Oxford University Admissions Interview Question: What is the optimal size for a tin of cat food?

University of Oxford Mathematician Dr Tom Crawford explains how to solve an optimisation problem used as a maths undergraduate admissions interview question.
"For a cylinder of fixed volume V=1, what aspect ratio between the radius and height gives the smallest surface area?"
This question was used by Tom for the Oxford admissions interviews in 2018.
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Other videos in the Oxford Calculus series can be found here: • Oxford Calculus
Finding critical points for functions of several variables: • Oxford Calculus: Findi...
Classifying critical points using the method of the discriminant: • Oxford Calculus: Class...
Partial differentiation explained: • Oxford Calculus: Parti...
Second order linear differential equations: • Oxford Mathematics Ope...
Integrating factors explained: • Oxford Calculus: Integ...
Solving simple PDEs: • Oxford Calculus: Solvi...
Jacobians explained: • Oxford Calculus: Jacob...
Separation of variables integration technique explained: • Oxford Calculus: Separ...
Solving homogeneous first order differential equations: • Oxford Calculus: Solvi...
M.I.T. Integration Bee Question: • M.I.T. Integration Bee...
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[@TomRocksMaths]

Try another Oxford admissions interview question from 2021 here: kzbin.info/www/bejne/Y6aZn5egpax4Z5Y

[@Nickle314]

Or mean average curvature is zero, and hence the tin should be a sphere!

[@GEMSofGOD_com]

If this is Oxford's level then I'm happy with Putin

[@JH-Wood]

Perfect timing and fantatsic video! I am teaching differentiation with my A-level classes at the moment. Next week we will be doing modelling with differentiation - I will be sure to include this lovely problem!

[@sneakerslab2771]

We did that 2 weeks ago

[@johnchessant3012]

The Heinz tin at the end has surface area 353.37 cm^2, about 1.46% more than the optimal tin of the same volume which has surface area 348.27 cm^2. The Tesco tin is much worse: 210.74 cm^2, about 10.10% more than the optimal tin at 191.40 cm^2. But even 1.46% is significant. Let's assume 100 million cat owners in the world; each spends $300 per year on cat food; Heinz has 10% of the market; they have a 30% profit margin; and 10% of the production cost is the tin. Then they could save $3.4 million by switching to the optimal tin!

[@iain8938]

The real loss is you feeding your cats heinz beans.. Poor kitty

[@tariqm6083]

meanwhile in the real world the height and radius of my stomach remain unrelated to packaging costs.

[@peterwilliams6633]

Actual manufacture may look at other aspects such as the waste material from a sheet of metal after the tins are cut out. This may change the economics, or possibly packaging/shipping pre-requisites

[@TomRocksMaths]

very true. I have no doubt this is an over-simplification of the actual manufacturing process!

[@reinhardtscheepers2349]

I love these types of questions. I’m a Chemical Engineer, and we typically use these types of questions when we are trying to maximise surface area for heat transfer inside a heat exchanger/multitubular reactor, while keeping the pressure drop along the reactor length within an acceptable margin.

[@ay-pj8co]

What an amazing video. At first I was puzzled because I had assumed that same volume = same surface area but I was wrong. The utility of Mathematics is something that has always amazed me.

[@JoePortly]

Dr Tom, thanks for making this solution remarkably clear and relevant On leaving school at an early age - after being told I'd never be good at maths - I'd anyway say that the seemingly less-than-optimal shape of food cans may be to-do with minimizing the ratio of empty space between them to their volume, and rendering them strong enough to resist denting or collapsing. And let us bear in-mind that cans of corned beef or Spam are made in a non-cylindrical form - so as to stack easily, I believe. And I (beelieve or) fancy that cans with a hexagonal cross-section would be the best of all

[@selimmecanna1074]

Great Video! This is a great problem for introducing Lagrange Multipliers, I remember my calculus professor used it as an example when doing so.

[@benjaminlennon]

Great video. Love the channel. It's a very interesting problem. One consideration may be the cost of retooling for the top and bottom. Sides are cheap to cut and the height can be varied easily. I can't imagine it's easy to vary the radius measurements of their whole production line.

[@shen7728]

Love it, Tom! Very interesting and fun problem

[@piotrpustelnik3109]

ngl, did not expect such an easy question for Oxford's interview

[@monismazhar6581]

Absolutely love your personality and demeanor while explaining maths. I showed your Gabriel’s horn video to my friends and teachers and they were elated by it

[@TomRocksMaths]

[@michael42158]

Got it right! Not bad for a chemistry teacher! Love these videos.

[@two697]

I had the same question at my mechanical engineering interview for imperial

[@manswind3417]

Imperial conducts interviews too? Wow, interesting to hear

[@a_aw1070]

@@manswind3417 they conduct interviews and admissions tests

[@mannydossantos9603]

Thanks, Dr Tom. It was a great fun and interesting problem. Clearly explained! The tip on using the Maple Calculator is highly appreciated.

[@themathsgeek8528]

Thank you Prof, this was a great video!

[@iteerrex8166]

Sounds like a derivative question.

[@wqltr1822]

Yeah i agree

[@curtisclement5467]

As an alternative method, you could also use AM-GM inequality (although I doubt many candidates would know this one). Using h = 1/(pi*r^2) you can minimize the surface area by minimizing r^2 +1/(pi*r) = r^2 +1/(2*pi*r) + 1/(2*pi*r) >= (r^2 * (1/(2*pi*r))^2)^{1/3} = (1/(2*pi))^{2/3} with equality when r^2 = 1/(2*pi*r)...

[@cdpm9982]

That's clever! I thought about it, but gave up because I was starting to get some horrible expressions. Is there some clear way how you can see the right use of AM-GM inequality in this problem? Or why to go for the third root instead of the second in AM-GM?

[@akshatkumar6180]

This was a great video, I really liked it!

[@MohamedYasser-km9rm]

I accually have this as a full lesson called the maximize and minimize application of derivative and it's one of the trickiest lessons because forming that equation with the one unknown can get quite frustrating

[@Dr.StructuralAnalysis]

Nice explanations, Thanks.

[@callmedeno]

I'm a cat-food mogul and this is a game changer for my business

[@electronickal_8598]

my brain finally understands and it's so simple for me now i love it

[@vimuthabeysinghe6]

That was a really fun question! Thanks for sharing 😊

[@TomRocksMaths]

Any time!

[@RC32Smiths01]

Very fun and interesting question for Oxford! Cool man!

[@iain8938]

"cool man" or "cool, man"

[@sahlansidjara6838]

Nice explanation tom, great

[@xavierwainwright8799]

Let h be the height of the tin and r the radius. h=cbrt(4/pi) and r=(4pi^2)^(-1/6) edit: Hey, I was right! Fun problem, would definitely like to see more problems like this Tom.

[@josejaramillo4346]

Excelente!!! Problema de Optimización.

[@qwerty_ytrewq4452]

I saw that h=2r, which is quite magical considering this would perfectly fit a 2r*2r*2r cube, so I thought it might have something to do with it: consider a rectangular prism with a square base where we can fit the tin perfectly inside. The ratio of the volume of cylinder:prism would be πr^2*h:4r^2*h=π:4 The ratio of the surface area of cylinder:prism would be 2πr^2+2πr*h:8r^2+4*(2r*h)=π:4 this problem then transforms to finding the minimum surface area of a rectanglar prism with a fixed volume, and the answer is a cube!

[@glennsaunders630]

Brilliant!!! And I understood this one!!! Albeit I have a dog not a cat!

[@zsoltlevente1413]

well explained ! even the cat's are getting it

[@benstart9827]

Hi Tom, would you ever consider doing a video where you sat an AQA GCSE Level 2 in Further Mathematics paper (similar to when you sat the GCSE Maths paper) - would like to see this 👍🏻

[@TomRocksMaths]

... this is coming soon :)

[@dylanmcnutt4250]

Had to figure this out for a project for AP Calculus I in Texas; assuming the amount of material is roughly equivalent to the surface area of the tin (ie negate the thickness of the tin), then the optimal volume is achieved if the height is twice the radius; h = 2 * r

[@The_man_himself_67]

Actually people would not believe the amount of analysis that has been done on this subject. Shaving a few tenths of a gramme off a coke tin has saved $millions!

[@oblivion591]

This is absolutely gorgeous problem, thanks for sharing it!

[@TomRocksMaths]

glad you enjoyed it!

[@mujtabaalam5907]

The real question is this: as the thickness of the can increases from 0, does the optimal shape become flatter and shorter, or taller and skinnier? Answer: Flatter and shorter. Consider M to be the amount of metal and Mt to be the theoretical amount of metal where Mt is just the surface area times the thickness of the metal, Now think of it his way: the only reason why the M isn't the same as Mt is because at the edge you would be double counting. There, part of the theoretical metal is subtracted to prevent overlapping metal. The shorter (and thicker) the can, the greater the overlap, the greater Mt-M is. Since the optimal can size is the one which minimizes M, and we just saw that shorter cans shrink M more than Mt, therefore M is minimized at a shorter can than Mt. This explains why tuna cans (which are quite thick) are shorter and fatter than expected. So what about the tall cans? Well, there's probably a psychological factor in play. It's well known that people tend to underestimate the circumference of a circle, therefore the height is given greater weight in a person's mental calculations of which can is the biggest, which biases cans with greater height.

[@UneCleUSB]

Hi, can you explain more about this 'double counting'? I'm really struggling to understand it and visualise it

[@mujtabaalam5907]

@@UneCleUSB Consider the top edge of the cylinder. If you were to extend a thickness inwards from both the side of the can and the top circle, they would both overlap at the top edge of the cylinder.

[@ericvosselmans5657]

You are an amazing guy

[@TomRocksMaths]

[@joecheng643]

Lagrange multiplier would also be great to solve this problem

[@amydebuitleir]

My cats, Professor Paws and Doctor Lal, approve of this question.

[@markhughes7927]

The size of the can has functional relation with a hole in the stomach the radius of which needs specifying…

[@edwinandemmanuel3399]

Nice maths teacher

[@slayz4708]

I literally did this in my yr12 maths mock last year

[@kashoot4782]

Interesting question. Could almost see this being asked in an a level exam

[@TomRocksMaths]

definitely

[@paulmaingi7382]

Awesome video. At 4:45, you mention that we could use partial differentiation to find the critical points of the function which should end up leading to h=2r. However, the system of equations from the partial derivatives with respect to h and r are: 2×pi×r and 4×pi×r+2×pi×h. Solving for this system yields a solution of h=0,r=0. How are we supposed to solve it using partial differentiation?

[@randomguy2169]

Im late but you are supposed to use larrange multipliers probably. h=0 and r=0 doesn't make sense in terms of the first piece of information which is pi*r^2 *h=1

[@augustecle9349]

super cool

[@vicar86]

I wonder (with a glint in my eye) while (at the moment, Friday afternoon) teaching matrix algebra via Zoom, would I qualify to Oxford with just a smart phone + Maple app?

[@happyhippo4664]

You might want to reduce the radius to minimize weld surface. Also, a lot of these foods are heated in the can, so perhaps a smaller height would make things heat faster. I don't know, I am just a poor country chemical engineer, dammit Jim. I think beer cans are narrowed on the top to reduce weld and the amount of thicker metal on the top, but of course those are pressurized.

[@happyhippo4664]

Would be an interesting optimization problem to take material cost, processing cost, energy costs, packaging cost, and other variables to see where the cost drivers are and what tge optimum design would be. Also, you'd have to deal with sales and marketing if you come up with a can of non-standard size.

[@ArduousNature]

I'm guessing the extra space occupied when packing many tins is worth making the individual tins slightly less 'optimal'

[@lechaiku]

That is the optimisation math problem. And the easiest solution is by this method. We know that the square has the largest area among all rectangles with a given perimeter. So all what we need is to equate both dimensions. __________ | | h diameter = d = 2r = h |_______| d h = 2r That's it.

[@bertblankenstein3738]

Excellent example of differentiation. I am going to guess that minimizing surface area for a given volume is not the only variable however. :)

[@justinharrison9521]

I think can sizes have originally been optimised on number of times the tin opener comes out of the cutting groove and hand injuries due to the sharpness of the jagged edge.

[@jakubhanel2478]

Could you do a similar video, but with cone?

[@donaldasayers]

Strewth I remember this as a question at A-Level. Although the question was posed the other way around: Given a certain area of tin plate what shape contains the most volume? The shape nearest a sphere was my first though then too.

[@TomRocksMaths]

spheres are always the way to go for optimisation problems!

[@sdlillystone]

Hence why bubbles are spherical - minimizing given surface area./surface tension for a given volume

[@AndreasEldhSweden]

I think the Baked Bean can is absolutely right, as most cans uses thicker material for the top and bottom circles. This means a slightly higher can is better. As for the Tuna can it's difficult to see in the video, and I also might have gotten it completely wrong, but what if you use the thinner material to make the bottom circle and the cylinrical walls in one piece?

[@TomRocksMaths]

I think all cans aren't 'actual' cylinders as the bottom and top pieces are slightly indented and have ring pulls etc. This is just a simplification of the problem as an excuse to do some cool maths!

[@JaseBach]

Thanks for the long winded answer. Instinctively and inductively, I solved the problem in two seconds. Think of the perimeter of a square vs. that of a rectangle of area 1, the perimeter is at a minimum when the shape is a square.

[@charlesarthurs]

Inductively?

[@JaseBach]

Autocorrection problem. Intuitively of course.

[@kriswillems5661]

There are other factors, like the costs of the metal "fold" on the top of the bottom. You can be pretty sure the engineers designing those cans have taken everything into account, including your calculation.

[@maxswatchbuilds8129]

Hah I'm pretty sure he knows that

[@kriswillems5661]

You can not cut circle out of metal without wasting the the space of the hexagon around it (or square depending on the production process).

[@ELLIPTICALWR]

how did you get really good at maths sir, is there any advice u would give for someone who wants to take a level maths next year

[@TomRocksMaths]

lots and lots of practice

[@conorbmcgovern]

Can you ignore that it’s a cylinder and just consider the vertical cross section? When I tried it for a fixed rectangular area, I also get square as the answer.

[@dekeltal]

This ratio is correct ONLY for a fixed volume of 1. If you parameterize this problem to use a general volume of V, you can see that the ratio between the height and the radius is actually 2/V. So the more voluminous the can is, it will actually need to be shorter and fatter...

[@robguyatt9602]

I like going back to first principles. I expect the Oxford examiners do too. So I expect that an answer such as "by examination, the most efficient surface area is where h = 2r" would not get me in. :(

[@benjamindalle2626]

Can you make 22 while optimizing the surface area in order for the slices to be as close as possible to equal in dimensions ?

[@TomRocksMaths]

I reckon so (definitely not tying this though after the disaster last time)

[@DATR01]

Shouldn't the phrase "as square as possible" be "as spherical as possible"? The can is "trying" to be a sphere not a cube.

[@chris-bi8vn]

The real tins are not perfectry cylindrical (e.g. may have more material in the perimeter of the circles), and may not even have the same thickness in the sides and up&down sections, so that probably explains it but also the variation of heights and radii.

[@TomRocksMaths]

exactly. this is just a simple model for the problem.

[@isaosauzedde5513]

admission interview at ulm: the same, but the examinator only ask you "What is the optimal size for a tin of cat food?", and you have to guess that minimizing the area is an answer

[@thegrandmuftiofwakanda]

Standard stuff for teaching calculus at 16-18, or at least it was back when I were a lad.

[@lexusuk7901]

When you have a mathematics degree but haven't studied for 15 years and you realise you can't remember how to differentiate a function or rearrange functions any more. Damn. Any recommendations on how to get a refresh? :D

[@donovanb8555]

I did it with a much simpler method and got the same result. You can solve it with the theorem that says that, for all equivalent solids, it is the regular one with the least sides which has the smallest surface area. Thus, you get that h=2r, and you can process as it was shown in the video to get the same answer.

[@marklondon9004]

Can you please make a video (not an advert) as to why a student would choose Oxford for maths, vs other Universities.

[@charlesarthurs]

A top flight student would not choose Oxford to read maths.

[@likemath.]

Hay đó

[@flsal27]

Well you solded for V=1. What if you keep the generic value of V?

[@dragoncurveenthusiast]

The ratio of h and r would stay the same. We calculated the minimal surface area for a given volume. If you change the target volume, the ideal h and r would change, but the ratio between them would stay the same.

[@enigma7791]

So in a nutshell or rather cat food tin...None of the manufacturers employ someone as smart as Tom!

[@CorvanEssen]

There must be other reasons to not make the height the same as the diameter. Maybe a wider tin is easier to get the product out? And I have no clue about materials, so maybe there is an engineering reason for the higher tins?

[@gdvissch]

They could have other factors playing a role like optimised cooling of the still hot cat food once inside the can, by increasing the surface area or a combination of better cooling while still trying to reduce material cost, so neither smallest surface nor largest but somewhere in between

[@gdvissch]

@Z. Michael Gehlke interesting! Thanks for the insight

[@shankarh6915]

Perhaps the manufacturers also look at other aspects than just the most optimal dimensions from the perspective of minimizing the production costs. Ease of use of is one such. For eg, if the cat food tin is too tall, the cat won’t be able to eat off of it directly (should there be a need). Similarly, if you are talking beans, I think it makes sense for the container to be a bit tall, just easier to pour into narrow containers etc, plus easier to hold I think. There are such bean counters at these places that we’d be hard pressed to not find a legitimate reason for why the manufacturing is what it is - they optimize the heck out of it across an array of variables. So thanks, but no thanks.. probably!

[@frenchy16785]

So why do manufacturers not use the ratio demonstrated in this video? They use too tall or to short. I don't get it

[@markphc99]

I'm sure I've seen this problem before and it was solved without calculus in antiquity -wasn't it Archimedes ? One would expect eg: Soviet , Cuban, etc tins to be square - they didn't care about marketing , in contrast to Red Bull with their tiny elongated cans at an inflated price .

[@markwilliams1555]

Or we could keep the objective function as a function of r and h then use Lagrange Multipliers.

[@marklloyd7833]

would it have been easier to leave the minimum surface area result as r2 = 1/2*pi*r then can be substituted into the height formula h = 1/pi * r2 ie h = 2r

[@TomRocksMaths]

nice spot - certainly easier than dealing with those cube roots!

[@marklloyd7833]

@@TomRocksMaths Exactly, p.s. your colleagues in Computer Science posed a maths questions sin squared alpha plus cos squared beta equals 1 what is the relationship between alpha and beta apart from the obvious alpha equals beta plus multiples of 2 pi I was struggling, is it something you could cover ?

[@lukasvolak8200]

In the math side of things u dont even need to solve this problem that way... The most efficient shapes are symetrical... Sphere then square etc.. its wort adding to short side and not to longer side to the point that the sides are equal. From business perspective you cut the sheets for minimal waste and it depends on what product you have in the can.

[@dr_ned_flanders]

So it begs the question, why do we not see square tins on the shelves of the supermarkets? This was a relatively simple piece of mathematics for any large corporation to do.

[@colinmccarthy7921]

If h =2r.I would the height is 4,and r = 2.Therefore h=2r(4 = 4).

[@willgordge6003]

If these are the interview questions I shouldv’e applied😅

[@dir2002usable]

this problem is for 9th grade in Finland

[@PMA_ReginaldBoscoG]

That one subscriber who failed due to his wrong answer to this question: time to meet your destiny Tom

[@TomRocksMaths]

hello my old friend...

[@angelojones4330]

I miss how we got the "multiply by 8 inside the cube root" around 9:17 that let us simplify h. Can someone eli5?

[@TomRocksMaths]

8^(1/3) = 2

[@akshatmishra8537]

Came here from Mike's video

[@TomRocksMaths]

Welcome :)

[@scttsang]

I don't get 9:00 step ?

[@Rk-ut8kp]

Is this a calculus optimization question?

[@TomRocksMaths]

sure is!

[@christianmeer6415]

Is that the biggest volume to surface area ratio??

[@TomRocksMaths]

That would be a sphere

[@smor729]

Interesting that h=2r obviously also means h=d, meaning a vertical cross section is a square. This was my 1 second intuition and it was very satisfying to work it out and see that it was correct and why

[@rayhaan7860]

Why am I watching this when I am doing my GCSE's right now!:( that's how sad my life is rn lol

[@skyscraperfan]

Although the numbers work out, the result still feels strange, because the optimal box shape would be a cube. Having a circular instead of square cross-section does not seem to change the fact that a 1:1 ratio of height and width is optimal, although a circle is much more efficient in enclosing a space than a circle. I wonder what would happen with other shapes of the cross section? What if it had to be a regular triangle or pentagon?

[@relike868p]

L = (2 pi r^2 + 2 pi r h) - lambda (pi r ^2 h - 1) dL/dr = 4 pi r + 2 pi h - lambda (2 pi r h) = 0 dL/dh = 2 pi r - lambda (pi r^2) = 0 2 - lambda r = 0 lambda = 2/r 4 pi r + 2 pi h - 2/r * 2 pi r h = 0 2 r + h - 2 h = 0 h = 2r

[@jamienorth1603]

Yup, Lagrangian methods would’ve been easier I think

[@BusWankers]

Please can you do a hair tutorial

[@michaelwangCH]

r=1/(2*pi)

[@Hamza_Khan_Journey]

Good old optimization

[@andyblackett]

I can't see any way this is an Interview question, it's middle of the road AS Maths question which any B/C grade candidate could solve. Is this really just an advert for the Maple Calculator app?

[@TomRocksMaths]

We use lots of different types of questions in a typical interview to test different knowledge/abilities. This is just one part of the process.

[@andyblackett]

@@TomRocksMaths In which case, apologies. I'd watched some of the others which were an order of magnitude more challenging and assumed this to be the outlier. Why would you use this sort of question in an interview - the candidate sitting there is going to get an A* at A Level so they will be able to do this question without any bother, it seems unnecessary?

[@markwilliams1555]

This is quite an easy question for Oxford isn't it?

[@TomRocksMaths]

This is just one part of the interview. I like to test various different areas of maths rather than focusing on just one long question.

[@yoyokojo651]

Okay, maybe Lagrangian multipliers are a bit over the top for this one

[@TomRocksMaths]

you could certainly use them though!

[@isobar5857]

Strange....I get the height to be twice the diameter, not the radius, for maximum volume and minimum surface area. But them I'm no mathematician.

[@tariqm6083]

mate the optimum shape for surface area vs volume is a sphere.