已知x^2+2y^2=1.求2x+5y^2的最小值
问题描述：
已知x^2+2y^2=1.求2x+5y^2的最小值
答
x^2+2y^2=1
x^2>=0 y>=0
x^2=1-2y^2>=0
02x+5y^2=2(x^2+2y^2)+y^2=2+y^2
最小值为2
答
x^2+2y^2=1
|x|y^2=(1-x^2)/2
2x+5y^2
＝2x+(1－x^2)(5/2)
=-(5/2)x^2+2x+(5/2)=f(x)
|x|f(-1)=min=-2