## Posts tagged ‘negative’

### 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 non-math problems, too, but we don’t have time for all that.)

On the plane, the interactive in-flight 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 = 1.8C2 + 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.

### Math Jokes from Reader’s Digest

All of the following jokes were borrowed from Reader’s Digest, which I’m sure they borrowed from elsewhere.

Did you hear about the mathematician who’s afraid of negative numbers?
He’ll stop at nothing to avoid them.

How easy is it to count in binary?
It’s as easy as 01 10 11.

A Roman walks into the bar, holds up two fingers, and says, “Five beers, please.”

How many bananas can you eat if your stomach is empty?
Just one. Then it’s not empty anymore.

What do you call a number that sleepwalks?
A roamin’ numeral.
(And a nun who sleepwalks?
A roamin’ Catholic.)

Knock! Knock!
Who’s there?
Convex.
Convex who?
Convex go to prison!

### Free Copy of My Book

A few days ago, a seventh‑grade math teacher and assistant baseball coach sent me the following request:

I would love to have a copy of Math Jokes for Mathy Folks, but I am financially unable to purchace it right now because my wife is unable to work and hasn’t been approved for disability.

Now, I like to think I’m a generous guy, but I am unable to send a free copy of my book to everyone who asks for it. A little‑known fact about the publishing industry: The majority of authors actually have to pay for copies of their own book. It’s an interesting percent problem. I pay 50% of retail price plus shipping to purchase copies of my book, but I then receive a 15% royalty on the discounted price of every copy I purchase. (You can do the math to figure out how many books I could give away for free before going bankrupt.)

So I sent the following reply:

Send me your favorite joke(s), mathy or not, and I’ll send you a copy of the book.

My correspondent responded quickly with three jokes, two of which I had never heard before. A copy of my book is in the mail to him, and his jokes are pasted below for your reading pleasure.

What’s a seventh grader’s favorite excuse for not doing homework?
I have a solar‑powered calculator, but yesterday it was cloudy.

The student’s second semester seemed so much like her first that she hoped she could graduate sooner by combining like terms.

How is an indecisive third‑base coach like multiplying or dividing by a negative integer?
In both cases, the sign changes.

Incidentally, you can download one chapter of Math Jokes 4 Mathy Folks for free by clicking the following button:

The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

## MJ4MF (offline version)

Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature.