【答案】
分析:(1)依題意當(dāng)銷售單價定為x元時,年銷售量減少
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/0.png)
(x-100),則易求y與x之間的函數(shù)關(guān)系式,進而由題意易得Z與x之間的函數(shù)關(guān)系.
(2)把z與x的關(guān)系式化簡,得出當(dāng)x=170時,z取最大值;即可得出公司是盈利了還是虧損;
(3)根據(jù)z=(30-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/1.png)
x)(x-40)-310=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/2.png)
x
2+34x-1510=1130進而得出當(dāng)120≤x≤220時,z≥1130畫出圖象得出即可.
解答:解:(1)依題意知,當(dāng)銷售單價定為x元時,年銷售量減少
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/3.png)
(x-100)萬件,
∴y=20-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/4.png)
(x-100)=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/5.png)
x+30,
即y與x之間的函數(shù)關(guān)系式是y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/6.png)
x+30.
由題意得:
z=y(x-40)-500-1500
=(30-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/7.png)
x)(x-40)-500-1500
=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/8.png)
x
2+34x-3200,
即z與x之間的函數(shù)關(guān)系是z=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/9.png)
x
2+34x-3200.
(2)∵z=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/10.png)
x
2+34x-3200,
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/11.png)
(x-170)
2-310.
∴當(dāng)x=170時,z取最大值,為-310,
即當(dāng)銷售單價為170元,年獲利最大,并且第一年年底公司還差310萬元就可收回全部投資.
(3)第二年的銷售單價定為x元時,年獲利為:
z=(30-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/12.png)
x)(x-40)-310=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/13.png)
x
2+34x-1510.
當(dāng)z=1130時,即1130=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/14.png)
x
2+34x-1510,
整理得x
2-340x+26400=0,
解得:x
1=120,x
2=220.
函數(shù)z=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/15.png)
x
2+34x-1510的圖象大致如圖所示,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101192605837878934/SYS201311011926058378789021_DA/images16.png)
由圖象可以看出:當(dāng)120≤x≤220時,z≥1130.
故第二年的銷售單價應(yīng)確定在不低于120元且不高于220元的范圍內(nèi).
點評:本題主要考查了二次函數(shù)的應(yīng)用和一元二次方程的解法以及二次函數(shù)圖象等知識,根據(jù)已知得出z與x之間的函數(shù)關(guān)系是解題關(guān)鍵.