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HMMT 十一月 2009 · GEN1 赛 · 第 6 题

HMMT November 2009 — GEN1 Round — Problem 6

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [ 5 ] Find the maximum value of x + y , given that x + y − 3 y − 1 = 0.
解析
  1. [ 5 ] Find the maximum value of x + y , given that x + y − 3 y − 1 = 0. √ 26+3 3 13 2 2 2 2 Answer: We can rewrite x + y − 3 y − 1 = 0 as x + ( y − ) = . We then see that the set 2 2 4 √ 2 2 13 3 of solutions to x − y − 3 y − 1 = 0 is the circle of radius and center (0 , ). This can be written as 2 2 √ √ √ √ √ 13 13 3 3 13 3 13 ◦ x = cos( θ ) and y = sin( θ )+ . Thus, x + y = + (cos( θ )+sin( θ )) = + 2 sin( θ +45 ), 2 2 2 2 2 2 2 √ 26+3 ◦ which is maximized for θ = 45 and gives . (We could also solve this geometrically by noting 2 that if x + y attains a maximum value of s then the line x + y = s is tangent to the circle.)