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What’s up folks? I hope you guys are doing fine in these trying times. Wherever you are, please stay safe and take care.

Today’s article marks the beginning of a new series called the Worked Problem series (because I’m so good at title), or WP in short, where we’ll work through some problems to deepen our understanding and appreciate the practicality of mathematics.

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Problem 1

Jacob is planning a family road trip (which is not going to happen any time soon, unfortunately). He is comparing two car rental companies as follows:

Company F – a flat fee of $30 plus $0.40 per mile driven.
Company G – a flat fee of $50 plus $0.15 per mile driven.

 Assuming that cost is Jacob’s only concern, how should he decide?

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I strongly recommend that you guys give this simple problem a go, and if possible, try solving it in more than one way. Continue reading when you’re done.

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Analysis

Alright, let’s analyse the situation. We first see that Company F has a lower flat fee than that of Company G. On the other hand, for every fixed mileage, Company F charges a higher rate. Intuitively, if Jacob does not plan to travel far, Company F seems to offer a better deal. For example, when the travelling distance is 20 miles, the fees are as follows:

However, in the long run, Company G wins by a mile (no pun intended) due to its lower mileage rate. For instance, when the travelling distance is 200 miles, we get the following:

Do you notice what’s starting to happen? The flat fee gets sort of amortized over the charge per mile if the car is driven long enough distance. Let’s go to the extreme, say when the mileage is 2000 miles, and see what happens:

Observe that every time we jack the mileage up, the average fees per mile for both companies get closer to their respective mileage rates ($0.40 for Company F and $0.15 for Company G).

To put it simply, Jacob would benefit more from Company F in the case of shorter road trip. Otherwise, he should go with Company G.

More precisely, we need to determine the mileage at which the car rental fee of Company F equals that of Company G. If Jacob does not plan to travel beyond that value, he should rent the car from Company F. Otherwise, he should pick Company G.

We will now solve this problem in 3 different ways and try to gain different insights from each of them; more often than not, the process which yields the answer is equally, if not more, important than the answer itself.

Solution

Method 1 (Trial and error)

Yes, we are doing this here. Often times we are discouraged from using this foolproof method – especially at school – for various reasons (like, you know, because it’s not elegant, duh).

But you can’t deny the fact that it’s still valid. In fact, if done properly, trial and error can help shed light upon the problem (see Remark 7.1).

At any rate, I’ve taken the liberty of listing down the rental fees for Company F and Company G for selected mileages in the table below:

We can clearly see that when the mileage is 80 miles, Companies F and G share the same rental fee, and so we’ve gotten what we’re after. Notice that for any mileage less than 80 miles, Company F charges cheaper rates, and the reverse happens beyond any mileage greater than 80 miles.

Method 2 (Logic)

Alright, here’s the second way of solving, in case you find the first one unsatisfactory.

Let’s think of it this way. If we compare the mileage rates, every time Jacob drives the car from Company G for one mile, he saves $0.25. On the other hand, Company G’s flat fee is $20 higher than that of Company F, which is the amount he has to make up to make Company G a better deal. So how far must he drive in order to break even? Exactly

which we already know is the answer from Method 1. However, the two routes the lead us to the answer are quite different in nature: Method 1 essentially requires basic arithmetic and a tad bit of patience, whereas Method 2 is mainly based on logical reasoning coupled with some simple calculations.

Method 3 (Algebra)

Finally, we are going to present the algebraic solution, which I believe is the more popular choice among high schoolers. To do so, we need to come up with an equation that involves some variable.

Let f and g be the total rental fees charged by Companies F and G, respectively, and x be the total miles driven. We have

Hence, the question of determining the x value when f = g is equivalent to solving the following linear equation:

Again, the result is nothing new. What’s interesting is the comparison between Method 2 and Method 3. Observe that in line (3), the two numbers that show up are 0.25 and 20. Do they somehow ring a bell? Back in Method 2, we saw that 0.25 is the amount of dollar Jacob saves by driving one mile if he chooses Company G, and 20 is the amount of the flat fee difference he needs to make up!

Hopefully we can appreciate the power of algebra now. In this case, algebra allows us to repackage our logical thought process in Method 2 and present it systematically and succinctly in the form of equations. For instance, line (1) is the algebraic translation of the English sentence “If the mileage is x, the rental fees of Companies F and G are the same”, and line (2) is where the evaluation of the differences occur.

However, it’s also due to the immense convenience of algebra that we often lose sight of the meaning or idea behind the algebraic operations we carry out, an unfortunate consequence of mindless drilling that pervades school curriculum.

What I advocate is to really view algebra as a language (of mathematics) that can be understood and interpreted.

Learn how to do algebra, and READ algebra.

Conclusion

Our discussion about the three methods delivers a simple yet important message: even when a problem has just one correct answer, there’s never really one correct way to do the problem!

Moreover, instead of asking which method is superior, a better question to ask is what we can learn from each method. The ability to look at a problem from different perspectives helps broaden our horizons, which is absolutely crucial in both mathematics and daily life.

Next time, we will embark upon the exploration of a central topic called functions; check out the remark below that serves as a prelude to it. Until then, stay acute, stay safe. See you!

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Remark 7.1

Recall that in Method 1, we have a table of values for the mileages and their corresponding rental fees charged by Companies F and G. To ease our discussion, let’s focus on Company F for the time being and borrow the notation from Method 3, i.e.

What the above table shows is that every value of x (input) is associated with one and only one value of f (output). This input-output relationship is in fact the centrepiece of the topic of functions, which we investigate in greater details here.

Also, a neat thing about functions is that we can sometimes visualize them with a little help from coordinate geometry. Without belabouring the details, we simply plot the points (input, output) on the xy-plane. For example, the table of f has the following picture:

Notice how the points seem to be aligned nicely. As a matter of fact, the graph of function f is indeed a straight line, which is why we called it a linear function (now you know how LINE-ar equation got its name).

Similarly, the graph of function g is also a straight line. Let’s go ahead and plot both functions together on the same xy-plane.

I encourage you guys to ponder how every bits of information that we’ve learnt from solving the problem manifests itself graphically. Here are some example questions you can consider:

• What does the intersection point of the two lines mean?
• What do the relative positions of the lines (one above the other) imply?
• Why is the line of f steeper than that of g?

Also, feel free to play with the graphing tool and see how the graphs are affected. Remember, play is the prime way to learn!