The total cost of production of spare parts of computer is given as C=4000x-100x²+x³ where x is the number of units of spare parts produced so that the average cost will be maximum?

50

Step-by-step explanation:

Ok, we have the equation:

C=4000x-100x²+x³

Notice that the leading coefficient in this equation is positive, this means that as the value of x increases, also will the value of C.

Then we will not have a maximum, because as x →∞, C(x) →∞.

By doing an online search i found this same question, but with the change that we want to find x such that the average cost is minimum (this makes more sense).

Then let's do that:

The total cost for making x pieces is:

C(x) = 4000x-100x²+x³

Then the average cost, or the cost per piece, will be:

F(x) = C(x)/x = 4000 - 100*x + x^2

Notice that F(x) is a quadratic function with a positive leading coefficient, then the arms of the quadratic function will open upwards, this means that the minimum will be at the vertex.

To find the minimum, we must look at the zeros of the first derivation of F(x):

F'(x) = dF(x)/dx = -100 + 2*x

The zero is when:

F'(x) = 0 = -100 + 2*x

          100/2 = x

             50 = x

Then the number of x that minimizes the average cost is x = 50.

overpriced

step-by-step explanation: