Given cost function:

C(x)=0.002x3+8x+6912C(x)=0.002x3+8x+6912

First we will compute the average cost of the given cost function. As we know the formula for the average cost is:

AC(x)=C(x)x=0.002x3+8x+6912x(Plugging in the cost function)AC(x)=0.002x2+8+6912xAC(x)=C(x)x=0.002x3+8x+6912x(Plugging in the cost function)AC(x)=0.002x2+8+6912x

We will compute the derivative of AC(x)AC(x) to get the critical point. Take the derivative of AC(x)AC(x) with respect to xx:

ddx[AC(x)]=ddx[0.002x2+8+6912x]AC′(x)=0.002×2x+0−6912x2(Differentiating using the formula ddx[xn]=nxn−1)=0.004x−6912x2ddx[AC(x)]=ddx[0.002x2+8+6912x]AC′(x)=0.002×2x+0−6912x2(Differentiating using the formula ddx[xn]=nxn−1)=0.004x−6912x2

Plug in AC′(x)=0AC′(x)=0 for getting the critical point:

0.004x−6912x2=00.004x=6912x20.004x3=6912(Multiplying both sides by x2)x3=69120.004(Dividing both sides by 0.004)x3=1,728,000x=3√1,728,000(Taking the cube root both sides)x=1200.004x−6912x2=00.004x=6912x20.004x3=6912(Multiplying both sides by x2)x3=69120.004(Dividing both sides by 0.004)x3=1,728,000x=1,728,0003(Taking the cube root both sides)x=120

Thus, at x=120x=120 the average cost gives the minimum values.