Question

已知x,y属於R,且3x^2+2y^2=6x,求x+y的最大值与最小值(要求详细解答)

Answer 1

3(x^2-2x+1)+2y^2=3

3(x-1)^2+2y^2=3

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

令x-1=cosa,x=1+cosa

则2y^2/3=1-cos²a=sin²a

所以y=√(3/2)*sina

所以x+y=1+cosa+√(3/2)*sina

=√[(√3/2)^2+1^2]sin(a+z)+1

=√(5/2)sin(a+z)+1

所以最大值=（√10）/2+1,最小值=-（根号10）/2+1.