Given cost function:

C(x)=400+8x+0.03x2C(x)=400+8x+0.03x2

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=400+8x+0.03x2x(Plugging in the cost function)AC(x)=400x+8+0.03xAC(x)=C(x)x=400+8x+0.03x2x(Plugging in the cost function)AC(x)=400x+8+0.03x

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[400x+8+0.03x]AC′(x)=−400x2+0+0.03(Differentiating using the formula ddx[xn]=nxn−1)AC′(x)=−400x2+0.03ddx[AC(x)]=ddx[400x+8+0.03x]AC′(x)=−400x2+0+0.03(Differentiating using the formula ddx[xn]=nxn−1)AC′(x)=−400x2+0.03

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

−400x2+0.03=00.03=400x20.03x2=400(Multiplying both sides by x2)x2=4000.03(Dividing both sides by 0.03)x=√4000.03(Taking the square root both sides)x=115.47−400x2+0.03=00.03=400x20.03x2=400(Multiplying both sides by x2)x2=4000.03(Dividing both sides by 0.03)x=4000.03(Taking the square root both sides)x=115.47

Thus, at x=115.47x=115.47 the average cost gives the minimum values.

Maximum and Minimum Values:

Basically critical number gives maximum and minimum values of a function. So, first we have to understand the critical number to solve this problem:. Suppose a numbers is x where f′(x)=0f′(x)=0 or f′(x)f′(x) is undefined but f(x)f(x) is defined then that number x is called the critical number of f(x)f(x).