(In reply to
Analytical solution (spoiler) by Steve Herman)
Taking off from b=a/(a-1):
y = ab = a^2 / (a-1)
y' = (a^2 - (a-1)*2*a) / (a-1)^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.
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Posted by Charlie
on 2013-04-12 14:04:01 |