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 Optimal result (Posted on 2018-06-04)
A stick of length L is broken into n equal parts.

What is the maximal product of their lengths?

 No Solution Yet Submitted by Ady TZIDON No Rating

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 re(2): What Ady wants (spoiler?) | Comment 5 of 11 |

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(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.

 Posted by Daniel on 2018-06-05 11:08:00

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