Optimise a practical quantity with derivatives
Model a constrained practical quantity as a function, find its stationary value and justify whether it is a maximum or minimum.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Why should a practical domain be stated before optimising?
What you will learn
- Express a practical quantity as a function of one variable
- Use a derivative to find a candidate optimum
- Use a derivative test and contextual domain to justify the result
Use the constraint to reduce the model to one variable
Translate the practical condition into an equation, such as a fixed perimeter. Substitute it into the objective quantity so area, cost or another target is expressed using one variable.
Record the physical domain: lengths must be positive and may have a finite upper bound. A stationary point outside this interval cannot be the practical optimum.
Differentiate and justify the chosen extremum
Solve A′(x) = 0 for interior candidates, then use A″ or a first-derivative sign test to classify each candidate. Compare endpoints too if the allowed domain includes them.
State the resulting dimensions or settings and substitute them back into the original context. A stationary value alone does not explain what should be built or chosen.
Pause and explain
If A″(x) < 0 at an interior stationary point, what does that show?
Worked example
A rectangle has perimeter 40 m. Find the dimensions that give the greatest area.
Show the worked solution
- Let one side be x; the other is 20 − x, so 0 < x < 20.
- A(x) = x(20 − x) = 20x − x². Then A′(x) = 20 − 2x, so the stationary input is x = 10.
- A″(x) = −2 < 0, so this is a maximum; the other side is also 10 m and the area is 100 m².
Answer A square 10 m by 10 m gives the maximum area 100 m².
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A rectangle next to a river needs fencing on three sides. There are 24 m of fencing. Find the dimensions for maximum area.
Give me a hint
Let x be each side perpendicular to the river, so the third side is 24 − 2x.
Compare my reasoning
- A(x) = x(24 − 2x) = 24x − 2x², with 0 < x < 12.
- A′(x) = 24 − 4x = 0 gives x = 6 and the third side 12.
- A″(x) = −4 < 0, so the area is maximised.
Perpendicular sides 6 m each; fenced side 12 m; maximum area 72 m².
Look for these in your work
- Constraint used
- Stationary dimensions found
- Maximum justified and context restored
A wire 30 cm long forms a rectangle. Find the dimensions for maximum area.
Give me a hint
Use 2x + 2y = 30 to express y in terms of x.
Compare my reasoning
- y = 15 − x, so A(x) = 15x − x² for 0 < x < 15.
- A′ = 15 − 2x = 0 gives x = 7.5 and y = 7.5.
- A″ = −2 < 0, so this is the maximum; A = 56.25 cm².
A square of side 7.5 cm gives maximum area 56.25 cm².
Look for these in your work
- Perimeter condition applied
- Interior stationary point found
- Second derivative and area checked
An open rectangular display has fixed area 48 m². Express its perimeter P in terms of width x, find the stationary value, and justify whether it minimises or maximises P.
Give me a hint
Use length 48/x, with x > 0, before differentiating.
Compare my reasoning
- P(x) = 2x + 2(48/x) = 2x + 96/x.
- P′(x) = 2 − 96/x² = 0 gives x² = 48; since x > 0, x = 4√3 and length is also 4√3.
- P″(x) = 192/x³ > 0, so the stationary perimeter is a minimum.
Minimum perimeter occurs for a square: width and length 4√3 m; P = 16√3 m.
Look for these in your work
- Area constraint substituted
- Positive domain used
- Minimum justified with second derivative
Common mistakes
- Optimising before using the constraint
- Ignoring the physically allowed domain
- Reporting an unclassified stationary value
- Giving the variable value without translating it back to dimensions
What can you explain now?
What equation identifies an interior stationary candidate for A(x)?
Compare with the explanation
A′(x) = 0
A stationary value has zero first derivative.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How do a constraint, a one-variable model, a derivative test and the practical domain establish an optimum?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.