It is known that A > 0, B > 0 and a+b=2. Find the minimum value of y =1/a+4/b.

Analysis: ∫A > 0, B > 0, a+b=2.

∴a=2-b

∴y=f(b)= 1/(2-b)+4/b(0 & lt； b & lt2)

Let f' (b) =1(2-b) 2-4/B2 = 0 = > b1= 4/3, b2=4.

0<b< at 4/3, f' (b); 0

∴y takes the minimum value F(4/3)=3/2+3=9/2 at x=4/3.