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Basic Mathematics – I




                    Notes          (ii)  Find the value of L for which output is maximum.
                                   (iii)  Find the value of L at which the total product curve has a point of inflexion and verify that
                                       MP is maximum at this point. What is the nature of the point of inflexion?
                                          L
                                   (iv)  If the manufacturer sells the product at a uniform price of  10 per unit, find the maximum
                                       total revenue product and associated level of L.
                                   Solution:
                                                                   x          2
                                   (i)                      AP =      11 16L L
                                                               L   L
                                                                   dx           2
                                                            MP =       11 32L  3L
                                                               L   dL
                                                          d AP L
                                      We have                   = 16 2L   0    AP  is maximum at L = 8.
                                                            dL                  L
                                                         d AP
                                                          2
                                      Since                   L  =  2 0 , the second order condition is satisfied.
                                                           dL 2
                                                                            2
                                                The maximum AP = 11 + 168   8 = 75
                                                               L
                                               Further, MP  when L = 8, is 11 + 32   8   3   8  = 75
                                                                                    2
                                                        L
                                                        Thus, AP = MP , when AP  is maximum.
                                                               L     L        L
                                   (ii)  For maximum output:
                                                             dx
                                                                = 11 32L  3L 2  0
                                                             dL
                                      or            (11 – L)(1 + 3L) = 0, \ L = 11. The other value, being negative, is dropped.
                                             2
                                             d x
                                       Since   2  32 6L  34 0 , the second order condition for maxima is satisfied.
                                             dL
                                   (iii)  For point of inflexion:
                                                             2
                                                                                             3
                                                             d x                 16         d x
                                                            dL 2  = 32 – 6L = 0   L =   3  5.33  and   dL 3   = – 6 < 0.
                                       Thus the point of inflexion is of type I, i.e. the curve changes from convex to concave from
                                       below.
                                                             2
                                                                                3
                                                             d x    ( d MP L )  d x  d  2 (MP L )
                                                       Since   2  =       0  and   3   =   2  0 ,
                                                             dL      dL         dL    dL
                                              MP  is maximum at L = 5.33.
                                                 L
                                   (iv)                    TRP = p.x = 10 11L  16L 2  L 3  .
                                                               L
                                   Since TRP  is a constant multiple of the production function, therefore, maxima of TRP  will be
                                           L                                                            L
                                   at the same level of L where x is maximum. Thus, TRP  will also be maximum at L = 11. The
                                                                                L
                                                                   3
                                   maximum value = 10 11 11 16 11 2  11  =   7,260.
                                   14.3 Minimisation of Cost
                                                                            x
                                                                       C  F ( )         dC
                                                                                              x
                                   If total cost C = F(x), then we can define  AC  , and   MC  =   F  ( ).
                                                                        x   x            dx
                                   Very often we are interested in finding the level of output that gives minimum AC. For minima
                                   of AC, we have



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