(2) make Y=aX? +bX+c, substituting O (0 0,0), A (1, 1) and d (5/2,0), we get

c=0，b=5/3，a=-2/3

So Y=-2/3X? +5/3 times

(3) let p (x, y), 1 < x < 5/2, 0 < y < 1, and the vertical cross p is the OD cross q. ..

S=SOAB+SOAPQ+SPQD

= 1/2× 1× 1+ 1/2×(y+ 1)×(x- 1)+ 1/2×y×(5/2-x)

=0.5×( 1-x？ +5/2 times)

=-0.5×(x-√5/2)？ +9/8

Therefore, when x=√5/2, the maximum value of S =9/8.