eywrnj5rr
幼苗
共回答了13个问题采纳率:84.6% 举报
矩形MDNP的面积S = DM*DN(由于你没给图,所以我设点M在CD上,点N在DE上)
延长MP、NP分别交AF、BF于Q、R
AF/BF = 2,所以AQ/PQ = 2
设DM = X,则DN= DE-NE = DE-PQ = 4-AQ/2 = 4-(AF-QF)/2 = 4-(AF-MC)/2 = 4-[2-(CD-DM)]/2 = 4-[2-(4-X)]/2 = -X/2 +5
所以 S = X*(-X/2 +5)= -X²/2 +5X ,
由图得2≤X≤4,所以当X=4时,S=12最大
五边形ABCDE的面积为15,所以钢板的最大利用率为12/15 = 80%
1年前
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