已知x^2+xy+y^2=1,求x^2+y^2的最大值
1个回答
展开全部
x^2+xy+y^2=1
(x+y/2)^2+3y^2/4=1
令x+y/2=sina,根号3/2y=cosa
那么
y=2cosa/根号3
x=sina-y/2=sina-cosa/根号3
x^2+y^2
=sina^2-sin2a/根号3+cosa^2/3+4cosa^2/3
=sina^2-sin2a/根号3+5cosa^2/3
=1+2cosa^2/3-sin2a/根号3
=1+(cos2a+1)/3-sin2a/根号3
=4/3+cos2a/3-sin2a/根号3
最小值=4/3-2/3=2/3
最大值=4/3+2/3=2
(x+y/2)^2+3y^2/4=1
令x+y/2=sina,根号3/2y=cosa
那么
y=2cosa/根号3
x=sina-y/2=sina-cosa/根号3
x^2+y^2
=sina^2-sin2a/根号3+cosa^2/3+4cosa^2/3
=sina^2-sin2a/根号3+5cosa^2/3
=1+2cosa^2/3-sin2a/根号3
=1+(cos2a+1)/3-sin2a/根号3
=4/3+cos2a/3-sin2a/根号3
最小值=4/3-2/3=2/3
最大值=4/3+2/3=2
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询