How many candies are in a jar - packing efficiency and volume?

Factual Questions
41

I think my Trigonometry is mixed in my Geometry B and my pre-calculus. Although I’m not sure, this would be the best guess, because after this there is really no chance of Trigonometry. After Pre-Calculus next year and AP Calculus Junior year, I will be taking AP Statistics in Senior year. I think we have already done a lot of Trigonometry that’s mixed in with our Geometry because we’ve done a lot with triangles last year (congruent triangles, a lot of stuff with angles, etc.), and this year we started out with the Pythagorean Theorm.

So you think that I can start Calculus without Pre-Calclus for now? I want to know enough Calculus so that I can find out the volume of the jar at least. As you said, that’s a couple months of Calculus, right?

By the way, thanks for all the input. I haven’t had anyone help me out this much. Most are too lazy to even respond. I asked this on Yahoo! Answers, and didn’t get any support here (mainly because people are becoming so mathematically illiterate these days).

42

I’m pretty sure you can get started with Calculus without the Pre-Calc. But still, Pre-Calc is full of good stuff you’ll need to have, just like Trig.

Is that Geometry B class one you’ve already had (or currently in)? Or one you will take in the future? If you’ve had any Trig already, you’d definitely know it. It’s chock full of sines and cosines and arctangents, and whole lots of stuff with horrible-looking trigonometric equations to solve and identities to memorize. (ETA: And Greek letters.) You can’t possibly learn any Trig and not know it. It’s a full semester class. If they’re just mixing a little bit in with geometry, you might be getting a superficial coverage. Sooner or later, you GOTTA have a FULL semester of Trig.

Yeah, that’s right, you won’t get a quick answer to the Jar problem this way. There’s several months of calculus study to get to that point. And you will STILL need more information (than what you’ve given us so far) about the shape of that jar. There’s no solution in ANY book that you can do without knowing the mathematical detail about that jar.

Yahoo Answers isn’t The Straight Dope Message Board. There, you might (just might) learn how babby is formed, but for Calculus, fuhgeddabouddit.

Suggested extra-curricular reading:
Why Johnny Can’t Add: The Failure of the New Math by Morris Kline (1973). Critique of “New Math.” (I see from Wiki page on Morris Kline, he has another book titled Why Professors Can’t Teach, apparently referring to math profs.)
Innumeracy: Mathematical Illiteracy and its Consequences by John Allen Paulos (1988). Critique of declining mathematical education standards.

Both are well-known popular books that you can easily find for sale on-line. Libraries probably have them too.

43

A simple way that works for many jars is to insert an uninflated balloon and inflate it in the jar. Once you feel the balloon has taken the shape of the jar - tie it off - take it out and find the volume by water displacement.

The trick is to coax the balloon out of the jar once inflated - try a balloon at least 2-3 times the size of the jar.

44

Hard to picture how you’d hold the balloon underwater to do the Archimedes thingy, given how flexible and stretchy the balloon is and how compressible the air inside is. You’d have to carefully the air pressure inside the balloon while it’s still inside the jar, and again when you’re trying to hold it under water and make it the same, or calibrate you’re measures accordingly.

If you have access to the bottle and you’re allowed dump out all the candies to do this balloon thing, then why wouldn’t you be allowed to just, you know, fill the bottle with water?

45

You don’t have to put the balloon underwater - just insert it, albeit carefully, in a graduated cylinder.

46

What kind of measurements would you have to take though if I was to figure out the volume of a jar using calculus? Would I have to measure a lot of things, or would it just be a few measurements?

Also, maybe I might be able to trick some people into asking the weight/mass of the jar and then be able to figure out the volume from there using “chemistry” (the whole mass, volume, and density thing), but the problem with that is that a new variable is added. The volume of the jar. Because let’s say the people say the jar is 1200 grams (just an example), then a significant amount of the weight WILL be included in the glass part of the jar. Now the density of glass is 2.4965 g/mL, and I might be able to work out a system where you have two variables, the density of the inside of the jar (which is the average density of candies and air like I explained earlier), and the density of the glass part. The problem is that I wouldn’t know the ratio of glass to the inside part for the mass so technically I can’t figure it out using that if I don’t even know what percent is glass of that mass. However, I can tell you that it’s around 10% of the total mass. At least when I last did this activity in Science. The mass of the glass was around 100 and the mass of the inside was around 575. But the glass I think was a little thin and less amount compared to some jars. So I might be able to assume a ratio of 1 to 6 for the mass of the jar to the mass of the inside. It equals 16.66 %, however I’m going to round this to 20 because of the reasons I explained above.

So from there if you subtract 20% of 1200 from 1200, that would give you the mass of the inside which is 960. Then using the average density of the inside, you can divide the mass by the density so 960 divided by what ever the average density is, and then you have the volume (and the answer basically because you have all the numbers you need for the system I have shown above).

47

Integrating to find a volume of revolution could be easy or hard, depending on the shape. I think the real reason several people have suggested it is more to turn you on to higher math (since you seem to be interested in that), rather than to actually, you know, solve this jar problem.

Did you see the photo of the desk lamp I posted?

Suppose you wanted to get the volume inside the lamp shade. That appears to be a segment (frustum) of a cone. You would need to measure the radius at the top of the shade, and the radius at the bottom of the shade, and there’s a geometry formula to get the volume. Formally, you could use integration, but actually, that just amounts to deriving the basic formula you already got from your geometry textbook. (That’s how a lot of those formulas can get developed anyway.)

Now suppose you wanted to get the volume of the wooden stem. It’s a volume of revolution too, but much more complicated. If you were just doing it experimentally (by measuring things), you would need to measure the radius at every height from bottom to top. Or, to get an approximation, measure the radius at one-inch intervals from bottom to top. At each inch, you could use that radius to compute the volume to the one-inch-thick disk there, and add those up.

But suppose the profile of the stem was irregular, but not THAT irregular. Suppose you could find a mathematical equation, of the form: y = (some formula with x) that, when graphed, gives you the “profile” of that shape. Then you could find the radius at every inch, or even at every millimeter, without having to actually measure it at all those places. And, with integration techniques, you could develop a formula for the volume, just starting with that y = (some formula with x) equation.

If the cross sections of the bottle are not circles (say, they are ovals or rectangular, like some of the jars in that collection of jar photos you posted), you could still do something like that. Suppose the jar was shaped like a segment (frustum) of a pyramid (like that piece of pyramid on the back of a $1 bill). Again, if you knew a formula for the area of any horizontal cross section, you could use integration to find the volume. Or, if you just took some kind of measurements at 1-inch intervals (from bottom to top) to compute the areas of those cross sections, and then the volume of the 1-inch thick slices, you could get the entire volume.

So it may, ultimately, all boil down to how mathematically straightforward the shape of the bottle is. Once you get that, you still need to deal with the efficiency of packing the candy balls into it. (Although I still like my idea of just melting them all down to fill the jar. :stuck_out_tongue: )

48

Another (obvious?) thing to note: All this integration stuff is just theoretical, at some level. It ignores the thickness of the glass. All the talk about finding the volume of a solid assumes that the outer shell is zero-thickness. For a real life jar, if you took measurements around the outside, that still just gives you the volume of the whole jar, glass included. (What you would get if you dunked the whole jar into a tub of water to measure the run-off.) You would still need to have some knowledge of the thickness of the shell in order to deduct that, to get the space inside the jar.

Again, it’s not clear from what you’ve said, how much of this information you have (or expect to have), or how much of it you are just trying to make educated guesses about it.

In this contest you are contemplating, is it an “honest” contest for math students to show off their mathematical skills? Or is it some kind of “guess how many marbles” contest where the closest guess gets $10,000 for the winner and a trip to Disneyland? As I said above, those often use tricks – like hiding a tennis ball inside of all the marbles. Would they sock you with a really odd shaped jar, or one with really thick or irregular walls?

49

Well, I’m sorry to say this, but I’m more interested in solving the jar problem because I’m going to need an answer relatively quickly. Christmas is coming up, and there is quite a chance that I will be exposed to at least one of these “candies” in jar problems.

And it could be any kind of contest. For right now, just assume that it is an honest contest. We can worry about tricks later. Besides even if we can’t figure out a way around these “tricks”, we can still win some contests that are honest. And there actually might be a way to find out a way around those tricks, but for right now, let’s just assume it’s an honest contest.

The whole Calculus technique is last resort because like I said, I’d like to have this done within a couple of months max. And I just found out a way to simplify something a lot. I really over-complicated this with systems. In reality, you don’t need the system because even though it is a way of figuring this problem out, if I’m going to measure the density of all the candies (next week), then I am going to know the volume and mass (and density of course), for all the candies. So it would be easier and quicker to solve this by dividing the packing efficiency from the volume of the jar, and then dividing the volume of one candy from that number.

But the whole problem with this situation is the volume of the jar. That volume of the jar is what’s throwing off the answers. I am starting to turn away from making any measurements because in many contests, it’s not allowed. You could always estimate to nearest cup, however problems arise when the jar is very irregular. Let’s say you’re not allowed to measure the jar. Is there a way we can figure out the volume from there? I don’t see a way honestly.

50

Well, yeah, that’s true. This talk about Calculus isn’t likely to be immediately helpful for your December contest. I think it was harmonicamoon who first mentioned it, and I jumped onto the Calculus bandwagon because I thought (and still think) you’d get really interested in that. It’s of theoretical interest for the jar problem.

I still think we’re not at all clear on what kind of raw data you’re given to work with. We still don’t know anything, really, about the geometry of the hypothetical jar, and it sounds like you don’t know either. Is that part of the problem that you don’t get to see until, like, you actually see the problem?

Likewise, I’m not clear at all about what data you’re given about the candies. You believe they will be little sphere-shaped things. And you have some knowledge about the packing density of such things. I’m not getting why their weight or density has anything to do with this? And what information will you have about the size of these candies? Are they biggish (like large gum balls) or smallish (like M&M size, only spherical)? Obviously, that would seriously affect how MANY of them are in the jar. Are you saying, like, you will have some weight and density information and you need to compute the size of the balls from that? So it isn’t clear what data you will be given, and what chain of computations you will need to do from that.

So, not very clear what more we can do with this problem until you know what the actual problem is.

51

The point of a Candy in a Jar contest is that there are things you don’t know, but can only estimate - it makes it a challenge, and the rules are there to make it fair.

Given that you probably cannot measure the volume (or estimate weight by picking up), maybe could you use a cell phone to snap a photo or two. If the candy is a known quantity (i.e average diameter) you can then calculate height and width and external profile, and thus volume. From the image you can also get an idea of packing density (close or loose). If the jar is irregular or non-symmetric you will need photos from different angles, but that should not be too hard.

Si

52

Can you please start at the start, and describe the problem.

Your description will start like this:

“There is a contest taking place in December which is [open to the public; a competition for high-school students; something else], and which involves…”

53

I think hibernicus is overthinking the OPs motivation…

I believe the OP likes sweets, and is good at math. They think that by using their math skills (and a bit of research into sweet packing density) they can have a good shot at winning some of the many Candy in a Jar competitions running over the December period. However, as s/he has realised, most of these contests are set up to prevent these sorts of approach (no weighing or measuring). So the math skills count for very little.

Either that, or the OP has a homework assignment to come up with a method of winning said contests.

Si

54

Yes, you won’t know what the jar will look like. Basically I’m working on the formula before I even get to a situation like this. So there’s no data about the jar right now. And, don’t worry about the candies. When I go to the lab, I’m going to record their Mass, Volume, and Density. The reason for this is because this might be useful data when trying to solve problems later on, and will matter while calculating packing density.

Now, I think there really is no way of doing this without knowing the volume of the jar. This becomes a direct linear equation where y represents the number of candies, and x is the volume of the jar (the volume of the jar is the independent variable). The linear equation will be y=px/v where p is the packing effiinecy in decimal form and v is the volume of one candy. Don’t worry about p and v because I will have those numbers soon (couple weeks). And for p, it’s usually going to be around 0.7. Right now to set an example, let’s say the volume of one candy is 4 mL and the packing efficinecy is 70%. What you can do is y=0.7x/4 which will be simplified to y=0.175x. And this is logical because how else are you supposed to know how many candies are in there if you don’t even know how big the container is. All we can do is create a linear equation, nothing more, unless we’re given some other kind of data.

One thing you can do is get a raw estimate for the volume of the jar by counting how many are on the circular base and multiplying it by the number of candies going up. So let’s say you come out with 2,000 candies for the wild estimate. You can replace y with 2,000, 2,000=0.175x. And that would mean that x=11,428.57. The reason this helps is that you cna then fine tune the volume by converting it to ounces, and then if it comes out with something close to the nearest cup (for example it comes out with 33.5 ounces, then you know it’s 32, so you can fine tune the volume and set it to 32 ounces and convert it back to mL), you can fine tune the volume, and then replace x with that more accurate volume and find y. The problem arrises when the jar is so irregular that you can’t rely on it being set to the nearest cup. In fact, many jars are not 8, 16, 32, 64, 128 ounces jars anyways. And this would only be possible if the jar was a cylinder. If it’s a cylinder, this might be something to explore since most cylindrical jars are usually set to the nearest cup.

Or like I said, fool them into telling you the mass of the jar, and then take 0.8 of that and assume that that is going to be the weight of the inside (the 0.2 is eliminated for the glass weight). Then divide the average density (averaged by packing efficinecy, density of the air, and density of the candy are averaged) from the mass (see, this is where knowing the density is important). Then if you will come out with some volume which you can fine tune a little bit if it comes out close to a standard volume (8, 16, 32, 64 ounces, etc.). Then plug in that value for x. But the problem is, most hosts won’t know or won’t tell the mass of the jar.

So, if anyone has a way of figuring out the volume of the jar and could share it, I would be more than happy. That’s the most critical part of this problem. Like I said, that’s the independent variable. The number of candies DEPEND on the volume of the jar.

55

No, this isn’t a homework assignment. I’m not too obsessed with sweets. I just want to win contests by using mathematics because people always tell me I’m really good at math (and I am), and I really like math. It would be awesome if I could figure out the number of candies with simple mathematics. However I’m trying to overcome this preventions. I have to. But don’t worry, because there’s only one thing left to find out, a method for figuring out the volume of the jar.

56

The only thing to figure out is the volume of the jar like I said. See, this contest holders are pretty smart. They made it purposely so that the maximum we can make out of this is a simple y=mx equation. However, if anyone gives away the volume of the jar, it’s the end of the discussion. Just plug that in for x. But how can you figure out the measurements with just snapshots? You might have a point, but how would you figure out how big the actual thing is. Don’t worry about any of the density stuff. I’ll have that covered. I just need a way to figure out the volume of the jar.

57

Lets say you have a cylindrical jar of M&Ms. An M&M (to a rough approximation) is 1cm diameter. Load up your photo in a good image editor with a ruler. Find an end-on M&M pressed up against the wall of the jar, and put a selection round it, edge-to-edge. The ruler (or status box) will tell you how many pixels across (and high) the M&M is. Now you know the Pixels/cm for x and y. Do the same with the Jar, you now know the diameter and height of the jar. Some simple math (involving pi, left as an exercise for the reader) will give you the volume.

If the jar has a rectangular form, you need a photo from the front and the side. If your photos are on an angle, you need to do a little bit of Trigonometry to correct for that.

If the jar is irregular, divide it into sections and calculate each section separately.

Si

58

Who is sponsoring these contests?

And who is expected to enter these contests?

Is it a math competition for bright high school math students? (I went to one of those when I was in HS.) Sponsored by some college or mathematical society or something? If so, you might expect:
(a) An “honest” contest (no “tricks” like hiding a tennis ball inside the jar).
(b) However the problem is presented, it would be something you should be able to solve with mathematical skills. (Using algebra and geometry, and maybe estimation skills.) But not something that would just take guess work.
© The expected contestants would be HS math students who know how to solve algebraic problems with algebra.

If it is the type of contest where various candy stores put a big jar in their window with a sign saying “Guess how many gum balls for a free trip to Disneyland” then I imagine you might expect:
(a) Some kind of “trick” problem, like a tennis ball inside.
(b) Not designed for a mathematical approach, although any good math student might try that.
© The expected contestants would be the general public, who on the average couldn’t count to 10 and blow their noses at the same time, and would probably just take a wild guess. But algebra or wild guess, that tennis ball could result in some “experimental error”.

So the question here is, what sort of contests are you expecting to enter? Formal contest of math skillz or informal “guess the gumballs for idjits” contest? I’d expect those two kinds of contest might work in very different ways.

ETA: And I still really can’t see what air density would have to do with it. That would be such a minute factor, why not just ignore it?

59

Do not worry about it Senegoid. Assume it is a fair contest. We can find maneuvers for tricks later. For now, focus on finding out the volume of the jar. I know you really want to know about what kind of contest it will be, and let us say it is any situation asuming it is fair. You may not touch the jar. Try finding out a way with that restriction. If you cannot, then we will figure something out.

60

Also the air density will matter when there are mass problems.like I said if someone told you that the jar is 1200 grams then you can take 80 percent of that, and then divide if I the average density. Remembers mass divided by density meet equals volume. find the average density you will need to know the density of the candy and the density of the air. If you know that then you can take 70 percent of the density of the candy times and add 30 percent are the density of the air to it which will give you the average density. Now this example assumes that the packing efficiency is 70% but it can and will be different for others.