Let AP=m, AP: AB = Ah: AF = HP: BF.

m:√5=AH:2=HP: 1

Get AH=2m/√5 and HP=m/√5.

Then PM=4-(2-AH)=2+(2m/√5).

PN=4-PH=4-(m/√5)

So s rectangular PNDM

=PMxPN

=[2+(2m/√5)]x[4-(m/√5)]

=(-2/5)m^2 +(6/√5)m+8

When m =(-6/√5)÷(-4/5)=(3/2)÷5,

The maximum area is 12.

That is, when AP=(3/2)√5, the maximum area is 12.