Question

The total cost function of a firm is C = x² + 75x + 1600 for output x. Find the  output for which the average cost ls minimum. Is C_A= C_m at this output?  

Answer 1

Given C = x² + 75x + 1600         ..........(1)

Marginal cost C_m = `(dC)/dx`
Differentiating (i) w.r.t.x
C_m = 2x + 75
Average cost C_A = `C/x`
                           = x + 75 + `1600/x`
Diff (ii) w.r.t.x 
               `(dC_A)/(dx)  = 1 + 1600 x ((-1)/x^2)`

                             = 1 - `1600/x^2`

                             = `(x^2 - 1600)/x^2`

If `(dC_A)/dx = 0   then .(x^2  - 1600)/x^2 = 0`

         `x^2 - 1600 = 0
               `x^2 = 1600`
x = 40 and x = -40

Differentiating `(dC_A)/dx` w.r.t.x

`(dC_A)/dx = d/dx (1 - 1600/x^2) = 0 - 1600 xx (-2x^-3)`

`((dC_A)/dx^2) _(at x = 40) = 3200/(40)^3 = 3200/64000`

                                          = `1/20 > 0

C_m = 2x + 75
      = 2(40) + 75
      = 80 + 75 = 155
C_A = x + 75 + `1600/40` = 155

Average cost is minimum for output = 40 .
Since, C_m at this output= 2(40) + .95 = 155

C_m = C_A