AoPS Wiki talk:Problem of the Day/June 21, 2011

Solution

(2x+3y)^2 + (3x-2y)^2 = 13(x^2+y^2)=26 Hence (2x+3y)^2 <= 26 or 2x+3y = sqrt(26). If 3x-2y=0 and 2x+3y=sqrt(26), then 2x+3y attains this maximum value on the circle x^2+y^2=2.