Quadric Equtions question, 76 feet of fencing...

Question

Suppose you have 76 feet of fencing to enclose a rectangular dog pen. The function a=38x-x^2, where x = width, gives you the area of the dog pen in square feet. What width gives you the maximum area? What is the maximum area? Round to the nearest tenth as necessary.

Answer 1

 Width w and Length L

 

 2W + 2L = 76

  W + L = 38

 

  Area = Length x Width = WL

 

  Area = W ( 38 - W) = 38W - w^2   which is the same as given in the problem except the width is X

 

  The roots of this QUADRATIC function are 0 = 38w - w^2

                                                               0 = w^2 - 38w

                                                               0 = w(w-38)

                                                               w=0 or w-38=0

                                                               w=0 or w=38

 

   The axis of symmetry mirror is the average, so (0 + 38)/2 = 19 

 

   So the area is maximized at length 19. The max area is 38*19 - 19*19 = 361

 

   You can also complete the square: 38w - w^2 = 0

                                                      0 =-( w^2 + 38w

                                                      0 = -(w^2 + 38w + 361) + 361

                                                      0 = -(w + 19)^2 + 361

 

   The vertex or max point is (h,k) where H=19 and k=361