A stick of length L is broken into n equal parts.
What is the maximal product of their lengths?
(In reply to
re: What Ady wants (spoiler?) by Ady TZIDON)
this is the approach I had used as well
y=(L/n)^n
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(k1) and y(k)>y(k+1)
this gives us
(L/k)^k>=(L/(k1))^(k1) and (L/k)^k>=(L/(k+1))^(k+1)
simplifying the two equations gives us
k^k/(k1)^(k1)<=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/(k1)^(k1),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.

Posted by Daniel
on 20180605 11:08:00 