设x+2y=1,(x,y属于R),若x,y>=0求x2+y2的最大值.

x+2y=1

x=1-2y

x^2+y^2

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

=1-4y-4y^2+y^2

=5y^2-4y-1

=5(y^2-4y/5)-1

=5(y^2-4y/5+4/25)-1-4/5

=5(y-2/5)^2-9/5

x,y>=0

x=1-2y>0

y>0

0=<y=<1/2

当y＝1/2时取最小值

x^2+y^2

=5(1/2-2/5)^2-9/5

=自己算吧

求今年高考的答案

2002年估计还会做 现在不会咯

x=1-2y>=0, y>=0

==> 0<= y <=1/2

==> -2/3<= (y-2/3) <= -1/6

==> 1/36<= (y-2/3)^2 <= 4/9

==> -1/12<= (y-2/3)^2-1/9 <= 1/3

==> -1/4<= 3(y-2/3)^2-1/3 <=1

x^2+y^2==> 3y^2-4y+1=3(y-2/3)^2-1/3 <=1

答案是1