Cuboid challenge
Cuboid Challenge printable sheet
Take a square sheet of paper 20cm by 20cm, cut identical squares from each corner, and fold up the flaps to make a box (without a lid).
What is the volume of your box?
What different volumes can you make by varying the size of the squares you cut out?
What is the maximum possible volume of this type of box that can be made from a 20cm by 20cm square of paper?
Now try starting with different sized square sheets of paper.
Can you find a relationship between the size of paper and the size of the square cut-out that produces the maximum volume?
Click here for a poster of this problem.
It might be a good idea to work systematically:
Start by cutting out a 1cm square from each corner - what would the volume be?
Then a 2cm square, then a 3cm square, and so on...
The side length of the square doesn't necessarily have to be a whole number...
Perhaps it would help to use a spreadsheet to perform and record your calculations.
Siddhant from Singapore International School in India, Astrid, Anaira and Millie from CLSG in the UK and Ethan, Amaya and Lily from Pangkok Patana School in Thailand tested different cut sizes. This is Astrid, Anaira and Millie's work:
Ethan, Amaya and Lily used a spreadsheet to show that 3.333 was the best option with 3 decimal places. Click here to see their spreadsheet.
Arya from West Island School in Hong Kong got to the same answer using spatial reasoning:
I think the identical square to be cut from each corner has a length of 3.33 recurring cm.
Method
Every time you increase the height by 1 the length reduces by 2 each side.
Normally a cube will have a bigger volume than a cuboid that has lets say 2 cm more of height and 2 cm less than the original length.
But this time the ratio is 1:2 (because increasing the height by 1 cm decreases the length by 2 cm).
So the [ratio of] height to length is 1:2.
So 13.33333 and 6.66666
6.6666 has to be divided by two because it is minused from both sides.
Ethan, Amaya and Lily also tried starting from a 15 cm by 15 cm sheet of paper:
Year 9 Set 1 from British International School Phuket in Thailand tried a variety of different sizes. This is their work:
We used Desmos to plot the graphs and find the maximum volume.
Investigation 1
| Square size | Equation for Volume ($V$) |
Value of $x$ that maximizes $V$ |
|---|---|---|
| $10 \times 10$ | $(10-2x)(10-2x)x$ | $\frac53$ |
| $20 \times 20$ | $(20-2x)(20-2x)x$ | $\frac{10}3$ |
| $30 \times 30$ | $(30-2x)(30-2x)x$ | $5$ |
| $40 \times 40$ | $(40-2x)(40-2x)x$ | $\frac{20}{3}$ |
| $50 \times 50$ | $(50-2x)(50-2x)x$ | $\frac{25}{3}$ |
| $n\times n$ | $(n - 2x)(n - 2x)x$ | $\frac n6$ |
Conclusions: For an $n$ by $n$ square the value of $x$ that maximizes the volume is given by $\frac n6$
Investigation 2
| Paper size | Equation for
Volume ($V$)
|
Value of $x$ that maximizes $V$ |
|---|---|---|
| $10 \times 10$ | $(10-2x)(10-2x)x$ | $\frac53=1.667$ |
| $10 \times 20$ | $(10-2x)(20-2x)x$ | $2.113$ |
| $10 \times 30$ | $(10-2x)(30-2x)x$ | $2.257$ |
| $10 \times 40$ | $(10-2x)(40-2x)x$ | $2.324$ |
| $10 \times 50$ | $(10-2x)(50-2x)x$ | $2.362$ |
| $10 \times 1000$ | $ (10-2x)(1000-2x)x$ | $2.494$ |
| $10 \times 10000$ | $(10-2x)(10000-2x)x$ | $2.499$ |
Conclusions: For a $10$ by $n$ square the value of $x$ that maximizes the volume approaches $2.5$ as $n$ gets large (approaches infinity).
Investigation 3
| Paper size | Value that $x$ approaches as $n$ approaches infinity |
|---|---|
| $10 \times n$ | $2.5$ |
| $1=20 \times n$ | $5$ |
| $30 \times n$ | $7.5$ |
| $40 \times n$ | $10$ |
| $m\times n$ | $\frac m 4$ |
Conclusions: For an $m$ by $n$ square the value of $x$ that maximizes the volume approaches $\frac m 4$ as $n$ gets large (approaches infinity).
Hyochan from Brighton College Abu Dhabi in UAE and Scott, Jakob, Max and Timmy from Abingdon School in the UK used a different approach for the 20 by 20 sheet of paper. This is Hyochan's work:
Since the problem is about finding the maximum volume of a cuboid, I thought about using differentiation.
Max referred back to the original problem to explain why $x=10$ couldn't be a maximum:
I put $10$ back into the original cubic and it gave me the answer $0.$ This is because when $x=10$ there is no paper as $10\times2=20.$
Scott, Jakob and Timmy used more calculus to check that $x=\frac{10}3$ is a maximum. This is Jakob's work:
Max applied this method to differently sized sheets of paper:
I did this method again with $40$ cm of paper which gave me an $x$ of $\frac{20}3$ and volume of $4740.740$ recurring. I also did it with $65$ cm and got an $x$ of $\frac{65}{6}$ and volume of $20342.592$ recurring. From this the pattern for the maximum $x$ is the full edge of the paper divided by $6$.
I also found that the volume pattern is the previous volume times the multiplier of the edge cubed. For example, the volume of the $40$ cm piece of paper is $592.592$ (volume of $20 $cm piece) times $2^3=8$ which is $4740.740.$ The volume of the $65$ cm piece is $4740.740$ times $\frac{65}{40}=1.625$ cubed ($4.29$ (3sf)) which is $20342.592.$
Timmy and Charlie from Abingdon School and Ana from Bangkok Patana School generalised this. Ana wrote:
From this I can calculate that the formula for a box's volume constructed from a square with any length ($l$) is $$volume=(l-2x)^2\left(x\right)=l^2x-4lx^2+4x^3$$
Additionally the $x$ value which gives the largest volume is $\frac{\text{length of square}}{6}$
The maximum value is the maximum stationary point of the graph which can be found using differentiation. $$\frac{\text{d}}{\text{d}x}\left(4x^3-4lx^2+l^2x\right)=12x^2-8lx+l^2$$
The maximum point's derivative must equal $0$, thus at the maximum $12x^2-8lx+l^2=0.$
By using the quadratic equation, the root values can be found. $$\frac{8l\pm\sqrt{64l^2-48l^2}}{24}=\frac{8l\pm\sqrt{16l^2}}{24}=\frac{8l\pm4l}{24}$$
The maximum value cannot exceed half of the length of the square, thus the stationary point calculated is the smaller value. So the maximum point is at the $x$ value $$\frac{4l}{24}=\frac{l}{6}$$
Charlie checked that this is a maximum (not minimum) value by differentiating again (Charlie and Timmy use $y$ instead of $l$ for the side length of the sheet of paper):
At maximum $\dfrac{\text{d}^2V}{\text{d}x^2}<0$
$\dfrac{\text{d}^2V}{\text{d}x^2}=24x-8y$
$24\left(\frac y 6 \right)- 8y = -4y$ - this is the maximum
$24\left(\frac y2\right)-8y= 4y$
Charlie, Ana and Timmy found the maximum volume and checked that this agrees with the values found for the 20 by 20 sheet of paper. This is Timmy's work:
Why do this problem?
This problem is a great example of a "low threshold high ceiling" task that can be approached in different ways. As well as having an opportunity to practise visualising, 2D representations of 3D shapes, and volume calculations, students can be introduced to the idea of finding an optimal solution, a key idea in mathematics. As there are many ways to approach and solve the problem, students can be drawn into new ways of reasoning mathematically, using numerical, algebraic and graphical representations, and for the most curious student, it could provide a route into thinking about calculus!
Possible approach
This printable worksheet may be useful: Cuboid Challenge
"Imagine you have a 20 by 20 square sheet of paper. Now imagine cutting out a square of side 5cm from each corner, and folding up the flaps. What would the dimensions of the resulting box be? Check with your neighbour and see if you agree."
"What would the volume of the box be?"
Bring the class together and confirm the dimensions, and the resulting volume. Take time to discuss how students worked it out, perhaps using diagrams showing the net and the finished box.
"I wonder if we could make a box with a larger volume..."
Allow some time for the students to work in pairs to explore the effect of cutting out different squares on the volume of the resulting box.
Then collect the results so far in a table on the board or a spreadsheet.
"What's the maximum volume we've found so far? Can we do any better?"
Invite students to comment on patterns they can see in the results, and to make suggestions of methods that might lead to the optimum solution. This could include:
- Trial and improvement, possibly using calculators
- Using a spreadsheet and a formula to calculate the volume for a given size of cut out
- Finding a formula and using graphing software to find a maximum value
Encourage some students to try each method so that you can bring the class together to compare and discuss their results.
You could use these GeoGebra resources, created by Alison Kiddle, to shed light on students' results.
Once students have found the maximum for a 20 by 20 square, set different groups of students a different size of starting square. Can they use one of the methods to find the cut out that gives the maximum volume?
Finally, collect in a table the cut out that maximises the volume for each square, and invite students to look for a general pattern that will work for any size of square.
Key questions
How can you be sure you have found the maximum volume?
Possible support
Offer students 20 x 20 square grids and encourage them to make different sized boxes, working systematically and recording their results as they work.
Possible extension
Instead of starting with square sheets of paper, students may investigate rectangular ones. In order to make pattern spotting easier, you may wish to organise this in some way, for example giving different groups of students sets of rectangles (such as rectangles where the length is twice the width, three times the width, four times the width, etc.)
Students could use algebra to represent the relationships they find, and if they know some calculus, use that to maximise the volume.