问题
填空题
我市化工园区一化工厂，组织20辆汽车装运A、B、C三种化学物资共200吨到某地．按计划20辆汽车都要装运，每辆汽车只能装运同一种物资且必须装满．请结合表中提供的信息，解答下列问题：
（2）如果装运A种物资的车辆数不少于5辆，装运B种物资的车辆数不少于4辆，那么车辆的安排有几种方案？并写出每种安排方案； （3）在（2）的条件下，若要求总运费最少，应采用哪种安排方案？请求出最少总运费． |
答案
（1）装运A种物资的车辆数为x，装运B种物资的车辆数为y，则装运C种物资的车辆数为（20-x-y），
根据题意得，12x+10y+8（20-x-y）=200，
整理得，y=20-2x；
（2）当y=20-2x时，20-x-y=x，
则装运A、B、C三种化学物资的车辆数分别为x，20-2x，x，
由题意，得
，x≥5 20-2x≥4
解这个不等式组，得5≤x≤8，
因为x为整数，所以x的值为5，6，7，8．
所以安排方案有4种：
方案一：装运A种物资5辆、B种物资10辆、C种物资5辆；
方案二：装运A种物资6辆、B种物资8辆、C种物资6辆；
方案三：装运A种物资7辆、B种物资6辆、C种物资7辆；
方案四：装运A种物资8辆、B种物资4辆、C种物资8辆；
（3）设总运费为W元，
则W=12×240x+10×320（20-2x）+8×200x，
即：W=-1920x+64000，
∵W是x的一次函数，且W随x的增大而减小，
∴当x=8时，W最小，最少为48640元．
son would 36 him on the farm, enable it to expand, and eventually take it 37 . But Henry proved a 38 . He hated farm work and did everything he could to 39 it. It was not that he was lazy. 40 from it! Give him a mechanical job to do, from mending a gate to sharpening tools, 41 he would set to work eagerly. It was the daily life of the farm, with its dull tasks, 42 upset him.
Henry was excited by the development in technology that could __43 farmers like his father from wasteful and 44 labor. But these developments, in Henry’s boyhood, had touched farming 45 at all and farmers went on doing things in the way they had always done. So Henry 46 his attention elsewhere. When he was twelve, he became 47 in clocks and watches. Soon he was repairing them for friends, working at a bench he built in his bedroom. 
and watches as objects of Henry’s fascination. Making and installing them was the business of the Detroit workshop that he joined at the age of sixteen.
A. learn B. find C. Work D. join