已知绝对值Y小于等于1,且2X+Y=1,则2X平方+16X+2Y平方的最小值是多少

x=(1-y)/2
2x^2+16x+2y^2
=(1-y)^2/2+8(1-y)+2y^2
=y^2/2-y+1/2+8-8y+2y^2
=5/2*y^2-9y+17/2
=5/2*(y^2-18/5y)+17/2
=5/2*(y-9/5)^2+17/2-5/2*81/25
=5/2*(y-9/5)^2+2/5
由于|y|