请教几道数学题 急!

2025-01-01 05:22:24
推荐回答(1个)
回答1:

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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