Sending a parcel

What is the greatest volume you can get for a rectangular (cuboid) parcel if the maximum combined length and girth are 2 metres?

Age

11 to 14

Challenge level

Secondary

GEOMETRY

Perimeter, Area and Volume

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Problem

Sending a Parcel printable sheet

The Post Office used to have a strange way of limiting the size of parcels. The maximum allowed was "6 feet combined length and girth". That meant you added together the distance round the middle of the parcel (that's the girth) and the length.

You could have sent a fishing rod very nearly 6 feet long and just a few inches round the girth!

Bringing the problem up-to-date, let's set the maximum combined length and girth to be 2 metres (which is a bit more than 6 feet).

What is the greatest volume you can now get for a cuboid parcel with whole centimetre dimensions?

Student Solutions

Ken from Blessed Robert Johnson School thought a cube would provide the optimum solution:

First we all clearly knows that a regular square will give the biggest area while the perimeter stays the same.

Therefore a regular square cube (length = width = height) gives the biggest volume.

To get the biggest volume I already know that the length, width and height have to be the same (to get a cube).

Let's call them A (use the same letter because they have the same value).

I know that 2(length + width) + height = or < 200,

so substituting A into the inequality I get:

5A = or < 200 which gives me A = or < 40

(this means if A is more than 40, then it will exceed the length + girth limit).

Say A is 40, then the volume will be 40 x 40 x 40 which gives 64000cm^3,

at the same time the length + girth is just 200cm which is just 2 metres.

So the biggest volume you can get is 64000cm^3 .

Elizabeth from Waverton CP School improved on this by finding two cuboids where the length is approximately twice the width and breadth (think of them as made up from two cubes stuck together):

I think the greatest volume is 74052cm^3 .

I think the dimensions of the cube would be either

66cm long x 33cm width x 34cm breadth or

68cm long x 33cm wide x 33cm deep.

I got my answer by picking a length, subtracting it from 2 metres, dividing the girth by 2 and dividing it by two again for width and breadth. (A spreadsheet is ideal for doing this - it allows you to home in on the optimum solution quickly.)

Ruth from Manchester High School for Girls came to the same conclusion:

To maximise the area of the cross section for a given girth, the width and breadth are equal.

The length is 200 - the girth, which is 2 times (width + breadth), which is 4 times the width,

so the volume is

w^2 x(200 - 4w) = 200 xw^2 - 4w^3

4w^3 < 200 xw^2 because the volume has to be a positive number.

Therefore w^3 < 50 xw^2

w > 0 because it is an actual measurement so the w^2 can be cancelled out giving w < 50

Trying the possible values of w gives the biggest volume as 74052cm^3 when w is 33 so the width and breadth are each 33 cm and the length is 68 cm.

The same volume could also be achieved by the width being 33 cm, the breadth being 34 cm and the length being 66 cm. A spreadsheet can also help here.

Ravanan found that if you ignore the condition that measurements are all made in whole centimetres you can improve slightly on the volume with a cuboid made from exactly two cubes (the length is exactly twice the width and breadth):

Length = 200 / 3

Width = 100 / 3

Breadth = 100 /3

Volume = (200 / 3) x(100 / 3) x (100 / 3) = 74074 cm^3

Well done to you all.

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