hanulyeongsam 26.10.2024 18:18 For reals $x,y$, find the maximum of A. $ A=\frac{-x^2-y^2-2xy+30x+30y+75}{3x^2-12xy+12y^2+12} $
JH_K2IMO 14.12.2024 04:44 (-x²-y²-2xy+30x+30y+75)/(3x²-12xy+12y²+12) =1/3{-(x+y-15)²+300/(x-2y)²+4} ≤ 1/3 x 300/4=25. The answer is 25 (If x = 10 and y = 5 are substituted, then A = 25.)