(In reply to
Analytical solution (spoiler) by Steve Herman)
Taking off from b=a/(a1):
y = ab = a^2 / (a1)
y' = (a^2  (a1)*2*a) / (a1)^2
which is zero when
a^2 = 2*a^2  2*a
a^2  2*a = 0
a=0 or a=2
a=0 is not positive, and also for y itself 0<a<1 makes b negative.
a=2 can be shown to be a minimum for y rather than a maximum, so b = 2 and ab=4.

Posted by Charlie
on 20130412 14:04:01 