83. Answers will vary. – (−2) 2a 2 84. = 2x2 82x + 720 y – −4 When the height is 3, the width is x (3) 12 2 12 2 12 6 6. x − = − = − = (3) (3)2 (3) (9) A = −2 +12 = −2 + 36 36 18 = −18 + = The maximum cross-sectional area is 18 square inches. This occurs when the gutter is 3 inches deep and 6 inches wide. a. You can only see a little of the parabola.
– 83. Answers will vary. 2a 2 (−2) −4 84. y = 2x2 – 82x + 720 When the height x is 3, the width is 12 − 2 x = 12 − 2 (3) = 12 − 6 = 6. A (3) = −2 (3)2 +12 (3) = −2 (9) + 36 = −18 + 36 = 18 The maximum cross-sectional area is 18 square inches. This occurs when the gutter is 3 inches deep and 6 inches wide. a. You can only see a little of the parabola.
Precalculus-Essentials-5th-Edition-Blitzer-Solutions-Manual.docx