HMMT 二月 2001 · CALC 赛 · 第 5 题

英文解析

5. Same as question 4 , but now we want one of the rectangle’s sides to be along thehypotenuse.
   Solution: Put the hypotenuse along the x -axis, with the short leg starting at the originso that the right angle is at the point (9 / 5 , 12 / 5). For notational convenience, let’s just scaleeverything by a factor of 5 and then remember to divide the final area by 25, so now the toppoint is at (9 , 12).
   Let ( a, 0) be the point where the edge of the rectangle along the hypotenuse starts. Thenthe height is h = (4 / 3) a since the leg of length 3 is along the line y = (4 / 3) x . The leg oflength 4 is along the line x = 25 − (4 / 3) y , so the horizontal edge of the rectangle ends atb = 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
   a )(4 / 3) a = a − a . The derivative of this is 0 when = a , or a = 9 / 2. Thus the 2
   3 27 3 27
   (100 / 3)(9 / 2) − (100 / 27)(81 / 4)
   maximum area is = 3 .
   2 225