实数XY 满足2x+4y=1,则x平方+y平方最小值是

2x+4y=1,
x=(1-4y)/2,
x^2+y^2=[(1-4y)/2]^2+y^2
=(1-8y+16y^2)/4+y^2
=5y^2-2y+1/4
=5(y^2-2y/5)+1/4
=5[y^2-2y/5+(1/5)^2-1/25]+1/4
=5(y-4/5)^2-1/5+1/4
=5(y-4/5)^2+1/20≥1/20,
所以x^2+y^2的最小值是1/20,
答案选d.