Because DN=x, NH=40-x, NA=60-x,
Because NHHG=NAAM,
So 40? x 10=60？ XAM, so am = 600? 10x40？ x，
Pass m as MT∨BC, hand in CD at t, and then
SMBCDW=SMBCT+SMTDN=(40？ AM)×60+ 12(x+60)×AM，
So y = (40? 600? 10x40？ x)×60+ 12×(x+60)(600？ 10x)40？ 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),
(ⅱ)y = 2400？ 5(60? x)240？ x=2400？ 5[(40? x)+40040？ x+40]，
What about when and only when you are 40? x=40040？ When x, that is, x = 20 ∈ (0 0,30), y gets the maximum value of 2000.
Answer: When DN=20m, the public fitness square has the largest area, with the largest area of 2000m2. ..