By the end of this module, you should be able to:
- Revise the essential results of Modules 1–7.
- Connect concepts across topics (sequences ↔︎ interest, functions ↔︎ optimisation).
- Identify the appropriate mathematical method without a chapter label.
- Practise on mixed problems under exam-like conditions.
1 Diagnostic Activity
Attempt these identification questions without consulting your notes. For each scenario, name the mathematical method or concept you would use and the module it comes from. Do not solve — only identify.
- A retailer applies a 15% discount to a marked price. What method determines the final price and the overall percentage reduction?
- A firm’s total cost consists of a fixed component and a variable component proportional to output. At what output does revenue equal cost?
- A monthly salary grows by the same fixed amount each year. What type of sequence describes the salary?
- A firm’s profit function is quadratic in the quantity produced. What tool finds the profit-maximising quantity?
- Percentage change / successive discounts — Module 1. The factor applied is \((1 - 0.15) = 0.85\).
- First-degree equation / break-even analysis — Module 2. Set \(R(q) = C(q)\) and solve for \(q\).
- Arithmetic sequence — Module 4. The constant addition per period is the common difference \(d\).
- Derivative and optimisation — Module 7. Set \(P'(q) = 0\) and verify it is a maximum (since the parabola opens downward).
Use your results to identify which topics need review before the exam.
2 Opening Problem
In the following problems, no chapter title or formula sheet is provided. The challenge is not to calculate — it is to recognise.
Before reading any solution, write down for each problem below: (1) which mathematical method applies, and (2) which module it comes from. Only then should you verify by expanding the solution.
3 Core Concepts
This module contains no new theory. The table below summarises the seven methods covered in the course and the conditions under which each applies.
| Method | When to apply | Key formula or concept |
|---|---|---|
| Percentages & indices (M1) | Price changes, market-share shifts, index tracking | \(\Delta\% = (V_f - V_i)/V_i \times 100\); factors multiply |
| Algebra & break-even (M2) | Fixed + variable costs; revenue vs. cost crossover | \(Q^* = F/(p-v)\); set \(R(q) = C(q)\) |
| Lines & systems (M3) | Two linear relationships; equilibrium or crossover | \(y = mx+p\); intersect at \(x^* = (p_2-p_1)/(m_1-m_2)\) |
| Sequences (M4) | Quantities evolving by constant addition or ratio | Arithmetic: \(u_n = u_0 + nr\); Geometric: \(u_n = u_0 r^n\) |
| Simple & compound interest (M5) | Savings, loans, investment comparisons | Simple: \(V_s = P(1+in)\); Compound: \(V_c = P(1+i/k)^{nk}\) |
| Functions & equations (M6) | Modelling, domain analysis, quadratic roots | Discriminant \(\Delta = b^2 - 4ac\); quadratic formula |
| Derivatives & optimisation (M7) | Finding the maximum or minimum of a polynomial | \(f'(x^*) = 0\); sign of \(a\) determines max or min |
4 Worked Example
Integrated problem — financing and break-even
A firm borrows €50,000 at a compound annual rate of 6% to cover its fixed launch costs. The loan is repaid after 3 years. The firm sells a product at €25/unit with variable costs of €10/unit.
Step 1 — Total repayment (Module 5).
\[ V_c = 50{,}000 \times (1.06)^3 = 50{,}000 \times 1.191016 \approx 59{,}550.80 \text{ €}. \]
The firm must repay approximately €59,551 — these become the effective fixed costs to cover over the product’s life.
Step 2 — Break-even quantity (Module 2).
\[ Q^* = \frac{59{,}550.80}{25 - 10} = \frac{59{,}550.80}{15} \approx 3{,}970 \text{ units.} \]
Step 3 — Interpretation. The firm must sell at least 3,970 units to recover the financed launch cost. Selling fewer means the interest cost alone makes the venture unprofitable.
Methods used: Module 5 (compound interest) to find the repayment, then Module 2 (break-even equation) to find the minimum sales volume.
5 Guided Practice
A company’s revenue (thousands of euros) is modelled by \(R(t) = -0.5t^2 + 8t + 10\), where \(t\) is the number of years since launch. The company’s annual cost is a flat €20,000.
- Find the year in which revenue is maximised.
- What is the maximum annual revenue (in €)?
- Is the company profitable in year \(t = 5\)? By how much?
For Q1 and Q2: use the derivative (Module 7) — compute \(R'(t)\), set it to zero, and check that it is a maximum (leading coefficient \(< 0\)). For Q3: convert to euros and compare with the cost.
- \(R'(t) = -t + 8 = 0 \Rightarrow t = 8\) years. Since \(a = -0.5 < 0\), this is a maximum.
- \(R(8) = -0.5(64) + 64 + 10 = -32 + 74 = 42\) thousand euros $= $ €42,000.
- At \(t = 5\): \(R(5) = -0.5(25) + 40 + 10 = 37.5\) thousand euros $= $ €37,500. Cost is €20,000. Profit $= 37{,}500 - 20{,}000 = $ €17,500 — profitable.
6 Interactive Exploration
The following problems are drawn from across Modules 1–7. Identify the appropriate mathematical method before expanding each problem.
Identify the method and module before expanding any problem.
Method: Successive percentage changes — Module 1
A winter jacket costs €180. The retailer applies a +20% mark-up for the premium season, then a −25% discount during the clearance sale.
Step 1 — Price after mark-up. \[180 \times 1.20 = 216 \text{ €}.\]
Step 2 — Price after discount. \[216 \times 0.75 = 162 \text{ €}.\]
Step 3 — Overall change. \[\Delta\% = \frac{162 - 180}{180} \times 100 = \frac{-18}{180} \times 100 = -10\%.\]
The final price is €162 — a 10% reduction despite a 25% nominal discount, because the mark-up raised the base first.
Method: First-degree equation / break-even — Module 2
A workshop producing custom furniture has fixed monthly costs of €4,200 and variable costs of €85 per item. Each piece sells for €220.
Break-even: \[Q^* = \frac{4{,}200}{220 - 85} = \frac{4{,}200}{135} \approx 31.1 \text{ items}.\]
The workshop must sell at least 32 items per month (rounding up to whole units) to be profitable.
Method: System of two linear equations / line intersection — Module 3
A delivery firm considers two van-rental contracts. Contract A: €80/day fixed plus €0.15/km. Contract B: €30/day fixed plus €0.40/km.
Let \(d\) = distance driven per day (km).
\[C_A(d) = 80 + 0.15d, \quad C_B(d) = 30 + 0.40d.\]
Intersection: \[80 + 0.15d = 30 + 0.40d \Rightarrow 50 = 0.25d \Rightarrow d = 200 \text{ km/day.}\]
For fewer than 200 km/day, Contract B is cheaper. For more than 200 km, Contract A is cheaper.
Method: Arithmetic sequence — Module 4
An employee starts at €32,000 and receives a fixed raise of €1,200 per year.
\[u_n = 32{,}000 + 1{,}200\,n.\]
After 7 years: \(u_7 = 32{,}000 + 8{,}400 = 40{,}400\) €.
Cumulative salary over 8 years (years 0 to 7): \[S = \frac{8 \times (32{,}000 + 40{,}400)}{2} = 4 \times 72{,}400 = 289{,}600 \text{ €}.\]
Method: Compound interest — Module 5
€6,000 is deposited in an account offering a compound annual rate of 3%, compounded semi-annually (\(k = 2\)), for 5 years.
\[V_c = 6{,}000 \times \left(1 + \frac{0.03}{2}\right)^{5 \times 2} = 6{,}000 \times (1.015)^{10} \approx 6{,}000 \times 1.1605 \approx 6{,}963 \text{ €}.\]
Interest earned: \(6{,}963 - 6{,}000 = 963\) €.
Method: Second-degree equation / discriminant — Module 6
A firm’s monthly revenue model is \(R(t) = -2t^2 + 14t - 12\) (thousand €), where \(t\) is the months of operation.
Find the months when revenue is zero: \[\Delta = 14^2 - 4(-2)(-12) = 196 - 96 = 100 > 0.\] \[t = \frac{-14 \pm 10}{-4} \Rightarrow t = 1 \text{ or } t = 6.\]
Revenue is positive (and the firm earns) during months 1 to 6 of operation (\(t \in ]1, 6[\)).
Method: Derivative and optimisation — Module 7
A firm’s profit function is \(P(q) = -0.3q^2 + 24q - 200\) (thousand €).
\[P'(q) = -0.6q + 24 = 0 \Rightarrow q^* = 40 \text{ units.}\]
Since \(a = -0.3 < 0\), this is a maximum. Maximum profit: \(P(40) = -0.3(1{,}600) + 24(40) - 200 = -480 + 960 - 200 = 280\) thousand euros.
Methods: Compound interest (Module 5) + derivative / optimisation (Module 7)
A fund manager invests €10,000 at a compound rate that grows at \(i(t) = -0.01t^2 + 0.12t\) per year for \(t\) years of commitment (where \(t \leq 6\) years is the maximum).
Step 1 — Optimal investment period (Module 7). Maximise \(i(t)\): \[i'(t) = -0.02t + 0.12 = 0 \Rightarrow t^* = 6 \text{ years.}\] Maximum rate: \(i(6) = -0.36 + 0.72 = 0.36 = 36\%\) per year.
Step 2 — Final value (Module 5). At the optimal commitment of 6 years with annual compounding: \[V_c = 10{,}000 \times (1 + 0.36)^6 \approx 10{,}000 \times 6.328 \approx 63{,}280 \text{ €}.\]
The 6-year commitment produces the highest rate and the largest accumulated value.
7 Applied Problem
A firm launches a new service. Its weekly subscriber count (in thousands) is modelled by \(N(t) = 2 \times 1.15^{t}\), where \(t\) is the number of weeks since launch.
- Identify the type of sequence and give the first term and common ratio.
- How many subscribers (in thousands) does the service have after 8 weeks?
- A rival service models its growth as \(M(t) = 5 + 3t\) (thousands). After how many weeks does the exponential service overtake the linear one? (Use trial or estimate.)
- What module’s methods are needed for each sub-question?
8 Common Errors
Students often apply the wrong method because they recognise a surface feature (e.g., “there is a percentage, so I use the percentage formula”) rather than the underlying mathematical structure.
Diagnostic test: Before choosing a method, ask: - Does the quantity change by a constant amount each period? → Arithmetic sequence (Module 4). - Does the quantity change by a constant factor each period? → Geometric sequence / compound interest (Modules 4–5). - Is there a quadratic and I need a maximum or minimum? → Derivative (Module 7). - Is there a quadratic and I need the values where it equals zero? → Discriminant / quadratic formula (Module 6). - Are there two linear relationships I need to equate? → System of equations / break-even (Modules 2–3).
Distinguishing Module 4 (sequences) from Module 5 (interest) is the most common confusion: both involve geometric growth, but Module 5 focuses on the financial interpretation (principal, rate, period).
9 Check Your Understanding
- A company’s revenue increases by 8%, then falls by 5%. What is the overall percentage change?
- A sequence has first term €1,000 and grows by €150 each period. What is the term after 12 periods? Is this arithmetic or geometric?
- €20,000 is invested at a compound rate of 4% for 6 years (annual compounding). What is the acquired capital?
- A profit function is \(P(x) = -2x^2 + 48x - 128\). Without fully solving, explain whether the optimum is a maximum or minimum and why.
- A student encounters the sentence “fixed costs are €5,000 and variable costs are €12 per unit”. Which module’s method is needed to find the break-even quantity if units sell for €20?
10 Summary
- Method selection is the core examination skill: no problem in this course tells you which formula to use — you must identify the mathematical structure (linear, quadratic, exponential, arithmetic, geometric) from the economic context.
- The seven methods connect: compound interest is a geometric sequence; the derivative finds the vertex of a parabola; the break-even is a first-degree equation — understanding these links lets you transfer methods across unfamiliar problems.
- Verify your answer against the context: a negative break-even quantity, a profit that is actually minimised, or an interest calculation that ignores compounding frequency are signals that a method was misapplied — always check that the result makes economic sense.
11 Independent Work
- Work through the eight challenge problems again without expanding the callouts — time yourself to simulate exam conditions.
- Attempt the practice examination from your course materials.
- Review the formula sheet (
formula-sheet.qmd) and confirm you can derive each formula from first principles. - Correct any remaining errors in your error log and identify which method produced each error.
- Review the learning guide (
learning-guide.qmd) for the suggested two-week revision schedule before the final exam.