设应该将每件的售价定为x元(x﹥10)时,才能使每天获利最大；则每件的利润为x-10+10-8=(x-8)元,每件的销售价提高了(x-10)元,销售量将减少10(x-10)件,实际销售量是[100-10(x-10)]件,得方程：(x-8)[100-10(x-10)]
=(x-8)(200-10x)
=200x-10x²-1600+80x
=-10x²+280x-1600
=-10(x²-28x)-1600
=-10(x²-28x+14²)-1600+10×14²
=-10(x-14)²+360
当x=14时,利润为360元
加油!