Here is an amusing (and valid) variation on the OP. Given that the area and volume are numerically equal and that r = 5, find L and h. There is a unique solution and it is not, of course, the numbers in the OP.
L = 85/8 and h = 75/8.
In general, there is no solution unless r > 3 in the given units (and the fact that there is such a condition illustrates my second point). Then h = 6r^2/(r^2 - 9) and L = r(h/3 - 1).
Feynman talks about a similar kind of category error in school textbooks:
Finally I come to a book that says, “Mathematics is used in science in many ways. We will give you an example from astronomy, which is the science of stars.” I turn the page, and it says, “Red stars have a temperature of four thousand degrees, yellow stars have a temperature of five thousand degrees . . .” – so far, so good. It continues: “Green stars have a temperature of seven thousand degrees, blue stars have a temperature of ten thousand degrees, and violet stars have a temperature of . . . (some big number).” There are no green or violet stars, but the figures for the others are roughly correct. It’s vaguely right – but already, trouble! That’s the way everything was: Everything was written by somebody who didn’t know what the hell he was talking about, so it was a little bit wrong, always! And how we are going to teach well by using books written by people who don’t quite understand what they’re talking about, I cannot understand. I don’t know why, but the books are lousy; UNIVERSALLY LOUSY!
Anyway, I’m happy with this book, because it’s the first example of applying arithmetic to science. I’m a bit unhappy when I read about the stars’ temperatures, but I’m not very unhappy because it’s more or less right – it’s just an example of error. Then comes the list of problems. It says, “John and his father go out to look at the stars. John sees two blue stars and a red star. His father sees a green star, a violet star, and two yellow stars. What is the total temperature of the stars seen by John and his father?” – and I would explode in horror.
My wife would talk about the volcano downstairs. That’s only an example: it was perpetually like that. Perpetual absurdity! There’s no purpose whatsoever in adding the temperature of two stars. Nobody ever does that except, maybe, to then take the average temperature of the stars, but not to find out the total temperature of all the stars! It was awful! All it was was a game to get you to add, and they didn’t understand what they were talking about. It was like reading sentences with a few typographical errors, and then suddenly a whole sentence is written backwards. The mathematics was like that. Just hopeless!
It is well known that there are bad textbooks, but I maintain that the blame for assigning a stupid or nonsensical problem falls on the instructor. It is his or her fault for not even looking at the exercise before assigning it as problem № 8. Even, indeed especially, if the book written by somebody who didn’t know what the hell he was talking about was forced upon the class by a committee. If the book is known to be pants, you should think twice before picking problems from it sight unseen.
Well, you could argue that the ratio of surface area to volume is of interest when considering how a heated body cools.
But I concur that in this problem such interest appears contrived.
I agree.
Notwithstanding that the question is malformed:
If r = 6 cm and L = 10 cm, then there is a unique, “consistent” solution: h = 8 cm*
I’m going to wonder what happens if the units were metres not centimetres:
If r = 6 m and L = 10 m, then there is a unique, “consistent” solution: h = 8 m
No surprise there then.
Except I’ve just scaled the linear measures by 100, so the volume is scaled by 100^3 (ie 1 000 000) and the area by 100^2 (ie 10 000).
So it is not possible that both these results are true.
This is a consequence of the category error that Hari_Seldon mentions.
*I bet this was the original question that some careless author thought to modify to get more material for little work.
Yes, the ratio of volume to surface area might be of interest in some contexts. But that ratio has units of length. The proper way to pose the question in the OP would be to specify “a cone where the ratio of volume to surface area is 1 cm”. It’s not ignoring the units at all: Since the relevant quantity is dimensioned, it must be given with units. One could just as well have instead described it as being 0.01 m, or 0.3937 inches, and the fact that the quantity happens to be one of a relatively common unit isn’t really inherently relevant at all.
Yes but the result of such a ratio is not a number but some quantity of inverse cm. The coefficient by itself makes no more sense than to say that my height is 18.3.
That is technically true but the problem said “the surface area in square cm equals the volume in cubic cm” so it is clear they are just saying the numbers are the same. It’s a little different than saying “the volume equals the area.”
Which is what makes the problem garbage. Numbers are irrelevant; what matters are quantities. And when the quantities are lengths (or areas or volumes), they can only be specified with a combination of a number and a unit. 100 cm is the same quantity as 1 m, and so “the number part of that quantity, without the unit” can equally well be 100 or 1.
This reminds me of a graph I found in which the bottom axis was the base ten logarithm of the Fahrenheit temperature. At first glance one is tempted to assume there was some sort of printing error. As it happened, for that particular application, this way of transforming the temperature gave the graph more room pretty much exactly where it was valuable to show more detail.
Which is quite beautiful on top of being superficially stupid.
In that case.
But I think generally people who make a living understanding the quantitative world will pretty much agree the author of that math problem wasn’t one of us, or at least shouldn’t have been allowed to be.
So here would be an interesting question they could have asked
You have a right cone, of the size described above. An alien comes down and tells you that according to their measurement system the surface area of the cone in square smergs is equal to the volume of the cone is cubic smergs.
How long is a smerg in centimeters?
Of course, the correct response to that question would be: Holy crap we’re being contacted by intelligent aliens!!!
Yeah. All of this is about people that are trying to teach basic algebra, just manipulating equations.
And somehow they decide it needs a “real world connection” to be “less abstract” and “more relatable”. Along the way they demonstrate they don’t understand a damn thing about the real world or real applications of math to real problems. And end teaching, or at least exemplifying, ignorance that gets in the way of the subsequent learning of the students who do go on to deal with actual real world problems.
This always reminds me of those beginning algebra problems along the lines of:
Johnny has so many quarters, and so many nickels, and some dimes in his pocket. The total amount is $4.35. How many dimes does Johnny have?
Real world solution: Dump the whole pile out on the table and count them!
Hmm. I screwed that up. Let’s try it this way.
“Screams in dimensional analysis…”