问题 解答题

我市“上品”房地产开发公司于2010年5月份完工一商品房小区,6月初开始销售,其中6月的销售单价为0.7万元/m2,7月的销售单价为0.72万元/m2,且每月销售价格y1(单位:万元/m2)与月份x(6≤x≤11,x为整数)之间满足一次函数关系:每月的销售面积为y2(单位:m2),其中y2=-2000x+26000(6≤x≤11,x为整数).

(1)求y1与月份x的函数关系式;

(2)6~11月中,哪一个月的销售额最高?最高销售额为多少万元?

(3)2010年11月时,因会受到即将实行的“国八条”和房产税政策的影响,该公司销售部预计12月份的销售面积会在11月销售面积基础上减少20a%,于是决定将12月份的销售价格在11月的基础上增加a%,该计划顺利完成.为了尽快收回资金,2011年1月公司进行降价促销,该月销售额为(1500+600a)万元.这样12月、1月的销售额共为4618.4万元,请根据以上条件求出a的值为多少?

答案

(1)设y1=kx+b(k≠0),

由题意

6k+b=0.7
7k+b=0.72

解得:

k=0.02
b=0.58
y1=0.02x+0.58.

(2)设第x个月的销售额为W万元,

则W=y1y2=(0.02x+0.58)(-2000x+26000)

=-40x2-640x+15080,

∴对称轴为直线x=-

b
2a
=-
-640
-80
=-8,

∵当6≤x≤11是W随x的增大而减小,

∴当x=6时,

Wmax=-40×62-640×6+15080=9800(6分)

∴6月份的销售额最大为9800万元.

(3)11月的销售面积为:-2000×11+26000=4000(m2

11月份的销售价格为:0.02×11+0.58=0.8(万元/m2

由题意得:4000(1-20a%)×0.8(1+a%)+1500+600a=4618.4,

化简得:4a2+5a-51=0,解得:a1=3,a2=-

17
4
(舍)

∴a=3.

单项选择题
问答题

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