Solution: (1) is GH⊥EF, and the vertical foot is H.

Because DN=x, NH=40-x, NA=60-x,

Because NHHG=NAAM,

So 40? x 10=60？ XAM, so AM=600? 10x40？ x，

Cross m is MT∑BC, CD is t,

Then smbcdw = smbct+smtdn = (40-am) × 60+12 (x+60) × am,

So y=(40-600? 10x40？ x))×60+ 12(x+60)×600？ 10x40？ x=2400-5(60？ x)240？ x

Since AM=AF=30 is suitable for the condition when n and f coincide, x ∈ (0 0,30),

(2)y=2400-5(60？ x)240？ x=2400-5[(40-x)+40040？ x+40]，

So if and only if 40-x=40040? When x, that is, x = 20 ∈ (0 0,30), y gets the maximum value of 2000.

Therefore, when DN=20m, the public fitness square has the largest area, with the largest area of 2000m2. ..