Mathematical problem solvingAQA A-Level Maths: Revision notes

Mathematical problem solving follows a cycle:

1. Specify the problem: decide exactly what is to be found.
2. Collect information: gather the given data and any facts you need.
3. Process and represent the information: draw a diagram, define variables, form equations, calculate.
4. Interpret the results in the context of the problem. If the result is unreasonable or does not answer the question, you repeat the cycle with a better specification or more data. Written solutions should show the stages: define symbols, state equations, solve, then answer in words with units.

Many problems come in words. Find the underlying structure and simplify it by:

• choosing variables ("let $x$ be the width in metres");
• stating the quantities that are fixed (a total length, a rate);
• identifying the type of mathematics: linear, quadratic, trigonometric, and so on. Diagrams help you extract information and construct a mathematical picture, including sketches of shapes, forces or velocities in mechanics. Label every known and unknown length, angle or force. Example: a garden fenced on three sides with 40 m of fencing, using sides of length $x$. The side along the wall is $40-2x$, so $A=x(40-2x)=40x-2x^2=-2(x-10)^2+200$, with maximum area $200\ \mathrm{m}^2$ at $x=10$.

An unstructured problem gives no steps. Plan it:

1. Define variables and draw a diagram.
2. Write the constraint (for example $2L+4w=120$).
3. Write the quantity to find in terms of one variable (for example $A=w(60-2w)$).
4. Solve or optimise (completing the square or a graph).
5. Check the solution makes sense. Worked example: 120 m of fencing makes a rectangle with two internal dividers. $2L+4w=120$ gives $L=60-2w$ and $A=60w-2w^2=-2(w-15)^2+450$, so the largest area is $450\ \mathrm{m}^2$ at $w=15$. For a minimum area per pen, solve an inequality, such as $60w-2w^2\ge360$, giving $15-3\sqrt5\le w\le15+3\sqrt5$.

A mathematical answer must be interpreted:

• discard solutions that make no sense, such as a negative length or $x\ge20$ when the third side $40-2x$ must stay positive;
• give units and sensible rounding (usually 3 s.f.);
• state the domain that is realistic for the variable;
• say what the number means: "Firm A is cheaper by £0.90". Example: two taxi firms charge £3.50 plus £1.80 per mile and £2.00 plus £2.20 per mile. They cost the same when $3.5+1.8d=2+2.2d$, so $d=3.75$ miles. For longer journeys Firm A is cheaper, and for shorter journeys Firm B is.

Check that an answer is reasonable by estimating: round the numbers to 1 significant figure and compute roughly. For a tank with radius 1.5 m and height 2.0 m, $V=\pi r^2h\approx3\times2\times2=12\ \mathrm{m}^3$, close to $14.1\ \mathrm{m}^3$. Comment on limitations:

• rounded inputs lead to rounded outputs, so quote results to a suitable number of significant figures;
• assumptions such as constant rate, ideal shapes or flat ground may not hold;
• a result outside the sensible domain shows that the model or working needs repair, which means repeating the cycle. Example: filling a 14 137-litre tank at net 20 litres per minute takes about 707 minutes, but a leak that grows as the tank fills would make it take longer.