Page 116 - How To Solve Word Problems In Calculus
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EXAMPLE 16
                                A conical cup is constructed from a circular piece of paper of
                                radius 5 in by cutting out a sector and joining the resulting
                                edges. What is the maximum volume of the cup?

                                    Solution
                                    Step1
                                    Join AB to AC

                                                 B        C                r
                                                        5
                                                                        h
                                                     A                      5



                                    Step2
                                                                              π  2
                                    The volume of the resulting cone is V =     r h.
                                                                              3
                                    Step3
                                    From the diagram in step 1 it is apparent that
                                                             2
                                                                       2
                                     2
                                 2
                                h + r = 25. It follows that r = 25 − h .
                                                      π
                                                        2
                                                 V =   r h
                                                      3
                                                      π
                                                               2
                                                   =   (25 − h )h
                                                      3
                                                      π
                                                                3
                                              V(h) =   (25h − h )     0 ≤ h ≤ 5
                                                      3
                                    Step4
                                                             π
                                                                       2
                                                     V (h) =   (25 − 3h )
                                                             3
                                                             π
                                                                       2
                                                        0 =    (25 − 3h )
                                                             3
                                                        0 = 25 − 3h 2
                                                         2
                                                      3h = 25
                                                                                        103
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