Many of the problems in my previous post were instances of a general question:

Given three positive numbers V, S, and E, does there exist a cuboid with volume V in cubic inches, surface area S in square inches, and edge sum E in inches?

When I considered this question, I found that such a cuboid exists if and only if

**E^2 S^2 - 2 S^3 - 16 E^3 V + 36 E S V - 108 V^2 >= 0.**

The inequality was easy to obtain as soon as I realized that the desired cuboid exists if and only if the following polynomial in

*p*has three positive roots:

*p*^3 -

*E p*^2 + (

*S*/2)

*p*-

*V*.

(The left-hand side of the inequality is the discriminant of the polynomial.) Proof sketches are appended at the end.

***

The VSE inequality can be expressed in terms of two variables, the scaled surface area

*s*=*S*/(2*E*^2) and scaled volume*v*=*V*/(*E*^3). In terms of these parameters, the space of all possible cuboids looks like so:
Each light-blue or dark-blue point corresponds to a cuboid (except for the dashed line along the bottom, which is

*s*= 0). The two curved boundaries are where the discriminant vanishes, indicating a repeated root. Along the left boundary, there is a double root and its value is smaller than the other root; these are the rod-shaped cuboids. Along the right boundary, there is a double root and its value is larger than the other root; these are the tile-shaped cuboids. The right and left boundaries meet at a point right at the top; this cuboid is both rod-shaped and tile-shaped, i.e., it is a cube.
I put a little tour of the space up on YouTube here.

***

Well, it turns out that a cuboid can't have

*V*,

*S*, and

*E*numerically equal when measured in the same system of units. Seems a shame, really. Well, there's a desperate example in one of my notebooks with edge lengths 2,

*i*, and -

*i*.

But to keep it real, let's open things up to other shapes. Define the

*diameter*of a 3D shape to be the greatest possible distance between two of its points. Can you find a shape for which the volume, surface area, and diameter are all numerically equal when measured in the same system of units? (A few examples are on my whiteboard right now.)

***

It is nice to have an if-and-only-if condition, but the inequality is bit complicated. A few simpler, "only-if" inequalities (necessary conditions) are easy consequences of the inequality relating arithmetic and geometric means: Given

*n*nonnegative numbers

*c*_1, ...,

*c*_

*n*,

(1/

*n*)(

*c*_1+

*c*_2 + ...+

*c*_

*n*) >= (

*c*_1

*c*_2 ...

*c*_

*n*)^(1/

*n*),

with equality if and only if all the

*c*_

*i*are equal. (This inequality played a role in one of our previous puzzles.)

If we choose the

*c*_

*i*to be the edge lengths

*x*,

*y*, and

*z*, then we find

**E >= 3V^(1/3),**

with equality if and only if the cuboid is a cube.

This inequality verifies our intuition that a cuboid can't enclose a given volume using arbitrarily little edge length. For example, there is no cuboid with a volume of a million cubic inches and an edge sum of 1 inch.

The inequality can also be written

**(E/3)^3 >= V**

again with equality if and only if the cuboid is a cube.

The two forms of the inequality correspond to two optimal properties of the cube: Of all cuboids with a given volume, the cube has the least edge sum; and of all cuboids with a given edge sum, the cube has greatest volume.

(Proof of the second property: Let C be a cuboid with edge sum

*E*. Suppose K is any cuboid with edge sum

*E*that is not a cube. Denote the volumes by

*V*_C and

*V*_K. Applying the inequality to K, we have (

*E*/3)^3 >

*V*_K. But of course, (

*E*/3)^3 =

*V*_C. Therefore

*V*_C >

*V*_K.)

Now take the

*c*_

*i*to be

*xy*,

*yz*,

*zx*. This gives

**S >= 6V^(2/3)**

again with equality if and only if the cuboid is a cube.

This again establishes something intuitively obvious, which is that a cuboid can't enclose a given volume using arbitrarily little surface area. For example, there is no cuboid with a volume of 1 cubic inch and a surface area of a millionth of an inch.

The inequality can also be written

**(S/6)^(3/2) >= V**

again with equality if and only if the cuboid is a cube.

The two forms again correspond to two optimal properties of the cube: Of all cuboids with a given volume, the cube has the least surface area; and of all cuboids with a given surface area, the cube has greatest volume.

The last two inequalities imply a third:

**ES >= 18V.**

This can also be found by applying the inequality of arithmetic and geometric means, taking the

*c*_

*i*to be the nine expressions

*xyz*,

*xyz*,

*xyz*,

*x*^2

*y*,

*xy*^2,

*x*^2

*z*,

*xz*^2,

*y*^2

*z*, and

*yz*^2.

For a cuboid with measures

*V*,

*S*, and

*E*to exist, it is necessary but not sufficient that

*V*,

*S*, and

*E*satisfy these simpler inequalities. For example, the numbers

*E*= 8,

*S*= 78, and

*V*= 9 satisfy all of the "only-if" inequalities, although in fact there is no cuboid with these measures.

Another "only-if" inequality follows from applying the Cauchy-Schwarz inequality to the vectors (

*x*,

*y*,

*z*) and (

*z*,

*x*,

*y*). The result is (1/2)

*S*<=

*x*^2 +

*y*^2 +

*z*^2. But x^2 + y^2 + z^2 =

*E*^2 -

*S*, so we have

**S <= (2/3)E^2**

again with equality if and only if the cuboid is a cube.

This inequality can be used to show two more properties of the cube: Of all cuboids with a given surface area, the cube has the least edge sum; and of all cuboids with a given edge sum, the cube has greatest surface area.

***

By the way, we still need the if-and-only-if VSE inequality. None of the "only-if" inequalities is strong enough to rule out the existence of a cuboid with a volume of 80 cubic inches, a surface area of 122 square inches, and an edge sum of 16 inches (though in fact no such cuboid exists).

***

Proof sketch:

(1) Given that a cuboid exists with measures

*V*,

*S*, and

*E*, to show that the polynomial

*p*^3 -

*E p*^2 + (

*S*/2)

*p*-

*V*has three positive roots.

*Idea: the positive roots are none other than the edge lengths.*

Denote the cuboid's edge lengths by (positive)

*x*,

*y*, and

*z*. Then since

*x*+

*y*+

*z*=

*E*, 2(

*xy*+

*yz*+

*zx*)

*=*

*S*, and

*xyz*

*=*

*V*, the polynomial is

*p*^3 - (

*x + y + z*)

*p*^2 + (

*xy + yz + zx*)

*p*-

*xyz*. This however factors as

*(*

*p*-

*x*)(

*p*-

*y*)(

*p*-

*z*), hence the polynomial has three positive roots.

(2) Given that the polynomial

*p*^3 -

*E p*^2 + (

*S*/2)

*p*-

*V*has three positive roots, to show that a cuboid with measures

*V*,

*S*, and

*E*exists.

*Idea: the edge lengths are none other than the positive roots.*

Denote the positive roots

*by*

*r*1,

*r*2, and

*r*3. Since the polynomial is a monic cubic, the coefficients can be expressed in terms of the roots as symmetric functions, specifically as

*E*=

*r*1+

*r*2+

*r*3,

*S*/2 =

*r*1

*r*2+

*r*1

*r*3+

*r*2

*r*3, and

*V*=

*r*1

*r*2

*r*3. But these functions are also the formulas for the measures of a cuboid. So create a cuboid with edge lengths

*r*1,

*r*2, and

*r*3. The cuboid then has the desired measures.

(3) To arrive finally at the desired inequality.

It is obvious by inspection that every real root of

*p*^3 -

*E p*^2 + (

*S*/2)

*p*-

*V*is positive. (Try substituting a negative number or zero for

*p*.) Hence by (1) and (2), a cuboid exists having the specified measures if and only if the polynomial has three real roots. It now follows that the discriminant of the polynomial provides a necessary and sufficient condition for the existence of a cuboid with measures

*V*,

*S*, and

*E.*

***

*k*th coefficient of an

*n*th-degree monic polynomial is going to be an elementary symmetric polynomial function of the roots...and meanwhile, the

*k*-dimensional measure

*M*_

*k*of an

*n*-dimensional cuboid is going to be the same elementary symmetric polynomial function of the edge lengths, multiplied by an overall combinatorial factor

*C*_

*nk*that depends on the dimension of the measure and the dimension of the space. So if we define the reduced measures of a cuboid by

*m*_

*k*= M_

*k*/

*C*_

*nk*, then there will exist an

*n*-dimensional cuboid with measures

*M_*1, ...,

*M*_

*n*if and only if the polynomial

*p*^

*n*-

*m*_1

*p*^(

*n*-1) +

*m*_2

*p*^(

*n*-2) - ... +

*m*_

*n*

has

*n*real roots.

Even supposing this principle does generalize, however, the discriminant for higher polynomials doesn't tell the real/complex nature of the roots in the simple way that it does for

*n*= 2 and

*n*= 3. So for

*n*>= 4, I don't think the discriminant will give an immediately usable condition on the coefficients

*m*_

*i*.

***

Usual disclaimers...I assume everything here that is correct is well-known. There may be some interesting related readings out there - if anybody knows of good ones, please put them in the comments!

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