某公司准备把240吨白砂糖运往A、B两地，用大、小两种货车共20辆，恰好能一次性装完这批白砂糖，
推荐回答（2个）
解：（1）设大车x辆，恰能一次性装完。
15x+10（20-x）=240
解得：x=8
故使用8辆大车、12辆小车可一次性装完。
（2）15m+10(10-m)>=115
0<=m<=8
解得：3<=m<=8
（3）设当m辆大车去A地时，总运费为W（m）
W(m)=630m+420(10-m)+750(8-m)+550[12-(10-m)]
=10m+11300
由于3<=m<=8，故当m=3时，运费最少。
最少运费为W(3)=10*3+11300=11330(元)
收起
30某公司准备把240吨白砂糖运往A、B两地，用大、小两种货车共20辆，恰好能一次性装完这批白砂糖，
提问者： 芊幻5
相关数据如下
载重量 运往A地的费用 运往B地的费用
大车 15吨∕辆 630元∕辆 750元∕辆
小车 10吨∕辆 420元∕辆 550元∕辆
(1)求大、小两种货车各用多少辆？
（2）如果安排10辆货车前往A地，其中大车有m辆，其余货车前往B地，且运往A地的白砂糖不少于115吨。
①求m的取值范围；
②请你设计出使总运费最少的货车调配方案，并求出最少总运费
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