已知2x+4y=1,则x²+y²的最小值是( ) A.1/5 B.1/10 C.1/16 D.1/20
问题描述：
已知2x+4y=1,则x²+y²的最小值是( ) A.1/5 B.1/10 C.1/16 D.1/20
答
2x+4y=1
y=(1-2x)/4
x²+y²
=x²+1/16 (1-2x)²
=5/4x²-1/4x+1/16
=5/4 (x²-1/5x)+1/16
=5/4(x-1/10)²-1/80+1/16
=5/4(x-1/10)²+1/20
即
当x=1/10时,取最小值1/20,选D