PUMaC 2015 · 团队赛 · 第 16 题

PUMaC 2015 — Team Round — Problem 16

[ 13 ] Let p, u, m, a, c be real numbers satisfying 5 p + 4 u + 3 m + 2 a + c = 91. What is the maximum possible value of: 2 2 2 18 pumac + 2(2 + p ) + 23(1 + ua ) + 15(3 + mc ) ? 3

解析

[ 13 ] Let p, u, m, a, c be real numbers satisfying 5 p + 4 u + 3 m + 2 a + c = 91. What is the maximum possible value of: 2 2 2 18 pumac + 2(2 + p ) + 23(1 + ua ) + 15(3 + mc ) ? Solution: We have: 8 5 5 5 5 5 7402 = 48(5 p + 4 u + 3 m + 2 a + c ) + 3034 5 5 5 5 5 = (216 p + 54 u + 24 m + 27 a + 8 c )+ ( ) 5 5 5 16 + (16 p + 16 + 16 + 16 + 16) + (4 p + 4 p + 4 + 4 + 4) + ( ) 5 5 5 5 5 5 46 + (46 u + 23 a + 184 + 184 + 184) + (46 u + 46 u + 23 a + 23 a + 184) + ( ) 5 5 5 5 5 5 270 + (60 m + 20 c + 540 + 540 + 540) + (30 m + 30 m + 10 c + 10 c + 270) ( ) 2 2 2 2 2 ≥ 5 · 36 pumac + (16 + 16 p + 4 p ) + (46 + 92 ua + 46 u a ) + (270 + 180 mc + 30 m c ) − 1328 2 2 2 873 ≥ 18 pumac + 2(2 + p ) + 23(1 + ua ) + 15(3 + mc ) 5 5 5 5 5 By AM-GM. Equality holds when p = 1 , u = 4 , m = 9 , a = 8 , c = 27. Thus, the maximum is 873 . EDIT: This problem had a typo that p, u, m, a , and c must be positive reals. As written, m = a = c = 0 and p taken arbitrarily large yields a valid u value in the original equation. Thus the maximum value of the expression is unbounded. This problem has been thrown out due to such a wide variety of possible answers. Author: Bill Huang 9