Find and prove optimal solutions
Separate feasibility from optimality and prove a minimum or maximum over all permitted choices.
Original StudyForA scenarios and formative model answers; no Cambridge marks, predicted grades or certified mastery. Alternative reasoned answers may be valid. Full-length unseen papers, a wider mixed-problem bank and calibrated specialist teacher marking remain outstanding.
Before you begin
- Read tables and instructions in English
- Use arithmetic, ratios and whole-number constraints
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Before calculating, what should you identify?
What you will learn
- Compare all feasible cases using an objective
- Prove that no permitted alternative improves the result
Define the objective and allowed choices
An optimum is the best feasible value, not merely a good value. State whether you minimise price, time or waste; those objectives can conflict. Specify whether extra capacity is allowed and whether packages may be repeated before deciding the search bounds.
Seat six people using individual tickets at 9 credits or a four-person pass at 30 credits. Passes and tickets may be bought repeatedly and unused places are allowed. Zero passes cost 54; one pass plus two singles costs 48; two passes cost 60. Three or more passes cost at least 90. The minimum is 48.
Use a complete comparison or bound
A feasible plan gives an upper bound for a minimisation problem; to prove optimality, exclude cheaper plans. A capacity bound plus an attaining construction can prove a maximum. If equal best cases exist, report them all when asked.
‘Cheapest per seat’ does not guarantee cheapest total: a large pass may waste seats. Search the number of passes, filling remaining demand with singles, and check price at the final feasible integer choices.
Sources for this lesson
Pause and explain
Does one feasible low-cost plan prove it is cheapest?
Worked example
Find the minimum cost of six seats under the ticket rules.
Show the worked solution
- Zero passes cost 6×9=54.
- One pass plus two singles costs 30+18=48.
- Two passes cost 60; any additional pass costs still more.
Answer The minimum is 48 credits, attained by one four-person pass and two individual tickets. Every pass-count case that could beat it has been checked.
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.
Guided: Explain why buying two passes is not optimal for six people.
Give me a hint
Compare total cost, not price per seat.
Compare my reasoning
- Two passes cost sixty.
- One pass and two singles cost forty-eight.
- Both provide at least six places.
Two passes are feasible but cost 12 credits more than the 48-credit plan. Their spare capacity does not compensate for the extra price when minimising total spending.
Look for these in your work
- Checks feasibility
- Compares total costs
- Uses the correct objective
Independent: Under the same prices and rules, find the minimum cost for eight people.
Give me a hint
Check zero, one and two passes.
Compare my reasoning
- Zero passes cost 72.
- One pass plus four singles costs 66.
- Two passes cost 60; three or more exceed 60.
The minimum is 60 credits for two passes. The zero- and one-pass cases cost more, and any greater pass count starts at ninety.
Look for these in your work
- Enumerates relevant cases
- Finds the attaining plan
- Excludes improvement
Application: For six people, the pass price changes to 38 credits. Find the new optimum.
Give me a hint
Re-evaluate all relevant pass counts.
Compare my reasoning
- Zero passes still cost 54.
- One pass plus two singles costs 56.
- Two passes cost 76, and more cannot help.
Six individual tickets now minimise cost at 54 credits. The old optimum is no longer best because the pass price changed; a previous answer cannot simply be retained.
Look for these in your work
- Updates changed prices
- Compares alternatives
- States the new optimum
Common mistakes
- Confusing feasible with optimal
- Choosing by unit price alone
- Ignoring allowed unused capacity
What can you explain now?
Does one feasible low-cost plan prove it is cheapest?
Compare with the explanation
No, cheaper permitted cases must be excluded
Optimality compares the plan against the full allowed set.
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Make it stick
Does one feasible low-cost plan prove it is cheapest?
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.