Question

实数x、y满足3x^2＋2y^2=6x,则x2＋y2的最大值为

Answer 1

由3x^2＋2y^2=6x得：y^2 = (6x-3x^2)/2

x^2+y^2 = x^2 + (6x-3x^2)/2 = -1/2x^2 + 3x = -1/2(x-3)^2 + 4.5 ≤ 4.5

最大值4.5