用一定长度的绳子绑成一个矩形,如果矩形的一边长x(m)与面积y(m²)满足函数关系式y=-(x-12)²+144(0
人气:238 ℃ 时间:2019-12-16 02:52:05
解答
分析:本题考查二次函数最大(小)值的求法.
由函数关系y=-(x-12)2+144(0<x<24)可知,
∵二次函数的二次项系数即-1<0,
∴当x=12时,y最大值=144.
当x=12m时,矩形的面积最大,最大面积为144m²
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