Question 2077
1. Sol:Assume the first piece is rm, and the 2nd piece is (1-r)m,
       where 0<= r <=1.
       The sum of  two prices = a(rm)^2 + a((1-r)m)^2 = a(2r^2 -2r+1)m^2
          = 5/9 am^2.
       So, 2r^2 -2r+1 = 5/9, or 9r^2 -9r + 2  = 0.
       By factoring (3r -1)(3r -2) = 0, we have r=1/3 or 2/3.
       [Note: In my opinion, if you can factor out directly,
       don't use quadratic formula.]
      Hence,the ratio of the two pieces is r:(1-r) = 1/3:2/3 =1:2
      (or 2:1).  
2.Use complete square
     f(r) = 2r^2 -2r+1 = 2(r^2 -r+1/4) +1/2   = 1/2 + 2(r-1/2)^2 => 1/2.
    Since 2(r-1/2)^2 >= 0, when r = 1/2, f(r) has minimum value 1/2.
    That is, if the ratio = r:(1-r) = 1:1, the the sum of the prices of 
    these two pieces is lowest.
  
 Kenny