To learn Optimisation with Differentiation quickly and correctly, begin by asking a practical question: what is the largest or smallest value allowed by the conditions? A rectangle might need the greatest possible area for a fixed perimeter, or a cost formula might need its lowest possible value. Differentiation finds where a smoothly changing quantity stops increasing and starts decreasing, but you must also understand the constraints and check boundaries. These eduKateSG Mathematics Quick Notes teach how to turn an ordinary word problem into one function, differentiate it and verify the maximum or minimum.
- Find the objective: state exactly what must be maximised or minimised.
- Define one variable: choose a length, time or quantity and give its units.
- Use the constraint: express other quantities in terms of the chosen variable.
- Build a function: area, volume, cost or another target.
- Find candidates: solve derivative = 0 and consider any points where the derivative is undefined.
- Check the domain and boundaries: a stationary point is not automatically the required optimum.
- Conclude in context: write the actual maximum or minimum and the measurements or units.
First, practise identifying what may change and what stays fixed. Imagine a rectangular garden with a fixed perimeter of 40 metres. Its length l and width w can vary, but 2l + 2w = 40 must remain true. Divide by 2 to obtain l + w = 20. If the width is x metres, the length is 20 − x metres. The garden must have positive dimensions, so 0 < x < 20. The constraint is not an optional detail; it converts a problem with two changing quantities into one variable.
Second, build the quantity you want to optimise. The garden’s area is length × width, giving A(x) = x(20 − x) = 20x − x² square metres. Differentiate: A′(x) = 20 − 2x. The derivative is positive for x < 10, zero at x = 10, and negative for x > 10. So the area increases until x = 10 and then decreases. The only stationary candidate in the allowed interval is x = 10. Its corresponding length is 20 − 10 = 10, forming a square.
Third, verify that the candidate is really the optimum. At x = 10, the area is 10 × 10 = 100 m². The second derivative is A″(x) = −2, which is negative, confirming a local maximum. The domain and quadratic shape also show this is the greatest attainable area: as x approaches either 0 or 20, the rectangle becomes very narrow and the area approaches zero. The conclusion must name the dimensions as well as the area: a 10 m by 10 m square has the maximum area of 100 m² for the stated fixed perimeter.
Worked example: a rectangular garden with a 30 m perimeter
A rectangle has perimeter 30 m. Let its width be x, so its length is 15 − x. Then A(x) = x(15 − x) = 15x − x² for 0 < x < 15. Differentiate: A′ = 15 − 2x. Set to zero: x = 7.5 m, and the length is also 7.5 m. Area = 7.5² = 56.25 m². Since A″ = −2, it is a maximum. Both dimensions remain positive and satisfy the original perimeter: 2(7.5 + 7.5) = 30 m.
Worked example: minimise a quadratic cost
Suppose a mathematical cost model is C(x) = x² − 8x + 25 dollars, for a real decision variable x ≥ 0. Differentiate: C′(x) = 2x − 8. Solve C′ = 0 to obtain x = 4. Since C″ = 2 > 0, this stationary point is a minimum. The minimum cost is C(4) = 16 − 32 + 25 = $9. You can check without calculus by completing the square: C(x) = (x − 4)² + 9, which can never be smaller than 9.
What a stationary point means
A stationary point is a differentiable point at which the derivative equals zero. A function may flatten at a local maximum, a local minimum or a stationary point of inflection. Thus f′(x) = 0 finds candidates rather than automatically certifying the final answer. For example, f(x) = x³ has derivative 3x² and stationary point x = 0, yet the function keeps increasing on both sides. To classify a stationary point, inspect the derivative’s sign, a conclusive second-derivative result or the broader function shape.
Use the sign of the first derivative
- Positive to negative: the function changes from rising to falling, giving a local maximum.
- Negative to positive: the function changes from falling to rising, giving a local minimum.
- No sign change: the stationary point may not be a maximum or minimum.
- Derivative undefined: the point may still be an optimum candidate if it belongs to the domain.
- Boundary points: endpoints can provide the greatest or least value even when the derivative is not zero.
Worked example: a closed time interval
A toy’s height above a reference level is modelled by h(t) = −t² + 6t metres for 0 ≤ t ≤ 6 seconds. Differentiate to get h′(t) = −2t + 6. The stationary time is t = 3 seconds. The height there is h(3) = −9 + 18 = 9 m. Check the endpoints: h(0) = 0 and h(6) = 0. Therefore the maximum over the stated interval is 9 m at t = 3 s. The endpoints are essential in a closed-interval optimisation problem.
How to handle a fixed area instead
A rectangle must enclose 36 m² and you wish to minimise its perimeter. Let one side be x > 0; the other must be 36/x. Then the perimeter is P(x) = 2x + 72/x. Differentiate: P′(x) = 2 − 72/x². Set P′ = 0: x² = 36. Since x > 0, x = 6 m. The other side is 36/6 = 6 m, giving perimeter 24 m. The second derivative 144/x³ is positive for x > 0, and the perimeter increases without bound near either end of the positive domain. So the square provides the minimum.
Why a maximum and a minimum require a domain
Optimisation never takes place without conditions, even if the question hides them in words. A length cannot be negative, the number of whole objects may need to be an integer, and a physical process may exist only during a particular time interval. In the garden example, x = 30 would produce a length of −10 m, which is not a valid garden. In an exam answer, state 0 < x < 20 or the appropriate interval before using calculus. Otherwise you might accept an algebraic candidate that cannot happen.
When the best value is at an endpoint
Consider f(x) = x² on the restricted interval 1 ≤ x ≤ 3. The derivative f′ = 2x is never zero inside this interval, yet the function still has a smallest and greatest value there: f(1) = 1 and f(3) = 9. The values are attained at the boundaries. This simple example shows why “differentiate and set to zero” is only one part of a complete optimisation strategy. For a continuous function on a closed interval, check stationary points, any relevant non-differentiable points and endpoints.
Second derivative: a useful but limited test
If f′(a) = 0 and f″(a) > 0, then the function has a strict local minimum at a. If f″(a) < 0, it has a strict local maximum. If f″(a) = 0, the test is inconclusive, not evidence that there is no optimum. For instance, f(x) = x⁴ has f′(0) = 0 and f″(0) = 0, yet x = 0 is a minimum. You can inspect the sign of the first derivative or the expression itself to classify it. No single derivative shortcut replaces checking the function.
A reliable seven-step optimisation plan
- Write one sentence naming the quantity to maximise or minimise.
- Define the variable and its permitted domain.
- Use every constraint to form the objective as a function of one variable.
- Differentiate and solve f′(x) = 0 for interior stationary candidates.
- Consider any relevant points where the derivative is undefined.
- Check the derivative sign or curvature and evaluate relevant boundaries.
- State the answer using the problem’s measurements, units and real-world restrictions.
Common optimisation mistakes
- Optimising the wrong quantity: distinguish area from perimeter, and cost from profit.
- Leaving two independent variables: use the constraint to reduce the expression when possible.
- Ignoring an impossible negative length: state the domain before solving.
- Calling every stationary point a maximum: verify the type.
- Forgetting endpoints: boundaries may contain the true global optimum.
- Giving only x when asked for dimensions and maximum area: return to the story.
- Rounding too early: keep accurate intermediate values, especially near constraints.
Mini-test: work from the constraint
- A rectangle has perimeter 40 cm. Find the dimensions that maximise area.
- A rectangle has perimeter 30 cm. Find its maximum area.
- Find the minimum value of C(x) = x² − 8x + 25 for real x.
- On 0 ≤ t ≤ 6, find the maximum of h(t) = −t² + 6t.
- On 1 ≤ x ≤ 3, find the minimum and maximum of f(x) = x².
Answers with the decisive step
1. Width x, length 20 − x; A′ = 20 − 2x = 0 gives 10 cm by 10 cm, area 100 cm². 2. A = x(15 − x); stationary point x = 7.5, so area 56.25 cm². 3. Minimum at x = 4, giving C = 9. 4. Stationary time t = 3; maximum 9 at that time, greater than both endpoint heights. 5. Endpoint values give minimum 1 at x = 1 and maximum 9 at x = 3.
Your twelve-minute optimisation study routine
- Minutes 1–2: highlight the objective, fixed constraint and variable limits.
- Minutes 3–5: make the objective a function of one variable.
- Minutes 6–8: differentiate and locate candidate stationary values.
- Minutes 9–10: check derivative signs and relevant endpoints.
- Minutes 11–12: substitute into the original story and state the actual maximum or minimum with units.
The fastest correct optimisation method begins with understanding the story. The derivative only becomes useful after you know what is being optimised and which values are allowed. Define, model, differentiate, compare and explain the result.
Continue the Mathematics Quick Notes learning pathway
For connected learning, read Differentiation · Quadratic Equations · Mathematics Word Problems · Integration. Try a changed example without looking at the worked solution and revisit it tomorrow.
For the longer teaching approach, see Secondary 1 Mathematics Tutor Clementi | Small Groups Tutorials. This immutable reference remains unchanged.