(1)设件数为x,依题意,得3000-10(x-10)=2600,解得x=50,
答:商家一次购买这种产品50件时,销售单价恰好为2600元;
(2)当0≤x≤10时,y=(3000-2400)x=600x,
当10<x≤50时,y=[3000-10(x-10)-2400]x,即y=-10x2+700x
当x>50时,y=(2600-2400)x=200x
∴y=
600x(0≤x≤10,且x为整数)
-10x2+700x(10<x≤50,且x为整数)
200x(x>50,且x为整数)
(3)由y=-10x2+700x可知抛物线开口向下,当x=-
700
2×(-10)
=35时,利润y有最大值,
此时,销售单价为3000-10(x-10)=2750元,
答:公司应将最低销售单价调整为2750元.
http://www.***.com/math/ques/detail/f992c90a-67d4-4c88-96a7-0cd12151fb27?a=1