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.