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Unit 13: Maxima and Minima




                                                                                                Notes
                                               2
                 Example: Find the derivative of y  3  x  and show that it is infinite at x = 0. Draw a graph
          of the function and indicate its behaviour in the neighbourhood of origin. Deduce that y has a
          minimum value at origin which is not a stationary value.
          Solution.

                                        dy  2  1
                          y =  3  x 2        x  3   at x = 0
                                        dx  3
          To draw graph, we find

                         2
                        d y     2  4     2
                            =    x  3        0 x .
                        dx  2   9      9x  4/3
          Thus, the function is concave from below for all values of x.
                                                                              dy
          Further, since  lim x 2/3  lim x 2/3  f  0  0,  the function is continuous at x = 0. Since   ,  the
                      x  0    x  0                                            dx
          function is not differentiable at x = 0. This situation is shown in Figure 13.8.

          Note that, as we move away from origin on both sides, the value of y becomes greater than its
                                                        2
          value at x = 0. Thus f(0) = 0 is a minimum value of  y  3  x  which is not a stationary value.

                                               dy                 1 2  2
                 Example: By examining the sign of   dx  ,  show that   y  exp x  5  x   has a maxima at
          25 16 .
          Solution.
                                            1
                                            x 2 2 x
          The given function can be written as,  y e  5
                        dy     x  1 2 2 5  x  1  1/2  2
                        dx  = e    2  x   5  0

                   1  x  1/2  2     1   2   or  x 1/2  5   or  x  25
                   2     5  = 0 or   2x 1/2  5   4      16

                1                             dy                     1  1/2  2
          Since   x 2 2 5 x   for all values of x, the sign of    depends on the sign of   x
               e    0                         dx                     2     5
                                            Figure  13.8











                                              Fig. 5.8
                                 25    24
          When x is slightly less than   say   , we have
                                 16    16
                  1   16  2
                            = 0.408 0.04 0
                  2   24  5



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