Question

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

Answer 1

2x^2+16x+2y=2[(1-y)/2]^2+16[(1-y)/2]+2y=1/2(y^2-14y+17)=1/2(y-7)^2-16

由|y|≥1,得y≥1,或y≤-1

当x=7时取最小值,即-16