⑵ When q moves on BC, that is, when 0 < t￠5, let Q(a, b).

∫C( 1，4)B(4，0)

∴L(CB)=√▔(3？ +4? ) = 5, then l (BCD) = 5+ 1 = 6.

∫l(ao)= 4 and PQ move at the same time and reach the end point at the same time.

∴ ao/VP = BCD/ 1，VP = 2/3。

∫b again? +(4-a)？ =t？ Q satisfies the function on BC, and B =-4/3a+ 16/3 is brought in.

Get b = 4/5t.

∴s= 1/2op×h= 1/2(4-2/3t)×4/5t=-4/ 15(t？ -6t)+ 12/5

When t = 3, the maximum value of s is 12/5. When q is on the CD, that is, when 5≤t≤6, Q(c, 4).

s = 1/2op×QF = 1/2(4-2/3t)×4 = 8-4/3t

When t = 5, the maximum s is 4/3.

(3) When t = 3, the maximum value of s is 12/5.