ln(y)=n*ln(L/n)=n*ln(L)-n*ln(n)

d/dx ln(y) = d/dx [n*ln(L)-n*ln(n)]

y'/y=ln(L)-ln(n)-1

y'=y*(ln(L)-ln(n)-1)
now we want to know when y'=0

since L>0 and n>=1 we know y>0 thus we need

ln(L)-ln(n)-1=0

ln(n)=ln(L)-1

n=L/e

as was already determined and as was already pointed out we need n to be an integer so we simply need to look at the floor and ceiling of L/e and pick the one that results in the higher value of y

another interesting approach to this problem is to look at Y as a series for fixed L and seeing which value of n gives the max

So we have for a fixed L we have
y(1)=L
y(2)=(L/2)^2
....

y(n)=(L/n)^n

now if y(k) is a local max then we need
y(k)>y(k-1) and y(k)>y(k+1)

this gives us
(L/k)^k>=(L/(k-1))^(k-1) and (L/k)^k>=(L/(k+1))^(k+1)
simplifying the two equations gives us
k^k/(k-1)^(k-1)<=L<=k^k/(k+1)^(k+1)

So what this means is for there to be a local max at n=k then we need for L to be in the interval

[k^k/(k-1)^(k-1),k^k/(k+1)^(k+1)]


So another approach to solving this is to simply compute these intervals for various values of k and seeing which interval contains your value of L and that gives you the value of n for which the maximum occurs.