Design case of square fitness area

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…(2 points)

Cross m is MT∑BC, CD is t,

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…(7 points)

Because when n and f coincide, AM=AF=30 is suitable for the condition, so X ∈ (0 0,30], …(8 points)

(2)y=2400？ 5(60? x)240？ x=2400？ 5[(40? x)+40040？ X+40], …( 10 point)

What about when and only when you are 40? x=40040？ X, that is, when X = 20 ∈ (0 0,30), y gets the maximum value of 2000, …( 13 points).

Therefore, when DN=20m, the public fitness square has the largest area of 2000m2...( 14).