Mathematical Modelling

The result of such a translation is called a mathematical model, and the problem itself is called an applied problem.

The branch of mathematics concerned with the construction and study of mathematical models is called mathematical modelling.

Solving an applied problem actually consists of three stages:

1. Constructing the mathematical model — translating the problem into the language of mathematics (equations, inequalities, systems, etc.).
2. Solving the mathematical problem — finding values of expressions, solving equations, inequalities and their systems, constructing graphical objects, etc.
3. Analysing the result — writing the obtained result in the language of the applied problem and checking its consistency with the problem conditions.

Problem statement. A wooden beam has a mass of 120 kg, and an iron beam has a mass of 140 kg. The iron beam is 1 m shorter than the wooden one. Find the length of each beam if 1 m of the iron beam is 5 kg heavier than 1 m of the wooden beam.

Solution.

Stage I. Constructing the mathematical model.

Let the length of the wooden beam be $x$ m, then the length of the iron beam is $(x - 1)$ m.

The mass of 1 m of the wooden beam is $\dfrac{120}{x}$ kg, and the mass of 1 m of the iron beam is $\dfrac{140}{x-1}$ kg.

Then the difference $\dfrac{140}{x-1} - \dfrac{120}{x}$ shows how much heavier 1 metre of the iron beam is compared to 1 metre of the wooden beam.

By the problem conditions, this difference equals 5 kg. So we obtain the equation:

$\frac{140}{x-1} - \frac{120}{x} = 5$

This equation is the mathematical model of the given applied problem.

Stage II. Solving the equation.

$\frac{140}{x-1} - \frac{120}{x} = 5; \quad \frac{28}{x-1} - \frac{24}{x} = 1$

$28x - 24(x-1) = x^2 - x, \quad x \neq 0, \quad x \neq 1$

$x^2 - 5x - 24 = 0$

$x = 8 \quad \text{or} \quad x = -3$

Stage III. Analysing the result.

The root $x = -3$ does not satisfy the problem conditions, since length cannot be negative.

Therefore, the wooden beam is 8 m long, and the iron beam is 7 m long.

Answer: 8 m, 7 m.

Problem statement. From two points 18 km apart, two tourists set out simultaneously towards each other and met after 2 hours. Find the speed of each tourist if one of them needs 54 minutes more than the other to cover the entire distance between the points.

Solution.

Let the speed of the first tourist be $x$ km/h and the second be $y$ km/h, where $x < y$. Before meeting, the first tourist walked $2x$ km and the second walked $2y$ km. Together they covered 18 km:

$2x + 2y = 18$

The first tourist covers the distance between the points in $\dfrac{18}{x}$ hours, and the second in $\dfrac{18}{y}$ hours. Since the first tourist needs $54 \text{ min} = \dfrac{9}{10} \text{ h}$ more than the second:

$\frac{18}{x} - \frac{18}{y} = \frac{9}{10}$

We obtain the system of equations:

$\begin{cases} 2x + 2y = 18, \\ \dfrac{18}{x} - \dfrac{18}{y} = \dfrac{9}{10}. \end{cases}$

$\begin{cases} x + y = 9, \\ \dfrac{2}{x} - \dfrac{2}{y} = \dfrac{1}{10}; \end{cases} \quad \begin{cases} x = 9 - y, \\ \dfrac{2}{9-y} - \dfrac{2}{y} = \dfrac{1}{10}. \end{cases}$

Solving the second equation of the last system, we get: $y_1 = 5$, $y_2 = -36$. The root $-36$ is rejected based on the problem context. Therefore, $y = 5$, $x = 4$.

Answer: 4 km/h, 5 km/h.

Problem statement. Two workers can complete a job together in 10 days. After 6 days of working together, one of them was reassigned, and the second continued working. After 2 more days of independent work by the second worker, it turned out that $\dfrac{2}{3}$ of the entire job had been completed. How many days would it take each worker to complete the entire job alone?

Solution.

Let the first worker be able to complete the entire job in $x$ days and the second in $y$ days. In one day, the first worker completes $\dfrac{1}{x}$ of the job, and in 10 days — $\dfrac{10}{x}$ of the job.

The second worker completes $\dfrac{1}{y}$ of the job per day, and in 10 days — $\dfrac{10}{y}$. Since in 10 days of joint work they complete the entire job:

$\frac{10}{x} + \frac{10}{y} = 1$

The first worker worked 6 days and completed $\dfrac{6}{x}$ of the job, while the second worked 8 days and completed $\dfrac{8}{y}$ of the job. Since this resulted in $\dfrac{2}{3}$ of the job being done:

$\frac{6}{x} + \frac{8}{y} = \frac{2}{3}$

We obtain the system of equations:

$\begin{cases} \dfrac{10}{x} + \dfrac{10}{y} = 1, \\ \dfrac{6}{x} + \dfrac{8}{y} = \dfrac{2}{3}, \end{cases}$

whose solution is the pair $x = 15$, $y = 30$.

Therefore, the first worker can complete the entire job in 15 days, and the second in 30 days.

Answer: 15 days, 30 days.

Problem statement. Each participant in a maths quiz received one of the scores: 9, 10, 11, or 12 points. Scores of 9, 10, and 12 were given to equal numbers of participants, while the score of 11 was given to more participants than all other scores combined. Fewer than 10 participants scored above 10. How many participants scored 10 and how many scored 11, given that at least 12 people took part in the quiz?

Solution.

Let scores of 9, 10, and 12 each be received by $x$ participants, and score 11 by $y$ participants.

Since score 11 was given to more participants than all others combined: $y > 3x$.

Fewer than 10 participants scored above 10: $x + y < 10$.

At least 12 people participated: $3x + y \geqslant 12$.

We obtain the system of inequalities:

$\begin{cases} y > 3x, \\ x + y < 10, \\ 3x + y \geqslant 12. \end{cases}$

From this, $x + y > x + 3x = 4x$. Then $4x < 10$; $x < 2.5$.

Since $x$ is a non-negative integer, from the inequality $x < 2.5$ it follows that $x = 0$, $x = 1$, or $x = 2$.

For $x = 0$: the system $y > 0$, $y < 10$, $y \geqslant 12$ has no solutions.

For $x = 1$: the system $y > 3$, $y < 9$, $y \geqslant 9$ also has no solutions.

For $x = 2$: $y > 6$, $y < 8$, $y \geqslant 6$. Hence $y = 7$.

Therefore, 2 participants scored 10, and 7 participants scored 11.

Answer: 2 participants, 7 participants.

Problem statement. An airline connecting cities $A$ and $B$ operates three types of aircraft. Each plane of the first, second, and third type can carry 230, 100, and 40 passengers respectively. All the airline’s planes can simultaneously transport 760 passengers. How many planes of each type does the airline operate?

Solution.

Let $x$, $y$, $z$ denote the number of planes of the first, second, and third type respectively. We obtain the equation:

$230x + 100y + 40z = 760$

which must be solved in natural numbers, i.e., only solutions that are natural numbers are accepted.

$\begin{cases} 23x + 10y + 4z = 76, \\ x \in \mathbb{N}, \quad y \in \mathbb{N}, \quad z \in \mathbb{N}. \end{cases}$

We have: $23x = 76 - (10y + 4z)$. Since $y \geqslant 1$ and $z \geqslant 1$, then $10y + 4z \geqslant 14$. So $23x = 76 - (10y + 4z) \leqslant 62$. Hence $x \leqslant \dfrac{62}{23}$. From the equation it follows that $x$ must be even. Therefore, $x = 2$.

Substituting the found value of $x$:

$\begin{cases} 10y + 4z = 30, \\ y \in \mathbb{N}, \quad z \in \mathbb{N}. \end{cases}$

$\begin{cases} 5y + 2z = 15, \\ y \in \mathbb{N}, \quad z \in \mathbb{N}. \end{cases}$

From this equation it follows that $z$ must be a multiple of 5. Moreover, $2z = 15 - 5y \leqslant 10$. Hence $z \leqslant 5$, so $z = 5$, $y = 1$.

Answer: 2 planes, 1 plane, 5 planes.

Two motorcyclists set out simultaneously from cities $A$ and $B$ towards each other. After one hour they met and, without stopping, continued at the same speed. One of them arrived at city $A$ 35 minutes earlier than the other arrived at city $B$. Find the speed of each motorcyclist if the distance between the cities is 140 km.

A boat sailed along a river from pier $A$ to pier $B$ and returned in 6 hours. Find the speed of the river current if the boat covers 2 km downstream in the same time as 1 km upstream, and the distance between piers $A$ and $B$ is 16 km.

The delivery cost of one truckload of sand to a construction site is 1000 UAH, and one truckload of gravel is 1600 UAH. It is planned to make 50 trips per day, with transport costs not exceeding 59,000 UAH. How many truckloads of gravel can be delivered per day?

In a shooting competition, each participant makes 25 shots. For each successful shot, 4 points are awarded, and for each miss, 2 points are deducted. How many misses can a shooter make and still score at least 60 points?