Page 131 - How To Solve Word Problems In Calculus
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A



                                                                    10 − h


                                                          2r      C
                                                  B                          10



                                                                 h



                                             D                           E
                                                           6

                                   
ABC is similar to 
ADE. Therefore, their sides are proportional.

                                                   2r    10 − h
                                                      =
                                                    6      10
                                                   20r = 60 − 6h
                                                   6h = 60 − 20r

                                                             10
                                                    h = 10 −    r    0 ≤ r ≤ 3
                                                              3

                                   By substitution the volume as a function of r is

                                                               2
                                                        V = πr h

                                                                      10
                                                     V (r ) = πr  2  10 −  r
                                                                      3

                                                                      10
                                                                  2      3
                                                          = π 10r −     r
                                                                      3
                                                                       2
                                                     V (r ) = π(20r − 10r )

                                                        0 = 10πr (2 − r )
                                                        r = 0    r = 2



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