问题 填空题
已知△AOB,点P在线段AB上,已知
OP
=m
OA
+4n
OB
,则mn的最大值为______.
答案

由点P在线段AB上可得A,P,B三点共线

由向量共线定理可得,存在实数λ使得

AP
AB
(0≤λ≤1)

OP
=
OA
+
AP
=
OA
+ λ
AB
=
OA
+λ(
OB
-
OA
)

=(1-λ)

OA
OB

OP
=m
OA
+4n
OB
OA
OB
不共线

∴m=1-λ,4n=λ

mn=

1
4
(1-λ)λ≤
1
4
(
1-λ+λ
2
)
2
=
1
16

故答案为:

1
16

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