摘要：28．如图20.在平面直角坐标系中.四边形OABC是矩形.点B的坐标为(4.3)．平行于对角线AC的直线m从原点O出发.沿x轴正方向以每秒1个单位长度的速度运动.设直线m与矩形OABC的两边分别交于点M.N.直线m运动的时间为t(秒)． (1) 点A的坐标是 .点C的坐标是 , (2) 当t= 秒或 秒时.MN=AC, (3) 设△OMN的面积为S.求S与t的函数关系式, 中得到的函数S有没有最大值?若有.求出最大值,若没有.要说明理由． (08甘肃白银等9市28题解析)28． 本小题满分12分 解:, ·················································································· 2分 (2) 2.6, ········································································································· 4分 (3) 当0＜t≤4时.OM=t． 由△OMN∽△OAC.得. ∴ ON=.S=． ···································· 6分 当4＜t＜8时. 如图.∵ OD=t.∴ AD= t-4． 方法一: 由△DAM∽△AOC.可得AM=.∴ BM=6-． ···························· 7分 由△BMN∽△BAC.可得BN==8-t.∴ CN=t-4． ·································· 8分 S=矩形OABC的面积-Rt△OAM的面积- Rt△MBN的面积- Rt△NCO的面积 =12--(8-t)(6-)- =． ··························································································· 10分 方法二: 易知四边形ADNC是平行四边形.∴ CN=AD=t-4.BN=8-t．·································· 7分 由△BMN∽△BAC.可得BM==6-.∴ AM=．······ 8分 以下同方法一． (4) 有最大值． 方法一: 当0＜t≤4时. ∵ 抛物线S=的开口向上.在对称轴t=0的右边. S随t的增大而增大. ∴ 当t=4时.S可取到最大值=6, ················ 11分 当4＜t＜8时. ∵ 抛物线S=的开口向下.它的顶点是(4.6).∴ S＜6． 综上.当t=4时.S有最大值6． ······································································· 12分 方法二: ∵ S= ∴ 当0＜t＜8时.画出S与t的函数关系图像.如图所示． ······························ 11分 显然.当t=4时.S有最大值6． ··································································· 12分 说明:只有当第问只回答“有最大值 无其它步骤.可给1分,否则.不给分．

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