# What’s The Best Soup Can Size? Solved With Calculus!

### 100 thoughts on “What’s The Best Soup Can Size? Solved With Calculus!”

• October 24, 2016 at 3:36 am

you can explain this for children who cant derivate if the goal function is only quadratic because in these case the extreme point is always at the (-b/2a) point you dont need the derivate make equal to zero

• October 24, 2016 at 3:44 am

Yayyyyy! I feel so smart for getting this right, because I tend to suck at calculus. My jam is more like combinatorics, graph theory, discrete stuff in general (since I'm a computer science major), so when I get a calculus question right, it's a big deal to me (although it might not be a big deal to other, more calculus savvy people).

• October 24, 2016 at 4:13 am

I figured it out by looking at the thumbnail

• October 24, 2016 at 4:14 am

2

• October 24, 2016 at 4:16 am

I take it the author has never physically held or even SEEN a can of soup?

• October 24, 2016 at 4:26 am

I got the derivative equation set up perfectly right and then I gave up cuz too much work to solve for r. I don't get why you solved for V. We need to find the r (in terms of V, a constant) that gives us the minimum.

• October 24, 2016 at 5:38 am

Couldn't do the math in my head, but guessed the answer would be the can that best fits inside a cube.

• October 24, 2016 at 5:58 am

I transformed the problem from 3D to 2D. Look from the side and the cylinder looks like a rectangle. Then I optimised for the rectangle. A square is an optimal solution. One side of the square is h and the other is 2r. The have to be the same length. h = 2r.

• October 24, 2016 at 6:04 am

Your channel refueled my love for all things math that seemed to have died years and years ago. Thank you!

• October 24, 2016 at 6:13 am

I probably should have remembered this. My guess was that the square root of pi r squared would be H.

• October 24, 2016 at 6:14 am

Sounds like preparing for World War III

• October 24, 2016 at 6:15 am

What is the procedure to find out the ratio of h/r for max surafce area??

• October 24, 2016 at 6:16 am

so the height should be the same as the diameter.

• October 24, 2016 at 6:21 am

You should mention that this is motivated by trying to minimize the amount of sheet metal and therefore cost of a can.

• October 24, 2016 at 6:44 am

YASS I'VE DONE IT! I'VE SOLVED ONE OF YOUR PROBLEMS WITHOUT WAITING THE ANSWER! I'M ACTUALLY NOT THAT PROUD OF MYSELF BECAUSE THIS IS KINDA EASY!

• October 24, 2016 at 7:06 am

I suspected the height*diameter to be a square.

• October 24, 2016 at 7:59 am

I'm a bit disappointed you haven't discussed the reasons why actual cans do not have that ratio. There is a ton of interesting stuff to learn from that.

• October 24, 2016 at 8:16 am

How bout a can with a circumference of 1

• October 24, 2016 at 8:38 am

Theoretically correct, but in practice, how many cans have you seen that have that ratio? Most cans are taller than they are wide, including most soda and beer cans. In the case of soda cans its because they are made using using deep die drawing and for a single cycle of the punch they get a larger volume when they go deeper. So for each cycle in time T they get more volume to sell. Time is money!

• October 24, 2016 at 9:18 am

I love riddles that don't require you to think

• October 24, 2016 at 9:26 am

In other words, the optimal solution is where the diameter and height are equal, and the cylinder more approsimates a sphere

• October 24, 2016 at 9:38 am

you can understand the logic of this solution by looking at the cylinder as a solid of revolution of a square. Since the square has the optimal ratio of circumference to area, it seems obvious that it's body of revolution maintains an ideal ratio of area to volume (just an extra dimension)

• October 24, 2016 at 9:39 am

The red one!

• October 24, 2016 at 9:54 am

2h:1r

so i knew the nswer without going into any mathemtics

simple logical explanation

what shape is a bubble?
a sphere because spheres have the best volume to surface area ratio
so we want out can to be as close to a sphere as possible
there for we want our diameter to be equal to our height
therefor our radius (half of a diameter) must be also half the height giving us the simple answer of h/r =2

• October 24, 2016 at 10:10 am

so the height is equal to the diameter, meaning the cylinder would fit perfectly inside a cube. I didn't work it out myself but that's what I thought it would be intuitively.

• October 24, 2016 at 10:40 am

its proportionate to h, but to r squared. make it as wide and short as possible.

• October 24, 2016 at 11:02 am

So in the end the height needs to be equal to the diameter. Why don't they make cans like this though?

• October 24, 2016 at 1:06 pm

d=h, but peaple like the golden cut, so…

• October 24, 2016 at 1:13 pm

Cylinders have only 2 sides

• October 24, 2016 at 2:07 pm

I used dummy numbers to figure it out, but the algebraic way is far more elegant.

• October 24, 2016 at 2:15 pm

This means that the vertical cross-section of the cylinder is a square. Surely there must be a simpler proof!

• October 24, 2016 at 4:07 pm

I figured it out without using derivatives.

• October 24, 2016 at 4:49 pm

That is exactly what we did today in math class…

• October 24, 2016 at 5:05 pm

ohh looks it forms a square when looked at it from the side!

we did this in school. with a bomus question for the super motivated 😉
what ration h/r is best if you want to minimize the weld lenght (weld lenght = h + 2pir + 2pir)?

• October 24, 2016 at 5:05 pm

I just figured it out by imagining that a sphere is the best way to get the most volume for the least surface area and the soup can should be as similar to a sphere as possible and if you think about then you get the same answer either way. I also don't know derivatives so I had to get creative.

• October 24, 2016 at 5:33 pm

It's interesting, that you get the same result when the surface area is fixed and you have to find the ratio h/r, so that the volume becomes a maximum.
Here's one for you Presh: Find the radius of a circle that partially covers a fixed square, so that the "error-area" (4 segments + 4 triangle-like shapes) becomes a minimum. Then switch the roles of the circle and the square.

• October 24, 2016 at 5:47 pm

A nice problem. A little conventional for those who know calculus, but still nice to flex the brain a bit. Going through the geometry involved would be fun too, approximating a sphere or forming a square in cross section, for example. A little more involved than your usual puzzle/answer format, but you've explored different perspectives before.

Thanks again!

• October 24, 2016 at 6:12 pm

it perfectly fits into a cube. cool..

• October 24, 2016 at 6:28 pm

what is a calculus? :p

• October 24, 2016 at 6:50 pm

I figured it out differently:

I looked at prisms around cylinders and found out that there is a fixed ratio between volume of the prism and the cylinder '4/pi'. Cube is the prism with the least surface area which means that the cylinder with least surface area must fit into a square thus diameter and height must be equal and thus 2r = h.

• October 24, 2016 at 6:58 pm

my intuitive guess was 2 so yay

• October 24, 2016 at 7:05 pm

Enfulykthvieo please check out my books

• October 24, 2016 at 7:48 pm

kinda obvious that h=2r because of all cylinders like that, this one is the closest to a sphere which has the lowest surface/volume ratio

• October 24, 2016 at 8:55 pm

The width of the can/the diameter should be the same value as the height edit after video so I was right because 2 radius is the diameter=1height.

• October 24, 2016 at 9:00 pm

Yes! Went with intuition: a circle from the top and a square from the side. (so height = diameter) My math is not good enough to actually show this… but I got it! I know this is no real victory for math, on a test I could not back it up and would probably get zero marks but it still feels awesome. Woohoo!

• October 24, 2016 at 9:54 pm

I just went an measured actual soup cans from my pantry

I got h/r=4 (the cans are about twice as tall as their diameter)

I figured the companies that make these cans would have already optimized it, guess I was wrong

• October 25, 2016 at 7:03 am

This reminds me of a favorite problem I had at school. You cut a sector of radius theta out of a circle, then fold the sector into a cone. What value of theta maximises the cone's volume?

• October 26, 2016 at 2:32 am

It's simpler to take the derivative of the area equation w/ resp to r — gives the result in 1 step

• October 26, 2016 at 6:40 am

Within just a couple of seconds I surmised that the answer would be about h=2r, but I wasn't really thinking that it would be exactly that.

• October 26, 2016 at 9:46 am

But you ignore the extra material required for the seams and the side walls not always being the same thickness as the top and bottom.

• October 26, 2016 at 2:41 pm

I like it

• October 27, 2016 at 5:37 pm

yes…I did.

• October 27, 2016 at 11:03 pm

the second can

• October 28, 2016 at 11:14 am

Literally just learned optimization in math last week o.o

• October 29, 2016 at 1:03 pm

I got to the same result using a different method. I considered a sphere (the shape with the smallest surface area to volume ratio) and made a cylinder which fit most closely to it. R of the cylinder is therefore equal to R of the sphere and h is equal to the diameter of the sphere, or 2 x R. Therefore, h must be equal to 2R.

• October 30, 2016 at 1:02 pm

Two weeks ago I was so frustrated that I couldn't solve this. I happened to start reading a calculus textbook out of curiosity.. I finished the first chapter about limits and I am on derivatives, so this video actually makes sense now, and it's a really beautiful way to solve things

• October 31, 2016 at 9:38 am

To make it perfect, it should be shown that the 2nd derivative at the extremum is positive (which is indeed the case here).

• November 3, 2016 at 10:30 pm

Can you force solve this problem "experimentally"? like..set 3 different variations of h/r ratios: r<2 h>2, r=2 h=2, and r>2 h<2 Then I solved the volume formula three time using exact SAME number choices while following the ratio rules above, and which ever yielded the highest volume, that was the correct answer. Highest volume using same amount of materials = optimum surface area. I got the exact same answer.

• November 4, 2016 at 9:18 pm

Here's how I did:

The volume of the cylinder is 𝐕=𝛑𝐫²𝐡 ⇔ 𝐡=𝐕/(𝛑𝐫²) (remember that 𝐕 is constant, which makes 𝐡 a function of 𝐫).

The area of the cylinder is 𝐀 = 2𝛑𝐫² + 2𝛑𝐫𝐡 = 2𝛑𝐫² + 2𝐕/𝐫 ⇒ 𝐀' = 4𝛑𝐫 – 2𝐕/𝐫².

𝐀' = 0 ⇔ 4𝛑𝐫 = 2𝐕/𝐫² ⇔ 𝐫³ = 𝐕/(2𝛑) = 𝐫²𝐡/2 ⇔ 𝐫 = 𝐡/2.

𝐀' ' = 4𝛑 + 4𝐕/𝐫³, which is always positive for any 𝐕>0 and 𝐫>0, meaning that 𝐫 = 𝐡/2 gives a local minimum for 𝐀.

In other words: for any given volume 𝐕, the smallest surface area 𝐀 will be obtained when the ratio 𝐫:𝐡 = 1:2.

• November 5, 2016 at 5:41 am

When taking the derivative of SA=2πr^2+2V/r (in respect to r), he neglected to consider the quotient rule when getting the derivative of 2V/r, so it should've been 2[(r(dV/dr)-V)/r^2]+4πr=dSA/dr=0.
I'm sure I'm doing something wrong though, please explain to me the many mistakes I made in this.

• November 5, 2016 at 6:52 am

Actually it was pretty intuitive to me that the height should be equal to the diameter, I solved it without making any calculations…

• November 8, 2016 at 10:33 am

Hey man! This problem is "old but gold". I always give it to people, because it's fun yet it shows off how math can be useful, without having to get too complicated. Cheers!

• November 10, 2016 at 2:53 am

This is a calculus problem.

• November 10, 2016 at 8:25 pm

Aight, so if two cans have the same volume, and the height of the first one is equal to the radius of the other one, what is the ratio of the height of the second one and the radius of the first

• November 14, 2016 at 5:53 am

Are these the dimensions of a soda?

• December 5, 2016 at 6:39 am

Nooooo optimization problems. I'm so haunted by it

• December 12, 2016 at 9:25 pm

what is pi

• January 11, 2017 at 3:31 pm

When he took the derivative of 2V/r, shouldn't it be -2Vr^-2 instead of -2Vr (like how it is shown in the video)
This is how I got my answer, please tell me where I went wrong.
2V/r
2Vr^-1
(Take the derivative) -1(2Vr^-1-1)
-2Vr^-2
I'm new at the whole calculus thing, how did I get this wrong?

• March 23, 2017 at 11:50 am

To the tune of the Talking Heads "IIIIIIIIIIIIIII figured it out".

• May 7, 2017 at 9:57 pm

nice but you can solve it easier by simple concept
now cut the cylinder vertically​ to get a rectangle 2r*h now we knew the max area at 2r=h ( squred )
then you can revolve the squred around itself to get maximum volume or minimum area

• May 28, 2017 at 7:20 pm

You can do it with derivatives, it's easy

• July 7, 2017 at 5:49 pm

Lol. My intuition said that the height should be equal to the diameter. That seemed like the most compact shape.

• March 10, 2018 at 10:16 pm

pretty sure it's .5, didn't watch yet

• March 23, 2018 at 9:31 pm

A classic problem in differential calculus.

• April 12, 2018 at 9:05 pm

I was doing home work, and this helped A LOT!(:

• April 14, 2018 at 5:47 am

Presh,
If you consider how these cans are stored, they will typically be in a box, with a certain number of cans, 4*3, 6*5 or similar. If you want to minimize storage space, you also have to consider the volume between the cans, as well as considering if certain ratios of rectangles are more efficient than others, 5*5, 5*4, etc. I.e making sure h=2r is still optimal, for different rectangle configurations or see if and how these soup storage factors relate. Do you know if this problem has been described?
I would suppose so…
Anyway, if could be an interesting video to make, if you haven't already done so.

• April 22, 2018 at 10:14 pm

OMG, I solved it without a lid. Then r=h. Dammit, waste of time for me 🙁

• April 23, 2018 at 7:33 pm

We had that exact problem in Calc I, 40 years ago … and I'm so glad to revisit it! Somehow, the joy of it never gets old. Thanks.

• April 29, 2018 at 10:06 pm

I still think the can w/ the least surface area would be 1 atom across. these types of cans could only contain space. so this proves the universe can fit into a can.

• May 13, 2018 at 2:44 am

I got the answer in quick succession by just imagine of a square in the cylinder therefore height is 2 times the radius. We know that square has largest ratio between area and perimeter.

• June 28, 2018 at 1:47 pm

I figured out these steps in my head but didn’t know what the equations would look like.

• July 13, 2018 at 4:43 am

I wish you had taken that further and explored max volume to see if it is also the same (h=d). I don't know, it's just the way I think. I'm always looking for thoroughness when I'm being taught. Maybe it's an OCD thing 🙂 EDIT So I just took a problem that I had worked out (a min surface area of a right cylinder)(can) and took the computed area and solved for max volume and got the same r and h values of the original problem. Therefore, a can with the least surface area is also a can with the maximum volume and h=d in both cases)(h=2r).

• July 21, 2018 at 4:42 am

There is a far simpler geometric solution that takes about 30 seconds to solve, and yields h=2r. This geometric approach when pursued further also solves the issue as to what general shape yields the smallest surface area for any given volume. [email protected]

• July 28, 2018 at 8:04 am

So why are real cans always higher than wide then, he?
Edit: (At least the ones you can hold with one hand)

• August 10, 2018 at 9:08 am

Curiosity.com brought me here and this is the best thing that's happened to me in months! This channel is a treasure for my brain! <3 I'm sooo glad to have found this channel I'm gonna binge watch a couple of videos now!

• November 4, 2018 at 11:01 am

Did you figure this out?

• November 20, 2018 at 5:01 pm

Turn on your captions and be welcomed by pressure locker…

• November 22, 2018 at 6:06 pm

Why you put 4pi r? Please explain in simple way.

• November 22, 2018 at 6:16 pm

Sorry I found that. Please don't mind.

• January 3, 2019 at 12:30 am

Yes I figured it out easy problem.

• January 29, 2019 at 8:45 pm

I decided to work it out with a spreadsheet. That was fun, got the same answer.

• April 7, 2019 at 7:12 pm

Just letting you know Presh, the captions said you were a Pressure Locker.

• May 12, 2019 at 7:10 pm

I am doing homework

• August 27, 2019 at 5:09 am

Why did you set the Derivitive equal to the original?

• September 3, 2019 at 4:46 pm

Yes – the diameter equal to height is the closest to a sphere which would have the least surface area, but spheres are not very easy to stack. There are obvious marketing considerations made concerning the ratios. I believe the closest I've seen to the area minimum are some chunky type soup cans (the traditional one is more golden ratio height to diameter for aesthetics) and a condensed or evaporated milk can (don't remember which or brand). Wish I had an old oil can as that is probably close to the optimal ratio and some sizes of metal coffee cans are undoubtedly close. Beer and soda cans must be able to be held in a hand comfortably, making ergonomic consideration of diameter override the optimum material use ratio for 12oz and 16oz cans. Once the marketroids get their hands on a prototype of the proposed optimal can, they will assuredly change dimensions to suit their idea of what will sell best (I've seen this happen with other kinds of packaging). 😉

Aesthetics and marketing aside, the more realistic problem is not least surface area, but lowest cost. Not only is there a cost of material, which is likely the largest component, there's a manufacturing cost to make the seam on the side and the crimp around the top and bottom related to how long the seam and crimps are. Cost to die stamp the side, top and bottom wouldn't change, but there could be an increased waste consideration for the top and bottom circular pieces. However, that waste would be recycled into more material for the can. The more real problem is finding the least cost for the metal material cost plus the variable manufacturing costs. As with the solution here, it's a first derivative of the total cost equation to find the minimum for the radius of a unit volume can. From that, the height of a unit volume can can be found along with the total cost per can. Using the height and radius, the ratio of the two can be found and you have an equation for the height and radius for any desired volume can, plus its cost, if that can be reasonably scaled for cans with a given size range. Obviously once one gets extremely tiny or extraordinarily large, cost factors change and it becomes a new problem. Would have been interesting to have seen it worked for a least cost. Not much more difficult. The cost of the metal is the original problem and all you're adding is a cost per linear measure for the side seam using the height and another one for the crimp (two times the side width).

• September 26, 2019 at 9:13 pm

I instinctively knew the answer. I am bewildered. 😳

• October 7, 2019 at 4:44 am

Superb