A free hanging cylindrical rope will break under its own weight if it exceeds a certain critical length. Suppose you have a well that's deeper than the critical length of your rope, and you need a rope that will reach the bottom of that well. A rope will break at any point if the stress at that point exceeds a certain critical value. The stress is the ratio of the weight below that point and the area of the cross section at that point. Thus it does not help to increase the radius of the rope by a certain factor. You'd increase the weight and the area of the cross section by the square of that factor. When computing the stress those squares would cancel. The critical length of a cylindrical rope is independent of its radius. However, you can increase the depth a rope can reach by increasing its radius towards the top. This problem explores that idea. The rope will break when the stress $s$ at a certain point exceeds a specific value $c$. That critical value $c$ depends on the material of which the rope is made. So we want to construct a rope which has a constant stress everywhere along the rope, and that carries a weight $w$, say. Putting this information into mathematical terms we obtain the equation $$\frac{u \int_0^x \pi r^2(t) \hbox{d}t + w}{\pi r^2(x)} = c$$ or $$u \int_0^x \pi r^2(t) \hbox{d}t + w = c{\pi r^2(x)}$$ for all $x$. Here

$u$ is the specific weight of the rope
$r(x)$ is the radius of the rope at a distance $x$ measured upwards from its bottom
$w$ is the weight carried by the rope.
$c$ is the critical stress (incorporating any safety factors).
Differentiate with respect to $x$ in the above equation, obtain a differential equation for $r$, determine $r(0)$ by the fact that $w$ is attached at $x=0$ and enter the radius of your special rope:
$r(x) =$ .
Of course, your answer will depend on $x$, $w$, $u$, and $c$.