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Unit 14: Business Applications of Maxima and Minima




                                             x
                                  d AC     xF  ( ) F ( )                                        Notes
                                                  x
                                        =      2     = 0
                                   dx         x
                                                      F ( )
                                                       x
                                   xF  ( ) = F ( )  or  ( )F x   or MC = AC.
                                            x
                                     x
                                                       x
           Thus, marginal cost is equal to the average at the minima of the later.

             Notes  The level of output at which AC is minimum is also known as the most economic (or
             capacity) output.


                 Example: The short-run cost function of a food manufacturer is given by
                                                            3
                                      C = 1,000 + 100x – 10x  + x .
                                                        2
          (i)  Find AC, AVC and MC functions.
          (ii)  Show that MC = min. of AC.

          (iii)  Show that MC = min. of AVC.
          (iv)  Show that total cost function has a point of inflexion at a level of output where  MC is
               minimum. Find min. MC.

          Solution:
                                          C   1,000         2
          (i)                       AC =           100 10x x
                                           x   x
                                                    ,
                                                   2
                                   AVC = 100 10x x MC    100 20x  3x  2
          (ii)  For minima of AC, we have
                                  d AC      1000
                                        =    2   10 2x  0  or  1000 10x 2  2x 3  0
                                   dx        x

              or             x 3  5x 2  500 = 0 or  x  10 x 2  5x  50

               Thus x = 10 is a stationary point. The other roots, being imaginary, are neglected.
                                  2
                                 d AC     2000
                      We note that   2  =   3   2 4 0,  at x = 10
                                   dL       x
                  Thus AC is minimum at x = 10.
                                          1000              2
                            Also min. AC =     100 10 10 10   200
                                           10
              and                   MC = 100 20 10 3 10  2  200,  at x = 10
                               Thus, MC = min. AC.
          (iii)  For minima of AVC, we have

                                 d AVC
                                   dx   =  10 2x  0  or x = 5, min.





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