Let MD = x and nd = y.

The side length of a square ABCD is 1.

∴ab=bc=ad=cd= 1,∠a=∠bcg=90

∴am= 1-x,cn= 1-y,,△bam≌△bcg(sas)

∴CG=AM= 1-y,BM=BG

∴gn=cn+cg=( 1-x)+( 1-y)=2-x-y

∫△MDN circumference is 2.

∴MN=2-x-y

∴MN=GN

BM = BG，，BN=BN

∴△BMN≌△BGN(SSS)

∴S△BMN

=S△BGN

=S△BCN+ S△BCG

= S△BCN+S△ bam

∴S square ABCD

= S△BCN+S△ bam+S△BMN+S△MDN

=2 S△BMN+S△MDN

∴S△BMN=(S square ABCD- S△MDN)/2

∫S squared ABCD =1*1=1,and find the minimum value of S△BMN.

∴S△MDN should take the maximum value.

∴DM=DN

That is x = y.

∴MN=x*√2

∴C△MDN=x+x+ x*√2=2

X=2-√2。

∴S△MBN

= S squared ABCD-S△BCN-S△ bam -S△MDN

= 1-[( 1-x)* 1]/2-[( 1-x)* 1]/2-(x * x)/2

= - 1+√2