S = (50-40 cos θ) (50-40 sin θ), when θ =0 or θ =, S m a x =500.

Solution: extend the cross CD of GH to n, then NH =40sin θ and CN =40cos θ.

∴ HM = ND =50-40cos θ，AM =50-40sin θ。

So s = (50-40 cos θ) (50-40 sin θ) =1

100[25-20(sinθ+cosθ)+ 16 sinθcosθ](0≤θ≤)。

Let t =sin θ +cos θ = sin( θ+),

Then sin θ cos θ =, and t ∈ [1,].

∴s = 100[25-20t+8(T2- 1)]= 800(t-)2+450。

T ∈ [1,] ∴ When t = 1, S m a x =500,

At this time, sin( θ+)= 1 sin( θ+)=.

∵≤θ +≤π, ∴θ+= or π,

I.e. θ =0 or θ =.

A: When the H point is at the terminal E or F, the fitness room has the largest area, with the largest area of 500㎡.