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Apparently I’m not a maths genius (or, On the Subject of Parcel Sizes)

This article on BBC News caught my eye because it has “maths” in the headline. Yes, I’m that easily pleased.

Somewhere in the middle, it says that myHermes requires the “volumetric area” of a parcel to be less than 225cm. That’s right: the “volumetric area” is neither a volume nor an area but a length. Anyway, the formula for volumetric area of a package with sides $a,b,c$, where $a \leq b \leq c$, is

\[ 2(a+b) + c \]

(Importantly, $a$ and $b$ are always the two shortest sides of the package)

So the constraint is

\[ 2(a+b) + c \leq 225 \]

In the next paragraph is the puzzling statement that the maximum allowable volume for a package is $82.68$ litres, or $82680$ cm3. How did they get that?

I decided to do some calculus of variations, or whatever it’s called.

First of all, let’s take as much “volumetric area” as we possibly can:

\[ 2(a+b) + c = 225 \]

Get $c$ in terms of $a$ and $b$:

\[ c = 225 – 2(a+b) \]

And sub into the formula for the volume of the package:

\[ V = abc = ab(225-2(a+b)) = 225ab -2a^2b -2ab^2 \]

Now, I reckon that if $a+b$ is constant, you get the best cross-sectional area $ab$ when $a = b$. For any value of $c$ you pick, $a+b$ is determined uniquely, so you get the biggest $abc$ when $ab$ is biggest, which happens when $a=b$.

(Proof that $a=b$ is best: Let $a+b = 1$, without loss of generality. Then $b = 1-a$, and $A = ab = a(1-a) = a – a^2$. Find stationary points by differentiating: $\frac{dA}{da} = 1 – 2a$. $\frac{dA}{da} = 0 \implies a = \frac{1}{2} \implies b = \frac{1}{2}$ so $a = b$.)

So, if we accept that $a=b$ is the best way to go, then the formula for $V$ becomes much easier:

\begin{align*} V &= 225a^2 – 2a^3 – 2a^3 \\ &= 225a^2 – 4a^3 \end{align*}

Find the maximum (and minimum) values of $V$ by finding the points where $\frac{dV}{da} = 0$:

\begin{align*} \frac{dV}{da} &= 450a – 12a^2 \\ &= a(450 – 12a) \end{align*}

So $a = 0$ or $a = 37.5$. I’ll guess that $a = 37.5$ is the maximum.

Now we can work out $c$:

\[ c = 225 – 2(37.5+37.5) = 225 – 150 = 75 \]

So the maximum volume $V$ is

\[ V_{max} = 0.375\text{m} \times 0.375\text{m} \times 0.75 \text{m} = 0.10546875\text{m}^3 \]

Or $105.4$ litres. That’s more than $82.68$, so apparently I’m not a maths genius.

I wanted to know where $82.68$ could possibly have come from, so I plugged it into Robert Munafo’s RIES. It gave two equations which produce numbers equivalent to $82.68$ to two decimal places:

\[ x = \frac{e}{\phi} + 9^2 = 82.6799905609888981 \]


\[ x = 9\left(\sqrt{2} + \sqrt{\pi}+6 \right) = 82.680006719507503 \]

I reckon the second one looks the most likely for “something someone in operations research made up to placate the journalist”. It’s still pretty weird though.

Can you come up with a better provenance for $82.68$? I’d be interested to hear it.

(No, really, I would.)

15 Responses to “Apparently I’m not a maths genius (or, On the Subject of Parcel Sizes)”

  1. Avatar Mark Taylor

    The maximum length of a MyHermes parcel is 120cm. (See

    The maximum volume of a 120cm parcel that fits the other constraints outlined above is 82.6875 litres (with both other dimensions at 26.25cm).

    But if you’re only willing to measure lengths to the nearest half a centimetre, the best you can do is one of 26cm and one of 26.5cm. That gives you a volume of 82.68 litres precisely.

      • Avatar Mark Taylor

        It seems entirely possible that the whole conversation was quite confusing for the spokesperson, and the journalist is as much to blame (cf. Paul’s comment below). After all, there’s no particular reason anyone should need to know the maximum volume of a MyHermes parcel: they don’t directly restrict it; their charges are based solely on weight; and most of the objects people post are a particular size and shape and can’t be repacked arbitrarily.

        In reality, it seems like the fixation with volume starts with the Royal Mail trying to demonstrate how much more generous their new parcel restrictions are and ought to have ended there, because it’s only really important to Brian Milligan.

        A more interesting question might be: if you take a narrow 50cm parcel into your local post office, will they accept it as a small parcel on the basis that it could be packed along the diagonal of a qualifying box? I’m quite tempted to try it.

          • Avatar Paul Taylor

            I feel a bit sorry now for the poor myHermes spokeswoman, who I imagine hesitantly coining the term “volumetric area” while trying patiently to explain the quite simple system to the annoying reporter who seems only to be able to think of things in terms of volumes.

        • Avatar Alison Scott

          Seeing as post office staff routinely reject parcels that are well within their own stated limits, it seems highly unlikely. But in any case, their limits are for the dimensions of the actual parcel, not of an imaginary box it could fit inside, so you are out of luck with your 50cm stick. Though you could probably blow their minds by additionally taking a maximal small parcel box and demonstrate that it’s possible to take a medium parcel and pack it in such a way that it becomes a small parcel.

  2. Avatar Nick Gardner

    MyHermes have used their maximum permitted parcel length to calculate the supposed maximum volume, even though this doesn’t give the true maximum.

    Maximum length is 120cm. Allows width and height to be 26.25cm, since

    “Volumetric area”: ((26.25 + 26.25) * 2) + 120 = 225

    Volume: 26.25 * 26.25 * 120 = 82.6875

  3. Avatar Paul Taylor

    The BBC article is exquisitely annoying. It’s the way they make a totally unfounded and obviously nonsensical assumption (‘its website declares its maximum volume is “225cm”, presumably meaning cubic centimetres’), directly refute it in the next paragraph, and then when the correct answer turns out to be different act like it’s the other guys who are being stupid.


  4. Avatar Luke Anjim

    Sorry to bring this down to earth, but 82.68 litres is not the same as 8268 cm3 !
    [Nice problem though.]

  5. Avatar Luke Anjim

    If we interpret CP’s original (algebraic) solution geometrically, then we have
    • the maximum volume is achieved when the rectangluar base is a square (a = b)
    • the maximum volume is achieved when the cuboid is formed of two identical cubes (a=b=0.5c).

    The first statement is ‘obvious’. Is the second? Not to me, at least not yet…

    It is interesting, though, that if we consider any of the rectangular faces with dimension a by c, then we know that the ‘partial’ perimeter a+c+a is constant [given: 2(a+b)+c=225], which means its area is a maximum when its shape is a ‘half square’ ie when c=2a [because if you imagine reflecting the face in one of the sides c, you want the resulting object+image to be a maximum, ie to be a square – this is the classic rectangular sheep pen task using a fixed length of fencing and a wall]. However, it is not obvious to me that having made the area of this face a maximum that the volume of the whole shape is also a maximum, as we might not now have an optimal value for a. What am I missing? Something ‘obvious’, no doubt!

  6. Avatar Simon Laurent

    A non-calculus approach can be used with the AM-GM inequality with 3 numbers. Since
    \[\frac{x+y+z}{3} \geq \sqrt[3]{xyz}\] we also have
    \[\left(\frac{x+y+z}{3}\right)^{3} \geq xyz\] For us, that means
    \[V = abc = \frac{2a\cdot 2b\cdot c}{4}= \frac{1}{4}\left(2a\cdot 2b\cdot c\right) \leq \frac{1}{4}\left(\frac{2a+2b+c}{3}\right)^{3}\]
    Using $2a + 2b + c = 225$, we get
    \[V \leq \frac{1}{4}\left(\frac{2a+2b+c}{3}\right)^{3} = \frac{1}{4}\left(\frac{225}{3}\right)^{3}\]
    with equality only if $c = 2a = 2b$.This implies $c=75$, $a=37,5$ and $b=37,5$.


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