作业帮平面直角坐标系中,四边形OABC是矩形,点B的坐标（4,3）.平行于对角线AC的直线m从原点O出发,沿看图吧,双击打开,
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平面直角坐标系中,四边形OABC是矩形,点B的坐标（4,3）.平行于对角线AC的直线m从原点O出发,沿看图吧,双击打开,
平面直角坐标系中,四边形OABC是矩形,点B的坐标（4,3）.平行于对角线AC的直线m从原点O出发,沿
看图吧,双击打开,
平面直角坐标系中,四边形OABC是矩形,点B的坐标（4,3）.平行于对角线AC的直线m从原点O出发,沿看图吧,双击打开,
(1)A(4,0) C(0,3) (2)当t=2秒或6秒时,MN=1/2AC（理由是,此时MN是AC的中位线）
（3）S＝1/2×t×(3/4t)=3/8t平方（t4)
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就是这样
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(1)（4，0），（0，3）； ·················································································· 2分
(2) 2，6； ···················································································...
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(1)（4，0），（0，3）； ·················································································· 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
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