HMMT 二月 2003 · CALC 赛 · 第 5 题

HMMT February 2003 — CALC Round — Problem 5

解析

Find the minimum distance from the point (0 , 5 / 2) to the graph of y = x / 8. √ Solution: 17 / 2 2 4 2 8 4 2 We want to minimize x +( x / 8 − 5 / 2) = x / 64 − 5 x / 8+ x +25 / 4, which is equivalent 4 2 2 to minimizing z / 4 − 10 z + 16 z , where we have set z = x . The derivative of this 3 expression is z − 20 z + 16, which is seen on inspection to have 4 as a root, leading to √ √ 2 the factorization ( z − 4)( z + 2 − 2 2)( z + 2 − 2 2). Since z = x ranges over [0 , ∞ ), √ the possible minima are at z = 0 , z = − 2 + 2 2, and z = 4. However, the derivative √ is positive on (0 , − 2 + 2 2), so this leaves only 0 and 4 to be tried. We find that the minimum is in fact achieved at z = 4, so the closest point on the graph is given by √ √ 2 4 2 x = ± 2, with distance 2 + (2 / 8 − 5 / 2) = 17 / 2. 1