y=3√( 1+2a^2)+2√[40+9( 1-a)^2]
=3√( 1+2a^2)+2√(49- 18a+9a^2),
y'=6a/√( 1+2a^2)+( 18a- 18)/√(49- 18a+9a^2)=0,
a√(49- 18a+9a^2)=(3-3a)√( 1+2a^2),①
Square a2 (49-18a+9a2) = (9-18a+9a2) (1+2a2),
18a^4-36a^3+27a^2- 18a+9
-9a^4+ 18a^3-49a^2=0,
9a^4- 18a^3-22a^2- 18a+9=0,
9(a^2+ 1/a^2)- 18(a+ 1/a)-22=0,
9(a+ 1/a)^2- 18(a+ 1/a)-40=0,
|a+ 1/a|=|a|+ 1/|a| >=2,
∴a+ 1/a= 10/3,
The solution is a= 1/3 (excluding a=3).
At this point, y = 5 √11;
When a=0, y =17; When a= 1, y=3√3+4√ 10≈ 17.8.
To sum up, the maximum value of y =3√3+4√ 10.