10.3 Mathematical Modelling and Process Skills
Key Takeaways
- MAT stems are unscaffolded: one question, four options, no "hence" steps of the kind common in NSC papers.
- The first modelling decision is the representation: equation, graph, or geometry — MAT will not name the topic for you.
- Translate every constraint into algebra, solve under the working rules for your test mode, then match; looking at options first invites distractors.
- A quadratic area model A = x(P − 2x) has its maximum at x = P/4 when three sides of fencing are used against a wall.
- Completing the square turns x^2 + y^2 + Dx + Ey + F = 0 into a circle; the same equation is not a line.
The 2015 CETAP MAT teachers' booklet describes MAT modelling as using mathematical process skills: translation from language to algebra, and solution of problems. The same document contrasts that demand with the National Senior Certificate (NSC) Mathematics papers. NSC questions are often scaffolded. A typical school item gives a sketch and then asks you to calculate a gradient, and hence to find the equation of a perpendicular line. The MAT version would still show a sketch, but the stem would read only "The equation of the perpendicular line is ..." with four options. There is no part (a) to tell you to start with the gradient, and no "hence" to tell you what to do next.
A second difference: NSC Paper 1 versus Paper 2 quietly signals algebra versus geometry. MAT does not. A stem may be solvable with similar triangles, with coordinate geometry, or with an equation you write yourself. You choose the representation. Independent OpenExamPrep practice below uses original situations to rehearse that choice.
The modelling loop
- Read the whole stem once without writing. Name the unknown in words.
- Choose a representation: an equation in one variable, a pair of equations, a graph, or a geometric figure.
- Translate every constraint. If a sentence is not in the algebra yet, you are not ready to compute.
- Solve with fractions and exact radicals, using only the working space permitted for your test mode.
- Only then open the four options. Distractors are usually the answers to the scaffolded sub-questions you were not asked, or the result of the representation you should not have chosen.
That loop is the process skill. The content (ratio, quadratic, circle, rate) changes; the loop does not.
Fully worked unscaffolded stem
Stem. A farmer has 72 m of fencing to enclose a rectangular vegetable bed that uses a long straight canal as one side, so only three sides are fenced. The greatest possible area of the bed, in square metres, is ...
An NSC paper would typically split this into (a) write the parallel side in terms of x, (b) hence write the area, (c) hence find the maximum. MAT will not. You must do all three steps unprompted, then pick from four numbers.
Representation. Let the two sides perpendicular to the canal be x metres each. The side parallel to the canal is then 72 − 2x. The area is the quadratic
A = x(72 − 2x) = 72x − 2x^2.
Solution. This is a downward parabola. The vertex is at x = −b/(2a) = −72 / (2 × (−2)) = 18. The parallel side is 72 − 36 = 36 m, so A = 18 × 36 = 648. Completing the square gives the same peak: A = −2(x − 18)^2 + 648. The maximum area is 648 m².
What the options are doing. 324 is the maximum if you fence all four sides (a square of side 18). 576 is three equal sides of 24 m, as if the canal side had to match the other two. 1296 is 36 × 36, using the parallel side twice. The item is easy once the model is written and brutal if you grab a four-sided rectangle formula from memory.
That is the whole MAT habit: the stem looks like a word problem from school, but you get one shot and four numbers, with no marks for a correct quadratic that you never used to produce an option.
Choosing graph versus equation versus geometry
Not every stem wants an equation in x.
Equation. "Two positive numbers differ by 8 and their product is 240. The larger number is ..." Let n be the larger. Then n(n − 8) = 240, n^2 − 8n − 240 = 0, (n − 20)(n + 12) = 0, so n = 20. Geometry will not help. A graph of y = x(x − 8) would work but is slower than factoring.
Graph / completed-square form. "The points (x, y) that satisfy x^2 + y^2 = 2x + 8y are ..." Rewrite x^2 − 2x + y^2 − 8y = 0, complete the square: (x − 1)^2 + (y − 4)^2 = 17. That is a circle with centre (1, 4) and radius √17, not a line through the origin and not a parabola. If you leave the equation in the uncompleted form you may misread it as "something squared equals a line."
Geometry. A 13 m ladder leans against a wall with its foot 5 m from the wall. The top is then 12 m up (a 5-12-13 triangle). If the foot is pulled 4 m farther from the wall, the new foot is 9 m out and the new height is √(169 − 81) = √88 = 2√22 m, so the top slides 12 − 2√22 m. Pythagoras is the representation; writing a linear equation for "distance" is the wrong model.
Rates as a linear equation. A tank holds 240 litres at t = 0 minutes. Water flows in at 8 litres per minute and out at 5 litres per minute. The tank holds 300 litres after how many minutes? Net rate 3 litres per minute; 60 extra litres needed; t = 20. The model is 240 + 3t = 300. A graph of volume against time is a straight line of slope 3, which is a valid second representation, but you do not need to sketch it to hit the option 20.
The decision is made in the first thirty seconds: if the conserved idea is a product, write n(n − d) = k. If the conserved idea is a sum of three sides, write P = 2x + L. If both x^2 and y^2 appear, complete the square or recognise a circle. If two similar triangles are staring at you, use a ratio of sides, not a quadratic.
Why NSC scaffolding is a poor MAT rehearsal
Scaffolding trains you to wait for the paper to name the next technique. MAT withholds that name. A writer who is fluent at "hence find the equation of the tangent" may still freeze when the same geometry is asked as a single multiple-choice stem. Practise by taking a familiar NSC three-part question, covering parts (a) and (b), and answering only the last numerical or algebraic request. Then check whether your own working independently produced the missing steps.
Cognitive demand on MAT is mixed. The 2015 CETAP MAT booklet has described a large share of items as knowledge, recall, and simple procedures, with a small share requiring insight. Process-skills items sit toward the complex-procedure and problem-solving end. Bank the items whose representation is obvious (a conserved acid volume, a same-base exponential) before you spend many minutes on an insight geometry that needs two representations at once.
There is no pass mark. Results are reported in Basic, Intermediate, and Proficient bands that universities interpret for admission and placement. Modelling that selects a strategy, integrates more than one piece of information, and uses more than one representation is the behaviour the 2015 CETAP MAT booklet associates with the upper bands. Recalling a formula and substituting once is the behaviour of the lower band. The unscaffolded stem is how the test observes that difference in a multiple-choice format.
Exam-day scrap-paper discipline
Write the variable definition in words ("x = width perpendicular to the canal"). Write the equation. Box the number or expression you intend to match. Then look at the options. If your boxed result is not there, you have a modelling error, not a rounding error — there is no calculator and the numbers were chosen to be exact. Re-read the constraint you might have dropped (three sides, not four; product, not sum; domain of a log).
Before you move on
- MAT will not say "hence." You choose the representation.
- Fence-against-a-wall area is A = x(P − 2x), maximum at x = P/4, not the four-sided square.
- Completed-square form distinguishes a circle from a line or a parabola.
- Work the stem to a boxed exact answer, then match. The distractors are the scaffolded sub-answers you were not asked for.
A farmer has 60 m of fencing to make a rectangular pen against a long straight barn wall, so the barn supplies one side. The maximum enclosed area, in square metres, is
The set of points (x, y) satisfying x^2 + y^2 = 2x + 8y is
Two positive numbers differ by 8 and their product is 240. The larger number is