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# A Better Cable Connection

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## Solution Notes

Problem 1152 was solved by Julien Beasley, Barry Cox, Joseph DeVincentis, and Stephen Dupal.

Solution by Julien Beasley:

Let S(x) be the weight that the cable can hold at point x; then S(x)=k π r(x)^2.
Let W(x), the weight of the system at x, be how much weight is at or below height x; W(x) is the weight of the cable below x, plus the mass of the box, so W(x) = 1 + d * (volume of cable below x) = 1 + d Integrate [π r(X)^2, {x,0,X}].

The cable we seek has S(x)=W(x).

Let's first find r(0), using the fact that the weight held at the bottom end of the cable is 1 unit.

1=W(0) = S(0) = k π r(0)^2, so r(0) = 1/Sqrt[k π].

This is our initial condition.

The general S=W condition gives: k π r(x)^2 = 1+d Integrate [π r(X)^2, {x,0,X}].

Differentiate both sides and simplify to get r'(x) = d r(x) / (2 k) , which implies r(x) = C e^(d x/ (2 k)).

The initial condition means gives C = 1/Sqrt[k π], and so r(x) is C e^(d x/ (2 k)) or Sqrt[ e^(d x/ (2 k))/(k π) ]