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Section 1.3 Translating Problems Into Algebra

Subsection 1.3.1 Word Problems

Figure 1.3.1. Translating Problems into Algebra

Word problems have historically plagued K-12 mathematics students, and with good reason. Many of the conventional word problems mocked in jokes and memes deserved to be mocked, with their strange first-name setups and contrived situations. However, the idea behind them is fundamentally sound. Mathematics, and algebra in particular, is useful for describing all the parts of the world which can be reasonable described with numbers.

Let me try to make the case with some “word problems” of my own. Hopefully, though, these are somehow less contrived that those historically found in many mathematics textbooks. In this section, I'm just going to state problems and talk about how to translate them into mathematical terms; I'm not also going to give solutions.

These notes were written in the summer of 2020, in the midst of the CoVid-19 pandemic. The following type of problem was extremely relevant.

A certain small community has 143 active CoViD-19 cases. The number of active cases in this community is doubling every 23 days. The hospital capacity for CoVid cases in this community is 47 beds. Assuming that 10% of cases require hospitalization, if the rate of transmission stays the same, how long until the hospital reaches capacity?

Solving this question would involved translating the problem into algebra. If \(t\) is time measured in days, then doubling every 23 days is the same as multipling by

\begin{equation*} 2^{\frac{t}{23}}\text{.} \end{equation*}

If we start at 143 cases and double from there, then the number of case after \(t\) days is

\begin{equation*} (143) 2^{\frac{t}{23}}\text{.} \end{equation*}

Finally, if we want to count hospitalization rate, we need 10% of this. To get that, we multiply by \(\frac{1}{10}\text{.}\)

\begin{equation*} \frac{143}{10} 2^{\frac{t}{23}}\text{.} \end{equation*}

The problem, then, is what is the smallest value of \(t\) such that this expression is larger than 47:

\begin{equation*} \frac{143}{10} 2^{\frac{t}{23}} > 47 \text{.} \end{equation*}

Such a problem can be solved with logarithms, as in Chapter 18.

(The numbers in this example are fabricated -- I haven't done the research to get the actual numbers). Say that 75% of employed Canadians commute by car and that the average commute is 7km one way. If some program could convince 5% of these commuters to use other modes of transport (walk, bicycle, transit), how much less fuel would be used for commuting purposes?

To translate this into mathematics, let \(p\) be the population of employed Canadians. \(75\)% of them commute by car, so we need to multiply by \(\frac{75}{100}\text{.}\) Doing this I get \(\frac{3p}{4}\) for the number of Canadian that commute by car. On averge, they drive \(7\)km one way, so \(14\)km in total. Then \(14\frac{3}{4}p = \frac{21}{2}p\) is the number of daily kilometers these Canadian drive. There is some average fuel comsumption for these kilometer, which we can call \(f\text{.}\) Then the fuel used up by these commutes will be \(\frac{21}{2}pf\text{.}\) Finally, we want to know what a 5% sating would bring, so we calcluate 5% of this fuel usage. 5% is the same as \(\frac{5}{100} = \frac{1}{20}\text{,}\) so I'll divide by \(20\text{.}\) The total fuel saved would be \(\frac{21}{40} pf\) where \(p\) is the number of employed Canadian and \(f\) is an average amount of fuel used per kilometer. Note that this gives us the saving per day; if we wanted the savings per year, we would have to multiply by the average number of working days per year.

In a very oversimplified model, let's assume that the cost of running a particular airline route consists of a fixed cost \(F\) (representing the plane, a share of the company infrastructure and other costs which don't depend on the number of flights) and a cost per trip \(T\) (representing the fuel costs, airport fees, staff costs for the pilots and crew, and other costs that are incurred per flight). The income from a flight (again, oversimplified) is given by the ticket price \(P\) per passanger. Over a period of 100 flights, if we know the numbers \(F,T,P\) how many passangers are required per flight for the flight to be profitable?

We can translate this into an algebraic equation (or inequality). The costs are \(F + 100T\) and the revenue is \(100Pn\) where \(n\) is the number of passanger (on average) per trip. If we want the break even point, we would solve this equation

\begin{equation*} F + 100T = 100Pn \text{.} \end{equation*}

If we wanted the range where \(n\) leads to profitable flights, we would solve the inequality

\begin{equation*} F + 100T \lt 100Pn\text{.} \end{equation*}

Hopefully, those were not contrived examples. Hopefully, they give a sense in which translating from quantitative real-world situation into mathematics is a reasonable and useful thing to do. Those these examples are simplified, the more complicated versions of these calculations are actually being done, in the world, by public health administrators, by environmental planners, or by airline executives.