we know that the average cost is minimized at a quantity of 2 items.