Posts tagged ‘volume’
Two the Hard Way (and an Easy Way)
During our trip to Arizona for winter break, two problems surfaced organically while we were on holiday. (Sorry. Two math problems. There were lots of nonmath problems, too, but we don’t have time for all that.)
On the plane, the interactive inflight map showed the outside temp, toggling between Fahrenheit and Celsius. That led to the following MJ4MF original problem, which I thought — and still think — is pretty good:
The conversion between Fahrenheit and Celsius temperatures follows the rule F = 9/5 C + 32. Sometimes, the temperature is positive for both Celsius and Fahrenheit; sometimes, the temperature is negative for both Celsius and Fahrenheit; and other times, Fahrenheit is positive while Celsius is negative. What is the least possible product of the Fahrenheit temperature and its corresponding Celsius temperature?
You can pause here if you’d like to solve this before I present a spoiler.
Before I reveal two different solutions, allow me to digress. It could be that the statement about the temps sometimes being positive and sometimes being negative is denying a teachable moment. The graph below shows the linear relationship between the two temperature scales. Perhaps a good classroom question is:
When will the product CF be positive and when will it be negative?
Or maybe a better question is:
When are C and F both positive, when are they both negative, and when do they have different signs?
So, to the problem that I posed. As I thought about it on the plane, I concluded that if F = 1.8C + 32, then the product CF = 1.8C^{2} + 32C. I then used calculus, found the derivative (CF)’ = 3.6C + 32, set that equal to 0, and concluded that C = ‑8.89, approximately. The corresponding Fahrenheit temperature is F = 16, so the minimum product is roughly ‑142.22.
Using calculus was like rolling a pie crust with a steamroller, though. I could have just as easily graphed the parabola and noted its vertex:
If I had used the other form of the rule, namely C = 5/9 (F ‑ 32), things might have been a little easier. Maybe. In that case, setting the derivative equal to 0 yields F = 16, which is arguably a nicer number. But then you still have to find the corresponding Celsius temperature, which is C = ‑8.89, and the product is still roughly ‑142.22. So, not much easier, if at all, and again graphing the parabola and noting its vertex would have done the trick:
The only real benefit to using this alternate version of the rule is that it provides a reasonable check. Since both methods — and both graphs — yield an answer of ‑142.22, we can feel confident in the result.
But there’s an easier way to solve this one.
Thinking this was a good problem — and because I like when my sons make me feel stupid — I gave it to Eli and Alex. Within seconds, Eli said, “Well, F is positive and C is negative between 0°F and 32°F, so the minimum will occur halfway between them at F = 16. That means C = ‑80/9, so it’s whatever ‑1280/9 reduces to.” (Turns out, 1280/9 = ‑142 2/9 ≈ ‑142.22.)
Eli hasn’t taken calculus, so he doesn’t know — or, at least, he hasn’t learned — that the minimum product should occur halfway between the x– and y‑intercepts of the linear graph. Yet, he had an intuitive insight that just happens to be true. As a result, what took me about five minutes of deriving and manipulating took him about five seconds.
The second problem arose at the grocery store. Among our purchases was a box of sugar cubes, which contained, surprisingly, 126 cubes.This number is surprising in the sense that it’s not a number you’ll see very often, except for an occasional appearance in the ninth row of Pascal’s Triangle, or maybe if you’re a chemist searching for stable atoms.
A question that could have arisen from this situation involves surface area and volume:
A rectangular prism with integer dimensions has a volume of 126 cubic units. What is the least possible surface area?
That’s not the question that was shared with Alex and Eli, though. (The answer, if you care, is 162 square units, which results from a 3 × 6 × 7 arrangement — which, in fact, is the exact arrangement of cubes in the box above. I suspect this is not a coincidence.)
The problem that I shared with my sons involved probability:
Imagine that the arrangement of cubes is removed from the box intact, and all six faces of the prism are painted red. Then one of the sugar cubes is selected at random and rolled. What is the probability that the top face of the rolled cube will be red?
The boys made an organized list, as follows:
Painted Faces  Number of Cubes 
3  8 
2  40 
1  58 
0  20 
Further, the boys reasoned:
 P(cube with 3 red faces, red face lands on top) = 8/126 x 1/2 = 8/252
 P(cube with 2 red faces, red face lands on top) = 40/126 x 1/3 = 40/378
 P(cube with 1 red face, red face lands on top) = 58/126 x 1/6 = 58/756
Therefore,
 (P of getting a red face) = 8/252 + 40/378 + 58/576 = (24 + 80 + 58) / 576 = 162/576 = 9/42
Wow! That seems like a lot of work to get to the answer. Surely there’s an easier way, right?
Indeed, there is.
Notice that the penultimate step yielded the fraction 162/576. The numerator, 162, may look familiar. It’s the answer to the question that wasn’t asked above, the one about the least possible surface area of the prism. That’s no coincidence. In total, there will be 162 faces painted red. And there are 6 × 126 = 576 total faces on all of the sugar cubes (that is, six faces on each cube). This again suggests that the probability of rolling a red face is 162/576.
Did you happen to notice that the volume and surface area use the same digits in a different order? Cool.
So there you have it, two problems, each with two solutions, one easy and one hard. Or as mathematicians might say, one elegant and one common.
It’s typical for problems, especially problems worth solving, to have more than one solution strategy. What’s the trick to finding the elegant solution? Sadly, no such trick exists. Becoming a better problem solver is just like everything else in life; your skills improve with practice and experience. It’s akin to Peter Sagal’s advice in The Incomplete Book of Running, where he says, “You want to be a writer? […] Just sit down and write. The more you write, the better a writer you will become. You want to be a runner? Run when you can and where you can. Increase your mileage gradually, and your body will respond and you’ll find yourself running farther and faster than you ever thought possible.” You want to be a problem solver? Then spend your time solving problems. That’s the only way to increase the likelihood that you’ll occasionally stumble on an easy, elegant solution.
And every once in a while, you may even solve a problem faster than your kids.
Where Would We Fit?
In the classic book 101 Puzzle Problems, Nathaniel B. Bates and Sanderson M. Smith make an astounding claim:
The volume of the 1970 world population is less than the volume of the Houston Astrodome.
That is, if you packed all of the people on Earth who were alive in 1970 — all 4 billion of them — as tightly as possible, they would’ve actually fit in the stadium, and they could’ve seen Evel Knievel jump 13 cars or Billy Jean King defeat Bobby Riggs.
Before you read any farther, you might wish to do a Fermi estimate to verify the reasonableness of this claim.
It’s an amazing fact!
Unfortunately, it’s also wrong.
Bates and Smith argue that the average human body has a volume of approximately 2 cubic feet. Their argument? That water weighs 62.5 pounds per cubic foot, and “the human body consists mostly of water.” Fair enough, if we think that an average human weighs about 125 pounds, which they probably didn’t, even in the 1970s.) Personally, I would’ve attacked the estimate a little differently. An average human can fit in a box that is 6 feet tall with a base that measures approximately 0.5 square feet — I know; I’ve been to funerals — and such a box has a volume of 3 cubic feet. But there’s a fair amount of wasted space in that box so, yeah, 2 cubic feet seems like a reasonable estimate.
Therefore, the world population in 1970 would have had a combined volume of 8,000,000,000 cubic feet.
So far, so good.
But Bates and Smith then make their statement about the Astrodome, with no calculations or estimates about the volume of the stadium to justify the claim. Please understand, I don’t fault Bates and Smith; as I understand it, the claim about everyone fitting in the Astrodome was a ubiquitous factoid during the decade of bellbottoms and disco. Still, you’d think that two mathematicians would have checked the stat before including it in a book.
The Astrodome has an outside diameter of 710 feet and a height of 208 feet. A cylinder with those dimensions would have a volume of
V = π × 355^{2} × 208 ≈ 82,000,000 cubic feet
Although this is surely an overestimate, it’s still an order of magnitude too small to fit the world’s population in 1970. There’s not nearly enough space to store all those buggers.
Bates and Smith state that the volume of the 1970 population “takes up about oneeighteenth of a cubic mile.” And I think therein lies the error. The height of the Astrodome is just shy of oneeighteenth of a mile, and the diameter is a whole lot more than oneeighteenth of a mile, so someone (incorrectly) assumed that a container with those dimensions would easily hold 1/18 cubic mile. But that’s not right. As any geometry student will tell you, shrinking each dimension to a fraction of its original length results in a volume that is the cube of that fraction. Oops.
The diameter of the Astrodome is about 1/7 of a mile, and the height is about 1/22 of a mile, so its volume would have been less than 1/1000 of a cubic mile.
So, what container would have been big enough to hold the 1970 population?
Honestly, who cares?
Things have changed. Today, we are bumping up against 8 billion people worldwide, the Astrodome was declared “unfit for occupancy” in 2016, and instead of calculating the volume of stadia, we can simply use Google to find the volume of some really large places.
In fact, Google claims that the volume of the Astrodome is 42,000,000 cubic feet. If we can believe what we read, then the calculations above were, sadly, unnecessary.
So, you may be wondering, what places in the world could hold all of us? That is, what has a volume of 16,000,000,000 cubic feet?
The building with the greatest volume in the world is the Boeing Everett Factory. It has a volume of almost 500,000,000 cubic feet — which means you’d need 32 of them to fit today’s world population.
The manmade structure with the greatest volume is the Great Wall of China, tipping the scales at just under 20,000,000,000 cubic feet. That would be large enough, though how would you get all of those people inside? For easier packing, you could turn to nature and just dump everyone into Sydney Harbour, which has a volume of 562,000 megaliters, or about 19,500,000,000 cubic feet.
How Would You Answer These Questions?
The following is one of my alltime favorite assessment items:
Which of the following is the best approximation for the volume of an ordinary chicken egg?

The reason it’s one of my favorites is simple: it made me think. Upon first look, I didn’t immediately know the answer, nor did I even know what problemsolving strategy I should use to attack it.
I used estimation, first assuming that the egg was spherical — and, no, that is not the start of a math joke — and then by attempting to inscribe the egg in a rectangular prism. Both of those methods gave different answers, though, and not just numerically; each led to a different letter choice from above.
Not satisfied, I then borrowed a method from Thomas Edison — I filled a measuring cup with 200 mL of water, retrieved an ordinary egg from my refrigerator, and dropped it into the cup. The water level rose by 36 mL. This proved unsatisfying, however, because although choice B is numerically closer to this estimate than choice C — only 29 mL less, compared to 34 mL more — it was five times as much as B but only half as much as C. For determining which is closer, should I use the difference or the ratio?
It was at this point that I decided the answer doesn’t matter. I had been doing some really fun math and employing lots of grey matter. I was thinking outside the box, except when I attempted to inscribed the egg in a rectangular prism and was literally thinking inside the box. And, I was having fun. What more could a boy ask for?
On a different note, here’s one of the worst assessment questions I’ve ever seen:
What is the value of x?
3 : 27 :: 4 : x
I can’t remember if it was a selectedresponse (nee, multiplechoice) item, or if was a constructedresponse question. Either way, it has issues, because there are multiple possible values of x that could be justified.
On the other hand, it’s a great question for the classroom, because students can select a variety of correct responses, as long as they can justify their answer.
The intended answer, I’m fairly certain, is x = 36. The analogy is meant as a proportion, and 3/27 = 4/36. (Wolfram Alpha agrees with this solution.)
But given the format, it could be read as “3 is to 27 as 4 is to x,” which leaves room for interpretation. Because 27 = 3^{3}, then perhaps the correct answer is 64 = 4^{3}.
Or perhaps the answer is x = 28, because 3 + 24 = 27, and 4 + 24 = 28.
Don’t like those alternate answers? Consider the following from Math Analogies, Level 1, a software package from The Critical Thinking Company that was reviewed at One Mama’s Journey.
If this analogy represents a proportion, then the correct answer is $10.50, but that’s not one of the choices. Instead, the analogy represents the rule “add $1,” and the intended answer choice is $10.00.
What amazing assessment items have you seen, of either the good or bad variety?
16 Math Problems for 2016
Yes, I know that I just posted some Math Problems for 2016 on December 19.
But I’ve decided to post some more for a variety of reasons:
 2016 is cool.
 It’s a triangular number.
 It has lots of factors. (I’d tell you exactly how many, except that’s one of the problems below.)
 After writing the problems for that previous post, I just couldn’t control myself.
 It’s my blog, and I can do what I want.
People who write math problems for competitions (like me) love to be cheeky and include the year number in a problem, especially when any sufficiently large number will do. When the year number is critical to the success of a problem, well, that’s just a bonus. With that in mind, there are 16 problems below, each of which includes the number 2016.
A fully formatted version of these problems, complete with answer key, extensions, and solutions, is available for purchase through the link below:
16 Problems for 2016 — just $1
Enjoy, and happy new year!
 What is the sum of 2 + 4 + 6 + 8 + ··· + 2016?
 Using only common mathematical symbols and the digits 2, 0, 1, and 6, make an expression that is exactly equal to 100.
 Find a fraction with the following decimal equivalent.
 How many positive integer factors does 2016 have?
 What is the value of n if 1 + 2 + 3 + ··· + n = 2016?
 Find 16 consecutive odd numbers that add up to 2016.
 Create a 4 × 4 magic square in which the sum of each row, column, and diagonal is 2016.
 Find a string of two or more consecutive integers for which the sum is 2016. How many such strings exist?
 What is the value of the following series?
 What is the units digit of 2^{2016}?
 Some people attend a party, and everyone shakes everyone else’s hand. A total of 2016 handshakes occurred. How many people were at the party?
 What is the value of the following expression, if x + 1/x = 2?
 A number of distinct points were placed along the circumference of a circle. Each point was then connected to every other point, and a total of 2016 segments were formed. How many points were placed on the circle?
 Let A = 1, B = 2, C = 3, …, Z = 26. Find a word for which the product of the letters is 2016. (This one may look familiar.)
 Each dimension of a rectangular box is an integer number of inches. The volume of the box is 2016 in^{3}. What is the least possible surface area of the box?
 What is the maximum possible product for a set of positive integers that have a sum of 2016?
UPDATE: Bonus Material!
Special thanks to my friend Harold Reiter, who created the following 2016 problems for use as MathCounts practice:
 What is the smallest number N such that the product of the digits of N is 2016?
 What is the sum of the divisors of 2016?
 What is the product of the divisors of 2016? Express your answer as a product of prime numbers.
 Solve the following equation:
 What is the binary representation of 2016?
 What is the base4 representation of 2016?
 What is the base8 representation of 2016?