Compare designs using capacity, material and cost constraints
Check feasibility before comparing cost · Optimise a simple geometric quantity over its permitted domain · State whether a minimum is global or among listed candidates
Section Paper 2 · Historical Extended · non-calculator
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Cuboid volume and open-box area
- Quadratic completed squares
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Check feasibility before comparing cost
- Optimise a simple geometric quantity over its permitted domain
- State whether a minimum is global or among listed candidates
Write the requirements first
For an open box with base l by w and height h, capacity is lwh and ideal sheet area is lw+2lh+2wh. Before choosing a cheaper design, check required capacity, allowed dimensions and what surfaces exist. A thinner or smaller design is not a valid saving if it violates a requirement.
Comparing two candidate boxes establishes the better of those candidates, not a minimum over every possible box. A real manufacturing comparison may also need thickness, folds and waste; use the assumptions actually supplied.
Prove an optimum within the domain
A rectangular enclosure using 36 m of fencing on all four sides has length 18−x when width is x, with 0<x<18. Area x(18−x)=81−(x−9)² is at most 81 m², attained at x=9. This proves the global maximum for this ideal fixed-perimeter rectangle model, without calculus. Different constraints, such as one unfenced side, change the equation.
Sources for this lesson
Pause and explain
Can the cheaper of two feasible boxes be declared the cheapest possible box without considering other designs?
Worked example
Compare open boxes A: 6×6×2 cm and B: 9×4×2 cm. Both must hold at least 72 cm³. Ignore thickness and waste.
Show the worked solution
- Both capacities are 72 cm³.
- Areas: A=36+24+24=84 cm²; B=36+36+16=88 cm².
Answer Both feasible; A uses 4 cm² less sheet among these two candidates.
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.
For the two open boxes in the example, check capacities and compare ideal sheet areas.
Give me a hint
Use lwh and lw+2lh+2wh separately.
Compare my reasoning
- Both capacities=72 cm³.
- Areas=84 and 88 cm², so A uses 4 cm² less.
Both hold 72 cm³; A uses 84 cm² versus B's 88 cm².
Look for these in your work
- Check feasibility
- Compare the requested material
Use 36 m of fencing for all four sides of a rectangle. Find its maximum possible area, with positive side lengths.
Give me a hint
Complete the square in x(18−x).
Compare my reasoning
- Area=81−(x−9)²≤81.
- Equality occurs at x=9, with the other side also 9 m.
Maximum 81 m² at a 9 m by 9 m square.
Look for these in your work
- State the permitted domain
- Prove the bound and its attainment
Sheet costs 0.02 currency units per cm². Boxes must hold at least 72 cm³. Find the costs of A and B above and their cost difference. A third open box C is 6×6×1 cm: can its lower cost justify choosing it? Does comparing these candidates prove a global minimum?
Give me a hint
Check capacity as well as material area.
Compare my reasoning
- A costs 84×0.02=1.68; B costs 88×0.02=1.76, a difference of 0.08.
- C has volume 36 cm³ and area 60 cm², costing 1.20; it fails the 72 cm³ requirement. Only the listed designs were compared.
A 1.68; B 1.76; difference 0.08. C costs 1.20 but holds only 36 cm³, so it is infeasible; no global-minimum claim.
Look for these in your work
- Reject a cheaper infeasible option
- Limit the conclusion to the evidence
Common mistakes
- A cheaper infeasible design is not an answer to the constraints.
- A candidate comparison alone does not establish a global optimum.
What can you explain now?
Can the cheaper of two feasible boxes be declared the cheapest possible box without considering other designs?
Compare with the explanation
No; the comparison covers only those candidates.
A global claim needs evidence over the full permitted design set.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Can the cheaper of two feasible boxes be declared the cheapest possible box without considering other designs?
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.