Question

已知：x的2次方+2y的2次方=4y,那么x的2次方+y的2次方的最大值是?

Answer 1

x^2+2y^2=4y x^2+2(y-1)^2=2 x^2/2+(y-1)^2=1 设x=√2cosa,y-1=sina x^2+y^2 =2(cosa)^2+(1+sina)^2 =2(cosa)^2+1+2sina+(sina)^2 =-(sina)^2+2sina+3 =-(sina-1)^2+4 (-1