Calculus and Optimisation
Mathematics
Calculus and Optimisation
Syllabus tag: KASNEB CPA | Foundation Level | CA15 Quantitative Analysis | Topic 3 Calculus and Optimisation
Lesson objectives
By the end of this topic, you will be able to:
- Differentiate polynomial functions
- Interpret a derivative as a marginal quantity
- Find turning points and classify them
- Maximise profit and minimise cost
- Apply integration to find total from marginal
Why this matters
Differentiation gives the rate of change of one quantity with respect to another — which in business is exactly what "marginal" means. Marginal cost is the derivative of total cost; marginal revenue is the derivative of total revenue.
Rules of differentiation
| Rule | Form | Derivative |
|---|---|---|
| Power | y = axⁿ | dy/dx = anxⁿ⁻¹ |
| Constant | y = c | dy/dx = 0 |
| Sum | y = f(x) + g(x) | f'(x) + g'(x) |
For y = 5x³ − 4x² + 7x − 12:
dy/dx = 15x² − 8x + 7
Note the constant vanishes. A fixed cost of 40,000 does not affect marginal cost, which is exactly right — it does not change when one more unit is made.
Marginal functions
Marginal cost = dTC/dQ Marginal revenue = dTR/dQ
Where TC = 200Q + 40,000, marginal cost is 200 — constant, because the total cost function is linear.
Where TR = 1,200Q − 2Q², marginal revenue is 1,200 − 4Q.
Compare that with the price function. If TR = PQ and P = 1,200 − 2Q, then MR falls twice as fast as price — the 4Q against 2Q. Selling one more unit means accepting a lower price on every unit, not just the last one, which is why MR declines faster.
Turning points
At a maximum or minimum the first derivative is zero. The second derivative classifies it:
| Second derivative | Type |
|---|---|
| Negative | Maximum |
| Positive | Minimum |
| Zero | Test further |
A candidate who finds dy/dx = 0 and stops has done half the work. Stating the second derivative test earns marks and is frequently required explicitly.
Profit maximisation
TR = 1,200Q − 2Q² and TC = 200Q + 40,000.
Method 1 — set MR = MC:
1,200 − 4Q = 200, so 4Q = 1,000 and Q = 250
Method 2 — differentiate the profit function:
Profit = TR − TC = −2Q² + 1,000Q − 40,000 dπ/dQ = −4Q + 1,000 = 0, giving Q = 250 d²π/dQ² = −4, negative, confirming a maximum
Both give the same answer, because MR = MC is the condition that the derivative of profit is zero.
| At Q = 250 | Working | KES |
|---|---|---|
| Price | 1,200 − 2(250) | 700 |
| Total revenue | 700 × 250 | 175,000 |
| Total cost | 200(250) + 40,000 | 90,000 |
| Maximum profit | 85,000 |
:::checkpoint A candidate computes Q = 250 by setting the derivative to zero and concludes profit is maximised. Explain what step is missing and why the examiner will deduct marks for its absence. :::
Break-even by quadratic
Profit is nil where −2Q² + 1,000Q − 40,000 = 0.
Using the quadratic formula with a = −2, b = 1,000, c = −40,000:
Discriminant = 1,000² − 4(−2)(−40,000) = 1,000,000 − 320,000 = 680,000
Q = 43.84 and Q = 456.16
There are two break-even points. Below 43.84 units the fixed costs are not covered; above 456.16 units the falling price has eroded the margin away. Profit is positive only between them, and is maximised at 250 — close to the midpoint, as it must be for a symmetrical quadratic.
Minimisation: the economic order quantity
Total inventory cost = ordering + holding = (DCo / Q) + (QCh / 2)
Differentiating and setting to zero gives the familiar result:
EOQ = √(2DCo / Ch)
With D = 24,000, Co = 500 and Ch = 15:
EOQ = √(2 × 24,000 × 500 / 15) = 1,264.91 units
The second derivative is positive, confirming a minimum — which is what the CA25 EOQ curves show graphically. Calculus and the diagram give the same answer by different routes.
Integration
Integration reverses differentiation: it recovers total from marginal.
∫axⁿ dx = axⁿ⁺¹ / (n + 1) + c
Where MC = 200 + 6Q, total cost is:
TC = 200Q + 3Q² + c, and the constant c is the fixed cost, recoverable only if a total cost at some output is known.
That constant is the examinable point. Marginal information alone can never determine fixed costs, because differentiation destroyed them.
:::checkpoint A firm's marginal cost is 150 + 4Q and total cost at 100 units is KES 60,000. Find the fixed cost, and explain why the marginal function alone was insufficient. :::