Mathematical modelling

Mathematical modelling means translating a real-life situation into equations, functions or inequalities and using them to answer a question. Most word problems in the exam are exactly this: choosing a tariff, motion, maximising an area, bank interest, minimising cost. The key skill is translating the condition correctly.

§1. Stages of building a model

1. Choose the unknown. Decide what is asked, denote it (or a convenient related quantity) by x, and write its unit.
2. Write the relations. Turn each sentence of the condition into a formula: "how much more" — a difference, "how many times" — a ratio, "per cent" — a multiplier.
3. Solve mathematically. An equation, a system, an extremum of a function, and so on.
4. Interpret and check. Does the answer fit the condition: is it positive, whole, within the allowed range, in the right units?

The fourth stage is often forgotten, yet that is where an extra root (for example, a negative length) gets rejected.

§2. Linear models

Constant speed, constant price, a fixed fee plus a unit price — all of these have the form y = kx + b.

Worked example

Tariff "A": a monthly fee of 30 000 sum plus 100 sum per minute. Tariff "B": no fee, 200 sum per minute. When is "A" better?

1. Number of minutes x. Costs: A(x) = 30000 + 100x, B(x) = 200x.
2. Equality: 30000 + 100x = 200x → x = 300.
3. For x > 300, A(x) < B(x). Answer: "A" is better for more than 300 minutes a month.

Worked example

Two cars leave at the same time from cities 300 km apart and drive towards each other at 70 and 80 km/h. When do they meet?

1. Closing speed: 70 + 80 = 150 km/h.
2. t = 300/150 = 2 h.

§3. Quadratic models

An area (a product of two dimensions), the height of a ball thrown upwards, revenue = price × quantity — these give quadratic functions. The largest or smallest value is at the vertex: x₀ = −b/(2a).

Worked example

A body is thrown upwards; its height is h(t) = 20t − 5t² (m). What is the maximum height and when does it land?

1. Vertex: t₀ = −20/(2·(−5)) = 2 s, h(2) = 40 − 20 = 20 m.
2. Landing: 20t − 5t² = 0 → t = 0 or t = 4. Answer: after 4 s.

Worked example

A rectangular plot by a river is fenced on three sides with 60 m of fencing (the river side is not fenced). What is the largest area?

1. Sides perpendicular to the river are x, the parallel side is 60 − 2x.
2. S(x) = x(60 − 2x) = −2x² + 60x, 0 < x < 30.
3. x₀ = 15, S = 15·30 = 450 m².

§4. Exponential models

If in each period a quantity changes by the same factor (the same percentage):

Worked example

1 000 000 sum is deposited at 20% per year (compound interest). How much will there be after 2 years?

1. q = 1.2.
2. 1 000 000·1.2² = 1 440 000 sum.

Worked example

A substance has a half-life of 5 days. How much of 80 g remains after 15 days?

1. 15/5 = 3 half-lives.
2. 80·(1/2)³ = 10 g.

§5. Optimisation with the derivative

If the function is not quadratic, the extremum is found with the derivative: build the model, solve f′(x) = 0, and compare the critical point with the endpoints.

Worked example

A river is 1 km wide. A cable must run from point A on one bank to point B on the other bank, 5 km downstream from the point opposite A. Laying 1 km under water costs 7500$, along the bank 6000$. What is the minimum cost?

1. Let the cable reach the far bank x km from the point opposite A. Cost: f(x) = 7500√(1 + x²) + 6000(5 − x), 0 ≤ x ≤ 5.
2. f′(x) = 7500x/√(1 + x²) − 6000 = 0 → x/√(1 + x²) = 0.8.
3. x² = 0.64(1 + x²) → 0.36x² = 0.64 → x = 4/3.
4. Under water √(1 + 16/9) = 5/3 km → 12500$; along the bank 11/3 km → 22000$. Total 34500$.
5. Endpoints: f(0) = 7500 + 30000 = 37500, f(5) = 7500√26 ≈ 38243. So the minimum is 34500$.

§6. Models of motion and joint work

A large share of word problems rests on one relation: ‘quantity = rate × time’. In motion problems this is distance, speed and time; in work problems it is the amount of work, the work rate and time.

The modelling procedure is the same as in §1: name the unknown, write an expression for the time or work of each part, and turn the total given in the problem into an equation. This often gives a fractional equation, which becomes a quadratic after multiplying by the denominators.
One of the two roots usually has to be rejected: a speed must be positive, the boat’s own speed must exceed the current (otherwise it cannot move upstream), and time cannot be negative. This check is the last, compulsory stage of the model.
The idea of a work rate applies not only to workers but also to pipes, pumps, printers — any process running at a constant rate.

Worked example

A boat travels 24 km downstream and 24 km back, taking 5 hours in total. The current flows at 2 km/h. Find the boat’s speed in still water.

1. Let the boat’s speed be v km/h (v > 2); downstream v + 2, upstream v − 2.
2. Model: 24/(v + 2) + 24/(v − 2) = 5.
3. Multiply by (v² − 4): 24(v − 2) + 24(v + 2) = 5(v² − 4) → 5v² − 48v − 20 = 0.
4. D = 48² + 4·5·20 = 2704 = 52²; v = (48 + 52)/10 = 10 or v = (48 − 52)/10 = −0.4.
5. Only v = 10 km/h satisfies v > 2. Check: 24/12 + 24/8 = 2 + 3 = 5.

Check yourself

1. A taxi charges 4000 sum to start and 1500 sum per km. The fare was 19000 sum. How many km?

2. What is the maximum height if h(t) = 30t − 5t²?

3. Bacteria double every hour. There are 500 at the start. How many after 5 hours?

4. Which rectangle of area 64 has the smallest perimeter?

5. Squares of side x are cut from the corners of a 12 × 12 cm sheet to make an open box. Maximum volume?

6. A price rose by 10% and then fell by 10%. How did it change overall?

7. One pipe fills a pool in 4 hours, another in 12 hours. How long do they take together?

8. A car covers the distance from A to B in 2.5 hours at 60 km/h. How long does it take at 75 km/h?

Common mistakes

• Not checking whether the answer fits the condition (negative length, a fractional number of people).
• Mixing units: minutes and hours, cm and m.
• Adding successive percentage changes (+10% − 10% ≠ 0).
• Not stating the allowed range of the variable in a quadratic model.
• Not checking the endpoints when optimising with the derivative.
• Adding or averaging the times in joint-work problems (4 + 12): it is the rates that add.