本试题主要是考察了函数在实际生活中的运用,借助于不等式的思想或者是函数单调性的思想,求解最值的实际应用。
(1)根据已知条件,设出变量,然后借助于面积关系,得到解析式。
(2)根据第一问中的结论,分析函数的性质,或者运用均值不等式的思想,求解得到最值。
解: (1)由题可得:xy=1800,b=2a
则y=a+b+3=3a+3, ··········· 4分
S=(x-2)a +(x-3)b=(3x-8)a=(3x-8)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221640637485.png)
=1808-3x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221640622386.png)
y. ········ 8分
(2) S=1808-3x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221640622386.png)
y=1808-3x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221640622386.png)
×
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221640731495.png)
=1808-3 (x+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221640747509.png)
) ······· 10分
≤1808-3×2
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221640762668.png)
=1808-240=1568, ·········· 12分
当且仅当x=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221640747509.png)
,即x=40时取等号,S取得最大值.此时y=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823221640731495.png)
=45,
所以当x=40,y=45时,S取得最大值. 15分