The Rope of Dreams Recut: Polynomials of the Third Order – Cubic Equations

The Rope of Dreams in 3D
The Rope of Dreams in 3D

The Rope of Dreams Recut – Cubic Equations

Did you make the most of The Rope of Dreams in my previous post “The Rope of Dreams : Polynomials of the Second Order – Quadratic Equations“. I hope so. Well now you have a chance to take your 120 meter rope and enter another Dimension with Cubic Equations.

What is the Scenario?

Once again you are given the Rope of Dreams, and a Golden Scissors which is the only thing that can cut the Rope of Dreams. You can cut the Rope of Dreams twice (cross section, no longitudinal cuts), which will give you 3 lengths. These lengths will be laid out one for each dimension X, Y, Z (i.e. left-right, backward-forward, up-down), and whatever volume you enclose anywhere on Earth is yours to keep, or do with whatever you wish. You can enclose only a single volume, a single time, and then must return the rope and the scissors. What would you do? You may not think of a Polynomial of the Third Order – a Cubic Equation, but you probably should. In this post we’re talking about cuboids, we’ll leave spherical shapes aside for now.

Do the Math!

This time we’ll do the mathematics first and then apply our findings to determine how best to maximise the volume we enclose. In the previous post we derived and proved the following quadratic equation:

Previously: \({ x^2 = (x-n)(x+n) + n^2 }\)

Multiply by \(x\): \({ x^3 = x ((x-n)(x+n) + n^2)}\)

Multiplying this out: \(\boxed{ x^3 =  (x-n) x (x+n) + xn^2}\)

And there we have it! The way to maximise volume is to have n=0. Why? Well may you ask. The equation is telling us that if we move n units from x, i.e. \((x-n)\), then the loss of volume will be \(xn^2\). If we don’t want to lose any volume then we should have \(n=0\) because then \(xn^2 = 0\).

With \(n=0\): \({ x^3 =  (x-0) x (x+0) + x0^2}\)

Which gives: \({ x^3 =  (x) x (x) + 0}\)

Let’s contrast that with moving 50 units from x:

With \(n=50\): \({ x^3 =  (x-50) x (x+50) + x50^2}\)

Which gives: \({ x^3 =  (x-50) x (x+50) + 2500x}\)

So by moving 50 units we lost a volume of \(2500x\). Ouch!

Prove It.

Let’s prove that: \({ x^3 =  (x-n) x (x+n) + xn^2}\)

Moving x: \({ x^3 =  (x( (x-n) \cdot (x+n) )) + xn^2}\)

Multiplying: \({ x^3 =  (x( x^2 +nx -nx -n^2) ) + xn^2}\)

Simplifying: \({ x^3 =  (x( x^2 -n^2) ) + xn^2}\)

Multiplying: \({ x^3 =  x^3 -xn^2 + xn^2}\)

Simplifying:  \({ x^3 =  x^3}\)


Which Shape should You use to Maximise Volume?

A cube will give the greatest volume, any other rectangular cuboid will be sub-maximal. So you should cut the 120m rope into three sections of 40m each – this will give you the maximum volume of 40 x 40 x 40 = 64000 m^3. Let’s look at some sample choices:

Two 10 meter sections and a 100 meter section: \( 10 \cdot 10 \cdot 100 = 10000m^3\)

Two 20 meter sections and an 80 meter section: \( 20 \cdot 20 \cdot 80 = 32000m^3\)

Two 30 meter sections and an 60 meter section: \( 30 \cdot 30 \cdot 60 = 54000m^3\)

Three 40 meter sections: \( 40 \cdot 40 \cdot 40 = 64000m^3\)

Two 50 meter sections and a 20 meter section: \( 50 \cdot 50 \cdot 20 = 50000m^3\)


A cube is the ideal shape (for cuboids) to maximise volume. The further you move from a cube the greater the loss in volume. The equation is simple to derive and we’ve proved it mathematically, and used sample values for our Rope of Dreams. I hope you found somewhere nice to use it. Send us a postcard!

The Cube is King of the Cuboids: \(\boxed{ x^3 =  (x-n) x (x+n) + xn^2}\)

BORG efficiency utilises the CUBE
BORG efficiency utilises the CUBE

Never one to shy away from a Star Trek connection I will point out that the stalwart of spacetravel for the BORG is their trusty BORG Cube, measuring 3km sides giving a total volume of 27 cubic km. Never one to miss out on an efficiency the BORG know their cubes and their cubic equations. (They also use Spheres – again an optimal shape for surface area to volume).


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