HMMT 二月 2001 · CALC 赛 · 第 5 题

HMMT February 2001 — CALC Round — Problem 5

Same as question 4 , but now we want one of the rectangle’s sides to be along the hypotenuse. 2 2

解析

Same as question 4 , but now we want one of the rectangle’s sides to be along the hypotenuse. Solution: Put the hypotenuse along the x -axis, with the short leg starting at the origin so that the right angle is at the point (9 / 5 , 12 / 5). For notational convenience, let’s just scale everything by a factor of 5 and then remember to divide the final area by 25, so now the top point is at (9 , 12). Let ( a, 0) be the point where the edge of the rectangle along the hypotenuse starts. Then the height is h = (4 / 3) a since the leg of length 3 is along the line y = (4 / 3) x . The leg of length 4 is along the line x = 25 − (4 / 3) y , so the horizontal edge of the rectangle ends at b = 25 − (4 / 3) h = 25 − (16 / 9) a . The area of the rectangle is ( b − a ) h = (25 − (16 / 9) a − 100 100 100 200 2 a )(4 / 3) a = a − a . The derivative of this is 0 when = a , or a = 9 / 2. Thus the 3 27 3 27 (100 / 3)(9 / 2) − (100 / 27)(81 / 4) maximum area is = 3 . 25 2 2