400 2 400 2 y x y x = − = − 3 3 3 2 400 2   40 y = − We need to maximize A xy x = =  . x − x 3 3  ( ) (40 ) A x x x = − = Rewrite as a function of A x. 40 x −x2 + ( ) 400 2 2 400   A x x x x2 x = − = − + 40 40 x b = − = − = − = 20. 2a 2 (−1) 3 3 3 3   −2  When the length is 20, the width is x y 40 40 20 20. y x = − = − = The dimensions of the rectangular region with maximum area are 20 yards by 20 yards. This gives an area of 20 20 400 square yards. ⋅ =
y = 400 − 2 x 3 y = 400 − 2 x 3 3 2  400 2  y = 40 − x We need to maximize A = xy = x − x  . 3 3  A ( x ) = x (40 − x ) = −x2 + 40 x x = − b = − 40 = − 40 = 20. Rewrite A as a function of x. A ( x ) = x  400 − 2 x  = − 2 x2 + 400 x 2a 2 (−1) −2  3 3  3 3  When the length x is 20, the width y is y = 40 − x = 40 − 20 = 20. The dimensions of the rectangular region with maximum area are 20 yards by 20 yards. This gives an area of 20 ⋅ 20 = 400 square yards.
Precalculus-Essentials-5th-Edition-Blitzer-Solutions-Manual.docx