推荐回答（1个）
1、设有x件玩具，y个小朋友
x=3y+59
5(y-1) ==>30≤Y<32
y=30,x=149或y=31,x=152
答：有149件玩具，30个小朋友或有152件玩具，31个小朋友
2、
1)设生产A产品x件
9x+4(50-x)≤360
3x+10(50-x)≤290
==>30≤x≤32
方案一、A产品30件，B产品20件
方案二、A产品31件，B产品19件
方案三、A产品32件，B产品18件
2)y=700x+1200(50-x)
=70000-500x
显然，x越小，y越大
当x=30时，ymax=55000元
3、
1)y=30x+50(6-x)+40(10-x)+80(8+x-6)
=20x+860
2)y=20x+860≤900
x≤2
方案一、从甲仓库调往A县农用车10辆，调往B县2辆，从乙仓库调往B县农用车6辆
方案二、从甲仓库调往A县农用车9辆，调往B县3辆，从乙仓库调往A县农用车1辆，调往B县5辆
方案三、从甲仓库调往A县农用车8辆，调往B县4辆，从乙仓库调往A县农用车2辆，调往B县4辆
3)当x最小时，y最小
即方案一，ymin=860元
4、设获奖人数x和书的本数y
y=3x+8
0 ==>5 ==>x=6
答：获奖6人，书有26本
5、 m(1+20%)(300-x)≥300m*80%
1.54mx>300m/2
==>97 x可取98、99、100，所以，有3种方案：
方案一、调配98人去生产新开发的B种产品
方案二、调配99人去生产新开发的B种产品
方案三、调配100人去生产新开发的B种产品
总年利润=m(1+20%)(300-x)+1.54mx
=0.34mx+360m
当x最大，总年利润最大
即方案三时，总年利润最大
小弟啊，这题目很简单啊，自己好好想想应该都能做出的、、（我很懒的，以后题目太多的话，别找我了,拜托~~^-^）
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