一道初中数学题

2024-12-18 15:03:54
推荐回答(5个)
回答1:

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回答2:

调配后,企业生产这种产品的年利润为(300-x)(1+20%)2;企业生产B种产品的年利润为1.54*2x;

设调配后企业全年总利润为y万元,则y关于x的函数解析式为
y=(300-x)(1+20%2m+1.54*2x.

Y=2.4(300-x)+3.08x
2.4(300-x)≥600*0.8
3.08x>300
x=100,99,98
x=100时最大

回答3:

1.调配后,企业生产A的年利润为_____________万元,企业生产B的年利润为__________万元,若设调配后企业每年总利润为y万元,则用含x的代数式表示y时,y=___________________.

解:A的年利润为2*1.2(300-X)=720-2.4X(万元)
B的年利润为3.08X(万元)
所以Y=720-2.4X+3.08X=0.68X+720

2.若要求调配后,企业生产A的年利润不小于调配前企年利润的4/5,生产B的年利润大于调配前企业年利润的一半,共有几种调配方案?其中哪种方案全年获利最大?
企业调配前年利润为600万元,
可列方程组
720-2.4X≥600*4/5 ①
3.05X≥600*1/2 ②
解得97.4≤X≤100
∴X=98 99 100
共有三种调配方案
Y随X的增大而增大,
所以当X=100时,Y最大=788

回答4:

1 调配后A产品平均每人每年创造利润:2×(1+0.2)=2.4万
所以调配后企业生产A的利润应为:(300-x)×2.4万元
B产品利润:3.08x 万元 y=(300-x)×2.4+3.08x

2 解不等式组:(300-x)×2.4≥0.8×(300×2)
3.08x>0.5×300×2
解出x的取值范围就是答案

回答5:

(1)第一个空:1.2*(300-x)*2,第二个空3.08*2x
(2)由题意得:(300-X)(1+20%)M>0.8*300*2
1.54*2>0.5*300*2

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