To maximize the product of lengths, note first that

all the pieces should be of equal length. (If you have pieces of lengths $a\pm h$ where $h\neq0$ then replacing them with two pieces of length $a$ increases the product of those two lengths from $a^2-h^2$ to $a^2$ without changing anything else.)

Now,

if you cut a rope of length $L$ into $n$ equal pieces then the resulting product is $(L/n)^n$. It'll be convenient to take its logarithm, $n\log L-n\log n$. If we imagine varying $n$ continuously (which of course we can't actually do) then the derivative is $\log L-\log n-1$, which is increasing for $L/n<e$ and decreasing for $L/n>e$. So the best $n$ is either $\lfloor L/e\rfloor$ or $\lceil L/e\rceil$.

Let the lengths of rope be $a$ (for Alice) and $5a$ (for Bob). Then the optimal numbers

are nearest approximations to $a/e$ and $5a/e$. I don't think it's possible for the ratio to be 6 (which is quite far from 5) unless we take $\lfloor a/e\rfloor$ and $\lceil 5a/e\rceil$. We also must have $a$ not too large (because as $a$ gets larger the ratio clearly $\rightarrow5$).

It turns out that

the only values of $a$ for which that ratio is correct are 3,6,13,16; and the only one of these for which the floor/ceiling choices work out the right way is $a=3$.

So the original length is

18 units.

(I may fill in some more details later. I need to be AFK for a bit starting about five minutes ago :-).)

To maximize the product of lengths, note first that

all the pieces should be of equal length. (If you have pieces of lengths $a\pm h$ where $h\neq0$ then replacing them with two pieces of length $a$ increases the product of those two lengths from $a^2-h^2$ to $a^2$ without changing anything else.)

Now,

if you cut a rope of length $L$ into $n$ equal pieces then the resulting product is $(L/n)^n$. It'll be convenient to take its logarithm, $n\log L-n\log n$. If we imagine varying $n$ continuously (which of course we can't actually do) then the derivative is $\log L-\log n-1$, which is increasing for $L/n<e$ and decreasing for $L/n>e$. So the best $n$ is either $\lfloor L/e\rfloor$ or $\lceil L/e\rceil$.

Let the lengths of rope be $a$ (for Alice) and $6a$ (for Bob). Then the optimal numbers

are nearest approximations to $a/e$ and $6a/e$. I don't think it's possible for the ratio to be 5 (which is quite far from 6) unless we take $\lceil a/e\rceil$ and $\lfloor 6a/e\rfloor$. We also must have $a$ not too large (because as $a$ gets larger the ratio clearly $\rightarrow6$).

It turns out that

the only values of $a$ for which that ratio is correct are 7 and 14; and the only one of these for which the floor/ceiling choices work out the right way is $a=7$.

So the original length is

7+42=49 units.