2, a, (0<X & lt9/5), X/(9/5) =EF/( 12/5) gives EF=4X/3, so Y=0.5*X*(4X/3)=2X2/3.

(9/5 & lt； = X & lt=5), (5-x)/(16/5) = ef/(12/5) gives EF=3(5-X)/4, so y = 0.5 * x * 3 (5-x).

So Y=( 15X-3X2)/8.

B, obviously when x; At 9/5, a parabola is the largest at the vertex, that is, when X=2.5, then Y=75/32. To sum up, the maximum Y =75/32, and X=2.5 at this time.

3. Triangle ABC: perimeter C= 12, area S=6.

So: the perimeter of AEF is 6 and the area is 3.

Namely: a, ef+x+AE = 6; b、0.5 * X * EF = 3； c、AE 2=X 2+EF 2

Simplification: 6/X+X=4: X2-4X+6=0 because B2-4ac.