HMMT 十一月 2009 · GEN1 赛 · 第 6 题

英文解析

6. [ 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 notingthat if x + y attains a maximum value of s then the line x + y = s is tangent to the circle.)2