I told you how to exclude 1.6.
Let the speed be v and the moving time be t.
From the meaning of the question: △OPQ bottom OQ=VT+2
The height can be expressed as (10-VT/2).
So the area can be expressed as:
s△OPQ = 1/2(VT+2)( 10-VT/2)
=-1/4 (vt) ˇ 2+9/2 (vt)+10 (process omitted)
Since v is a fixed value, it is considered a constant.
So when the area is the largest,
T=-b/2a=9/V (process omitted)
According to the image, when the area is the largest, t < 5.
Bring V=2 and V= 1.6 into T=9/V, respectively.
Available: when V= 1.6, t > 5, it should be discarded.
(2 stands for square)