Activity 2-8: V, S and E

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Autograph Activity link

E = total edge lengthS = total surface area

V = volume

There are six ways to write E, S and V in order of size.

Interesting question:can you find a cube

for each order?

If not, what about a cuboid?

E = 12x, S = 6x2, V = x3

Let’s try a cube, of side x:

We can plot y = 12x, y = 6x2, y = x3

together…

Autograph File 1

Only four regions!

log y = logx + log 12logy = 2logx + log 6

log y = 3logx

y = 12x, y= 6x2,y = x3

Or, taking logs with

gives us

and now we can plot log y v log x:

It’s clear that only 4 out of 6 orders are

possible.

0 < x < 2 V < S < E

2 < x < √12 V < E < S

√12 < x < 6 E < V < S

6 < x E < S < V

The four possible orders are:

What happens if we look at a cuboid instead of a cube?

Can we get the missing orders now?

Let’s start with a cuboid with sides x, x, and y.

V = x2yS = 4xy + 2x2

E = 8x + 4y

Autograph File 2

So we can work in 3D, plotting z = x2y, z = 4xy + 2x2, z = 8x + 4y.

Red < Green < PurplePurple < Red < Green Red < Purple < GreenPurple < Green < Red

It seems we can manage these orders, but no others:

E < S < VV < E < S E < V < SV < S < E

So we get the same orders that we had

with the cube...

There’s another way to look at this:

Autograph File 3

What happens as we vary k?

Take a cuboid with sides x, x, and kx

log y = logx + log (8+4k)logy = 2logx + log (2+4k)

log y = 3logx + log k

y = (8+4k)x, y = (2+4k)x2,y = kx3

Or, taking logs with

gives us

and now we can plot log y v log x,And we have three straight lines as before,

And only four possible orders.

So no new orders are possible!

Can we find a cuboid with sides x, y, z

such that S < E and S < V?

We need; xyz > 2xy + 2yz + 2zx

and 4x + 4y + 4z > 2xy + 2yz + 2zx

Now if a > b > 0 and c > d > 0,

then ac > bd > 0

So if

xyz > 2xy + 2yz + 2zx > 0

and 4x + 4y + 4z > 2xy + 2yz + 2zx > 0

then (4x+4y+4z)xyz > (2xy+2yz+2zx)2

So4x2yz+4xy2z+4xyz2 > 4x2yz+4xy2z+4xyz2+f(x, y, z)

Contradiction!

where f(x, y, z) > 0.

If x = 3, y = 4 and z = 5,

then V = 60, S = 94, E = 48.

Is there another cuboidwhere the values for V, S, and Eare some other permutation of

60, 94 and 48?

(2x-a)(2x-b)(2x-c)

= 8x3 - 4(a+b+c)x2 + 2(ab+bc+ca)x - abc

= 8x3 – Ex2 + Sx – V

where E, S and V are forthe cuboid with sides a, b and c.

The equation 8x3 – Ex2 + Sx – V = 0has roots a/2, b/2 and c/2.

y = 8x3 48x2 + 94x – 60,y = 8x3 48x2 + 60x – 94,y = 8x3 94x2 + 48x – 60,y = 8x3 94x2 + 60x – 48,y = 8x3 60x2 + 94x – 48,y = 8x3 60x2 + 48x – 94.

So our question becomes: which of the following six curves

has three positive roots?

Just the one. What happens if we

vary V, S and E?

Autograph File 4

a = 8, b = 30, c = 29.

Yellow: roots are 0.4123..., 1.2127..., 2, sides are double.

Green: roots are 0.5, 0.8246..., 2.4254..., sides are double.

V = 8, S = 30, E = 29.

V = 8, S = 29, E = 30.

With thanks to:Rachel Bolton,

for posing the interesting question at the start. Douglas Butler.

Carom is written by Jonny Griffiths, mail@jonny-griffiths.net