Rectangular boxes: handle with care

A rectangular box (aka parallelepiped) looks like a sturdy object:

Rainbow sheep inside
Rainbow sheep inside

But this particular box, with dimensions 7.147 by 6.021 by 4.095, took me the better part of an hour to figure out.

It was a part of a numerical methods assignment: find the dimensions of a box with given volume {V}, surface area {S}, and diameter {D} (i.e., space diagonal). Algebraic approach leads to pretty ugly expressions, which is the motivation for a numerical method. Specifically, the task was to apply the Newton-Raphson method to the map

\displaystyle    F(x,y,z) = \begin{pmatrix} xyz-V  \\ 2(xy+yz+xz)-S  \\ x^2+y^2+z^2-D^2 \end{pmatrix}

Of course, I understood that not every triple {(V,S,D)} is attainable. Also realized that the Jacobian of {F} is degenerate when two of the coordinates coincide, which is a problem for the method. So I thought: let’s generate some random {x,y,z} values that are not too close to one another, and give students the resulting parameters {V,S,D}.

With {x=7.147}, {y=6.021}, and {z=4.095} the parameters are {V=176.216}, {S=193.91}, and {D=10.203}. Sure, a little rounding can’t hurt when numbers are of this size and we are quite far from the critical points of {F}. So I put {V=176}, {S=194} and {D=10} in one of the versions of the assignment.

But the Newton-Raphson method would not converge… because no such box exists! The rounding did hurt after all.

This motivated me to describe all attainable triples {(V,S,D)} explicitly, which ended up being less of a chore than I expected. It helps to realize that {(x+y+z)^2 = D^2+S}, which reduces the search to the intersection of the sphere {x^2+y^2+z^2=D^2} with the plane {x+y+z=\sqrt{D^2+S}}. This is a circle (called {C} below), and the allowed range for {V} is between the minimum and maximum of {xyz} on {C}.

This goes into Calculus 3 territory. Using Lagrange multipliers with two constraints looks like a tedious job. Instead, I decided to parametrize {C}. Its center is {(c,c,c)} where {c=\dfrac13\sqrt{D^2+S}}. The radius is {\displaystyle r = \sqrt{ D^2 - \frac13(D^2+S)}=\frac13 \sqrt{6D^2-3S}}. We also need an orthonormal basis of the subspace {x+y+z=0}: the vectors

\displaystyle  \frac{1}{\sqrt{6}} \langle 2, -1, -1\rangle \quad \text{and}\quad \frac{1}{\sqrt{2}} \langle 0, 1, -1\rangle

do the job.

So, the circle {C} is parametrized by

\displaystyle    x  = c+\frac{2r}{\sqrt{6}} \cos t \\   y  = c-\frac{r}{\sqrt{6}} \cos t +\frac{r}{\sqrt{2}} \sin t \\   z  = c-\frac{r}{\sqrt{6}} \cos t -\frac{r}{\sqrt{2}} \sin t

This is not as bad as it looks: the product {xyz} simplifies to

\displaystyle xyz = c^3 - \frac{cr^2}{2} + \frac{r^3\sqrt{6}}{18} \cos 3t

which tells us right away that the volume {V} lies within

\displaystyle  c^3 - \frac{cr^2}{2} \pm \frac{r^3\sqrt{6}}{18}

In terms of the original data {S,D} the bounds for {V} take the form

\displaystyle    \frac{5S-4D^2}{54}\sqrt{S+D^2} \pm \frac{\sqrt{2}}{54} (2D^2-S)^{3/2}

(And of course, {V} cannot be negative even if the lower bound is.) It is easy to see that {2D^2-S\ge 0} with equality only for a cube; however {2D^2-S} can be relatively small even when the box does not look very cube-ish. For the box pictured above, the tolerance {\frac{\sqrt{2}}{54} (2D^2-S)^{3/2}} is approximately {1.4}; after rounding {S\approx 194 } and {D\approx 10} this drops to {0.38}, and the desired volume of {176} is way out of the allowable range {181\pm 0.38}.

Yes, the set of attainable triples {(V,S,D)} is quite thin. Parallelepipeds are fragile objects: handle them with care.

Attainable (V,S,D)
Attainable (V,S,D)


The diagonal {D=\{(x,x):0\le x\le 1\}} of the unit square {Q=[0,1]\times [0,1]} is a mysterious thing. For one thing, its length is an irrational number. But this is not what I’m writing about.

Behold the mystery
Behold the mystery

The diagonal has zero area. Lebesgue integrable functions on {Q} form the normed space {L^1(Q)} which, upon closer inspection, consists not of functions but of their equivalence classes. Two functions {f,g} are equivalent if the set {\{f\ne g\}} has zero area. In particular, changing all the value of {f} on the diagonal {D} does not change {f} as an element of {L^1(Q)}. The logical conclusion is that given an element {f\in L^1(Q)}, we have no way to give a meaning to the integral of {f} over {D}.

But wait a moment. If I take two square integrable function {u,v\in L^2[0,1]}, then the expression {\int_0^1 u(x)v(x)\,dx} makes perfect sense. On the other hand, it represents the integral of the function {f(x,y)=u(x)v(y)} over the diagonal {D}.

Pushing this further, if an element {f\in L^1(Q)} can be represented as {f(x,y)=\sum_{k=1}^n u_k(x)v_k(y)} for some {u_k,v_k\in L^2[0,1]}, then {\int_D f} is naturally defined as {\sum_{k=1}^n \int_0^1 u_kv_k}. I can even take infinite sums, assuming that everything converges.

This is confusing. If I pick up an element of {L^1(Q)} off the sidewalk, how will I know if it’s safe to integrate it over the diagonal? The existence or nonexistence of the decomposition into sum of products is not obvious.

I guess a satisfactory answer is given by the notion of a Lebesgue point. Given {f\in L^1(Q)} and a point {p} of {Q}, consider the following statement:

\displaystyle    \exists y_0\in \mathbb R \text{ such that } \lim_{r\rightarrow 0} \frac{1}{r^2} \int_{|x-p|<r} |f - y_0| =0   \ \ \ \ \ (1)

The validity of (1) and the value of {y_0} are not affected by the choice of a representative of {f}. If (1) holds, {p} is called a Lebesgue point of {f}, and we can think of {y_0} as the “true” value of {f(p)} (whether or not our representative agrees with that value). It’s a theorem that almost every point of the domain of {f} is a Lebesgue point. The meaning of “almost every” corresponds to the measure under consideration: on the plane it’s the area, on a line it’s the length.

The product {f(x,y)=u(x)v(y)} has a special feature. Since almost every {x\in [0,1]} is a Lebesgue point of {u}, and a.e. {y\in[0,1]} is a Lebesgue point of {v}, it follows that almost every point of the diagonal (in the sense of linear measure) is a Lebesgue point of the product {f}. (It helps to integrate over small squares instead of disks in (1), which does not change anything.) This makes it possible to define {\int_D f} unambigiously.

The sums of products also have the property that almost every point of {D} is a Lebesgue point. And other elements of {L^1(Q)} may also have this property: they are safe to integrate diagonally, too.