# Cuboids

Can you find a cuboid (with edges of whole number lengths) that has a surface area of exactly $100$ square units?

Can you find them all?

Can you provide a convincing argument that you have found them all?

Click here for a poster of this problem.

It may be simpler to focus on just three faces (the three different faces), rather than on all six.

This net of a cuboid may help.

Try to be systematic:

if the height is $1$, what are the possible combinations for the width and depth?

if the height is $2, 3, 4$... what are the possible combinations for the width and depth?

Justin from Mason Middle School found that a 1x2x16 cuboid also satisfies the conditions:

I back it up by saying that$1\times 2 = 2$,

$2 \times16 = 32$

and $16 \times1 = 16$

and add them all up and you get $50$, and then multiply by $2$ and you get a square area of $100$!

Megan was able to show that these were the only two possible solutions:

Call the lengths of the 3 dimensions (height, depth, width) $x$, $y$ and $z$.The surface area is $2xy+2xz+2yz$, as the area of each face is calculated by multiplying its two sides together, and there are $2$ of each face.

Hence, $2xy+2xz+2yz = 100$.

Dividing by $2$, $xy+yz+xz = 50$.

I will assume that $x$ is the shortest side and $z$ the longest to avoid repeating solutions. Therefore I must find integer solutions to the equation $xy+yz+xz = 50$ where $x < y < z$.

Rearranging the equation,

$yz+xz = 50-xy$

$z(x+y) = 50-xy$

$z = (50-xy)/(x+y)$.

I used an excel spreadsheet with 3 columns, 1 for each of $x$, $y$ and $z$.

In the $z$ column I write the rearranged formula.

I then started from $x = 1$, $y = 1,2,3...$ looking for integer values of $z$ until I reached a solution which had been repeated (as here $y$ is bigger than $z$, so I would be repeating values with $y$ and $z$ swapped around) or where $z$ became smaller than $y$.

For $x = 1$ I found only 1 solution, $(1, 2, 16)$.

Checking with the original formula this does agree to a total of $100$.

I then continued repeating the procedure with $x = 2$, $y = 2$,$3$,$4$... and found a solution of $(2, 4, 7)$.

With $x = 3$, $y = 3$,$4$,$5$... there are no solutions, as $z$ becomes smaller than $y$ where $y=5$ (and $z = 4.375$) and there are no integer solutions before this.

With $x = 4$, $y = 4,5,6...$ $z$ becomes smaller than $y$ when $y = 5$ $(z = 3.333)$ therefore there are no solutions where $x = 4$.

After $x = 4$, the intitial value of $z$ is always smaller than $y$, therefore there are no further solutions.

The only 2 solutions to the problem are: $(1, 2, 16)$ and $(2, 4, 7)$.

Fred from Albion Heights School also offered some non-integer solutions:

$h = 1$, $w = 1$, $l = 24.5$$h = 2$, $w = 2$, $l = 11.5 $

$h = 4$, $w = 4$, $l = 4.25$

$h = 2.5$, $w = 5$, $ l = 5$

$h = 1$, $w = 4$, $l = 9.2$

Well done to you all.

### Why do this problem?

This problem requires a lot of calculations of surface areas, within a rich problem solving context.

### Possible approach

*This printable worksheet may be useful: Cuboids.*

Work with a specific cuboid, eg $2 \times 3 \times 5$, or a breakfast cereal box, to establish how to calculate surface area of cuboids. Students could practise working out surface area mentally on some small cuboids made of multilink cubes.

It may be appropriate to draw a ladder on the board, with this on the steps (starting from the bottom):

- calculations going wrong

- no solutions yet

- one solution

- some solutions

- all solutions

- why I am sure I have all the solutions

- I'll change the question to...

Explain to students that wherever they are on the ladder, their goal is to move up to the next step. Circulate round the class and look out for students who are approaching the task systematically.

After students have been working for a little while, bring the class together and ask them for strategies that might help others move up the ladder.

Give them plenty of time to implement these suggestions.

This might be a good lesson in which to allocate five minutes at the end to ask students to reflect on what they have achieved, which methods and ideas were most useful, and what aspects of the problem remain unanswered.

### Key questions

- Have you found none/one/some or all of the solutions
- Is there a cube that will work?
- How might you organise a systematic search for the cuboids with surface area $100$?

### Possible support

Suggest students approach the task systematically:

if the height is $1$, what are the possible combinations for the width and depth?

In groups, or as a class, keep a record of all cuboids whose surface areas have been calculated.

Possible extension

The main extension activity could focus on the convincing argument that all solutions have been found. Once this has been answered, you might like to consider these extensions:

- Express the method for calculating surface area, algebraically.
- Which surface area values will generate lots of cuboids, and which give none or just one?
- Could you set up a spreadsheet to help with the calculations?