A mathematically correct answer can still be the wrong decision if it violates the conditions of the problem. This K310 workshop develops constraint reasoning: inequalities, domains, integer restrictions, bounds, capacity and feasibility.
It follows Vol 0070 and extends the threshold work in Vol 0067: Break-Even and Sensitivity. Use the G3 SEC learner hub for the wider route.
For 2027 school candidates, use the official K310 G3 Mathematics syllabus and SEAB G3 syllabus directory for assessment requirements. Every price, capacity and scheduling case below is invented teaching material.
A solution can be algebraically correct and still be unusable
Feasibility asks whether a mathematically valid result satisfies the conditions of the real problem. Negative lengths, fractional buses, capacities beyond a stated maximum and times outside the operating window are examples of results that can emerge from correct algebra but fail the context.
Constraints define the allowed region
A constraint limits the values a variable may take. It may be explicit, such as x ≥ 0, or implied by context, such as a number of students being a whole number. Strong K310 reasoning keeps these conditions visible from the start rather than checking them only at the end.
Domain is part of the model
The domain tells you which input values make sense. A formula may be valid only for 0 ≤ t ≤ 60, or a plan may apply only up to 100 units. Solving outside that domain can produce mathematically consistent but irrelevant answers.
Integer constraints matter
Counts of buses, boxes, people and tickets generally require whole numbers. If an equation gives 17.4 buses, the operational answer depends on the task. If 17.4 buses are needed to carry everyone, 18 may be required; if the question asks how many full buses can be filled, 17 may be the relevant count.
Non-negativity is a common hidden constraint
Time, physical length and many counts cannot be negative. An algebraic equation may have two roots, but only one may satisfy the physical meaning. Rejecting an impossible root is part of solving, not an optional afterthought.
Upper bounds matter too
A tank cannot hold more than its capacity; a room may have a maximum occupancy; a probability cannot exceed 1. Constraints often define both a lower and an upper limit. Feasible solutions sit inside the allowed interval.
Worked case 1: bus capacity
A school needs to transport 137 students. Each bus holds 40 students. 137 ÷ 40 = 3.425. The calculation is correct, but 3.425 buses is impossible. Four buses are required if every student must be transported.
Worked case 1: explain the rounding direction
Rounding to the nearest whole number gives 3, which fails the capacity requirement. The context demands enough buses, so the answer must be rounded up. This is why mathematical interpretation decides the rounding rule.
Worked case 2: boxes and leftovers
A warehouse packs 137 items into boxes holding 40 each. Three full boxes contain 120 items and 17 remain. If the question asks for full boxes only, the answer is 3; if it asks for the number of boxes needed to hold all items, the answer is 4.
Same arithmetic, different target
The bus and box examples use the same division but different response wording. Feasibility is not a fixed rounding rule. It comes from the target: enough capacity, full groups, remaining items or nearest estimate.
Inequalities describe feasible regions
If a room has capacity 120 and x students attend, x ≤ 120. If at least 30 participants are required, x ≥ 30. Together, 30 ≤ x ≤ 120. The feasible region is an interval, not one number.
Worked case 3: budget inequality
A club has at most $500. A venue costs $140 plus $18 per participant. If n is the number of participants, 140 + 18n ≤ 500. Solving gives 18n ≤ 360, so n ≤ 20. Since n is a non-negative integer, feasible participant counts are 0 through 20 unless another lower bound is stated.
Worked case 3: interpretation matters
The algebra n ≤ 20 is only useful after the variable is defined. “20” is not the budget; it is the maximum number of participants under the model. Strong solutions state the contextual meaning.
Worked case 4: minimum target
A fundraiser needs at least $900. Each ticket contributes $12 after fixed costs have already been accounted for. If n tickets are sold, 12n ≥ 900, giving n ≥ 75. Because tickets are discrete, 75 is the minimum whole-number count that meets the target.
Open and closed inequalities differ
x < 10 excludes 10, while x ≤ 10 includes it. In real contexts, wording such as less than, at most, more than and at least changes the boundary. Read the condition before choosing the symbol.
Compound conditions can narrow the answer sharply
A candidate may need a score of at least 60 but less than 80. The feasible interval is 60 ≤ x < 80. If the score must be an integer, the possible values are 60 through 79.
Worked case 5: rectangle feasibility
A rectangle has perimeter 30 cm. Let length be x and width be 15 − x. For both dimensions to be positive, 0 < x < 15. The algebraic expression for area, x(15 − x), should only be interpreted over this domain.
An expression can exist outside the physical model
The quadratic x(15 − x) can be evaluated at x = 20, but the corresponding width is −5 cm. The algebraic value exists; the rectangle does not. Domain filters mathematical outputs through meaning.
Worked case 6: probability constraint
If a calculation gives a probability of 1.2, something is wrong because probability must lie between 0 and 1 inclusive. Feasibility checking can therefore diagnose an earlier error even before the exact source is found.
Worked case 7: time window
A service operates between 08:00 and 18:00. A model predicts an arrival 11.5 hours after 08:00, which is 19:30. The calculation may be correct, but the service would be closed. The decision must account for the operating window.
Bounds can be explicit
If a measurement is given as 12.4 cm correct to the nearest 0.1 cm, the true value lies from 12.35 cm up to but not including 12.45 cm under the usual interpretation. Bounds preserve what the rounding statement actually tells you.
Bounds can be operational
A lift has a stated maximum load. A design may require a safety margin below that maximum. The feasible region may therefore be stricter than the physical capacity. Use the condition given by the task rather than inventing a safety rule.
Worked case 8: lower and upper bounds in a product
If a rectangle’s measured length and width each have bounds, the minimum possible area uses the lower bounds and the maximum possible area uses the upper bounds, provided the quantities are positive. This is a reasoning task about monotonicity, not just substitution.
Feasibility and optimisation are different
Feasibility asks which options are allowed. Optimisation asks which allowed option best meets a criterion. First identify the feasible set, then compare within it. A mathematically optimal point outside the feasible region is not a valid decision.
Worked case 9: cheapest feasible plan
Plan A costs less but handles only up to 80 participants. Plan B costs more but handles up to 120. For 95 participants, A is infeasible regardless of price. The decision must filter by capacity before comparing cost.
Constraints should be checked before optimisation
When a problem includes capacity, time and budget, check whether an option satisfies all of them. Do not optimise one criterion while ignoring another condition. Multi-constraint problems reward disciplined filtering.
Worked case 10: scheduling constraints
A workshop needs 90 minutes and must finish by 16:30. If setup takes 20 minutes, the activity must begin by 14:40 at the latest. A proposed 15:00 start is infeasible even if the venue is otherwise available.
Graphical feasibility
On a graph, inequalities can define a shaded region. Every point in the region satisfies the constraints. A decision point outside the region is invalid even if it looks attractive by another criterion.
Boundary lines matter
A solid boundary usually represents inclusion, while a dashed boundary can represent exclusion in a graph of inequalities, depending on notation. The learner should connect the graphical convention to the inequality symbol.
Intersection of constraints
If x ≥ 2 and x ≤ 7, the feasible set is the overlap [2,7]. If x ≥ 8 is added, the feasible set becomes empty. An empty feasible set means the stated constraints cannot all be satisfied simultaneously.
An empty feasible set is an answer
Learners sometimes assume every word problem must have a practical solution. If the constraints conflict, “no feasible solution” can be the correct conclusion. The response should explain which conditions make the set empty.
Worked case 11: impossible budget and minimum quality
A plan requires at least 10 premium units costing $60 each, but the budget is $500. Even before considering other costs, 10 × 60 = 600 exceeds the budget. No feasible plan satisfies both conditions.
Constraint redundancy
A constraint can be redundant if it does not narrow the feasible set. If x ≥ 10 and x ≥ 5 are both given, x ≥ 5 adds no new restriction. Recognising redundancy can simplify complex conditions.
Independent constraints add information
If x ≥ 10 and x ≤ 20, both matter. Neither can be removed without enlarging the feasible set. Strong reasoning distinguishes genuinely independent restrictions from repeated versions of the same one.
Worked case 12: age and capacity filters
Suppose participants must be at least 16 years old and there are at most 30 places. Age eligibility and capacity are independent constraints. One controls who may enter; the other controls how many can be accepted.
Piecewise rules create different feasible regions
A pricing rule may change after a threshold. The learner should use the correct formula for the relevant region. Solving with one branch outside its stated interval can produce an invalid answer.
Worked case 13: piecewise delivery fee
A fictional delivery fee is $8 for orders under $50 and free for orders of $50 or more. If the total before delivery is $49.50, adding $8 gives $57.50; the free-delivery rule is not triggered because the condition is based on the pre-delivery order value.
Piecewise models require condition checking
Each branch should be checked against the value that caused it to be selected. A solution obtained from the high-range formula must actually lie in the high-range domain. Otherwise the branch assumption is inconsistent.
Feasibility after a break-even result
Vol 0067 finds thresholds where models are equal. This workshop adds another question: are those thresholds allowed? A break-even at 120 units is irrelevant if one option is only available up to 80.
Near-boundary answers need careful interpretation
A value of 79.9 may be infeasible when the condition is integer x ≤ 79, but feasible when x is a continuous quantity bounded below 80. The variable type matters as much as the number.
Rounding can violate a bound
If x ≤ 12.4 and a calculation gives x = 12.36, rounding to 12.4 may make the reported value look feasible even though the exact value is already within the bound. Reporting rules and feasibility checks should not be confused.
Worked case 14: floor and ceiling thinking
A capacity problem often uses a ceiling idea: round up to ensure enough units. A complete-groups problem may use a floor idea: count only whole groups that fit. You do not need formal floor or ceiling notation to reason correctly about the context.
Worked case 15: negative root in geometry
A quadratic length problem yields x = 5 or x = −3. The negative root is rejected because a physical length in the model cannot be negative. State the reason rather than silently discarding the value.
Worked case 16: root outside a stated interval
A model is valid only for 0 ≤ t ≤ 20, but solving gives t = 7 and t = 32. Keep t = 7 and reject t = 32 for this model. The second root may solve the equation but lies outside the permitted domain.
Check feasibility before finalising units
A result may have the correct unit but still violate context. 150 people is a count with the right unit-like label, yet it is impossible if the room capacity is 120. Units and constraints check different aspects of correctness.
Constraint reasoning can expose copied-value errors
If a computed speed exceeds a stated maximum by a huge amount, inspect whether a time or distance was copied incorrectly. Constraints provide plausibility checks that can reveal earlier transcription mistakes.
Constraint reasoning can expose wrong formulas
A probability above 1 or an area larger than the enclosing region can signal that the chosen relationship is wrong. Feasibility is a diagnostic tool, not only a final filter.
Independent task A
A hall holds at most 280 people. Twelve seats are reserved and unavailable. How many additional attendees can be admitted? State the inequality and maximum feasible count.
Independent task B
A school needs at least 23 vans, but a parking area can hold at most 20. What does the feasible set tell you? Write the conclusion without trying to average the two limits.
Independent task C
A box holds 18 items. How many boxes are needed for 250 items? Then answer a different question: how many full boxes can be packed if no extra box is opened? Explain the different rounding.
Independent task D
A rectangle has perimeter 40 cm and length x. Write the width and the feasible domain for x if both dimensions must be positive.
Independent task E
A model gives p = 1.08 for a probability. State what the result tells you about the calculation before locating the exact algebraic error.
Independent task F
Two plans intersect at 140 units, but Plan A is offered only up to 100. Explain why the algebraic intersection does not create a usable break-even decision.
Worked feedback A
Available capacity is 280 − 12 = 268. If n is additional attendees, n ≤ 268 with n a non-negative integer. The maximum feasible value is 268.
Worked feedback B
The conditions require n ≥ 23 and n ≤ 20, so no value satisfies both. The feasible set is empty. The correct conclusion is that the current parking arrangement cannot meet the stated van requirement.
Worked feedback C
250 ÷ 18 ≈ 13.89. Fourteen boxes are needed to hold all items. Only thirteen full boxes can be packed without opening another, with 16 items left. The target decides whether the reasoning rounds up or counts completed groups.
Worked feedback D
If length is x, width is 20 − x. Both must be positive, so 0 < x < 20. Values outside this interval do not describe a physical rectangle in the model.
Worked feedback E
Probability above 1 is impossible, so the result exposes a model, arithmetic or interpretation error. Do not simply cap the answer at 1; find the reason the calculation violated the probability constraint.
Worked feedback F
The intersection lies outside Plan A’s availability domain. Within the feasible range, compare the plans directly. A threshold only matters when both options exist at that quantity.
Repair route
Begin with one-variable inequalities and integer contexts. Require the learner to state the allowed set in words. Then add domain restrictions and contextual rounding.
Stabilisation route
Mix continuous and discrete variables so the learner cannot use one automatic rounding rule. Ask what kind of quantity x represents before any calculation.
Extension route
Use two or three constraints, piecewise rules and empty feasible sets. Ask the learner to filter options before optimising cost, time or another criterion.
What progress should look like
Progress is visible when learners define domains early, reject impossible roots with reasons, handle integers correctly, recognise empty feasible sets and check whether algebraic thresholds are operationally available.
Frequently asked: should I always round up counts?
No. Round up when the task requires enough whole units to meet demand. Count down when the question asks how many complete groups fit. The context determines the direction.
Frequently asked: is every negative answer impossible?
No. Negative temperatures, coordinates and changes can be meaningful. Reject a negative value only when the variable’s definition makes it impossible, such as a physical length or number of people.
Frequently asked: do I need inequalities for every word problem?
No. Use them when the task involves a range, minimum, maximum or feasibility region. A direct equality may be sufficient for other questions.
Frequently asked: can a problem have no solution?
Yes. Conflicting constraints can produce an empty feasible set. State that clearly and explain which restrictions cannot be satisfied simultaneously.
Final operating rule
Define the variable, write the constraints, solve the mathematical relationship, then filter the result through domain, discreteness, capacity and context. A defensible K310 answer is not only mathematically valid; it is feasible.
K310 feasibility checklist
- define the variable and unit
- write explicit and implied constraints
- identify whether the variable is continuous or discrete
- solve the relationship
- reject values outside the domain
- test integer values around a boundary
- check capacity and operational conditions
- state the feasible conclusion in context
Continue with Vol 0072: Science — Measurement Uncertainty and Anomalies Workshop after this Mathematics workshop.
Advanced K310 feasibility laboratory
Advanced case: feasible does not mean optimal
An option may satisfy every constraint without being the best by the stated criterion. First filter infeasible options; only then compare cost, time or another objective. Mixing feasibility and optimisation too early can produce a low-cost answer that cannot actually be used.
Advanced case: one constraint can dominate
Suppose three options all satisfy budget and time, but only one meets capacity. Capacity becomes the decisive filter. The learner should identify which constraint actually removes alternatives instead of treating every condition as equally influential.
Advanced case: a redundant constraint adds no new restriction
If x ≥ 12 is already required, adding x ≥ 5 changes nothing. Recognising redundancy can simplify a crowded problem. But do not remove a condition merely because it looks similar; check whether it changes the feasible set.
Advanced case: two constraints can conflict silently
A model might require x ≥ 30 for quality and x ≤ 24 for budget. Solving the equations separately can hide the conflict. Write the feasible ranges together and inspect their overlap. No overlap means no feasible solution.
Advanced case: piecewise rules need branch validation
If a tariff uses one formula for n ≤ 50 and another for n > 50, a solution from the second branch is valid only if it actually exceeds 50. Always check a piecewise solution against the condition that selected the branch.
Advanced case: capacity plus minimum order
A supplier may require at least 20 units while storage holds at most 32. The feasible set is 20 ≤ n ≤ 32. If demand is 18, the order rule—not storage—creates the operational problem. Context determines which constraint matters.
Advanced case: integer boundary after inequality
If an inequality gives n > 14.2 and n counts people, the minimum feasible n is 15. If it gives n ≥ 14.2, the minimum integer is also 15. The exact inequality matters most when the boundary itself is an integer.
Advanced case: strict boundary at an integer
If n > 14 and n is an integer, 14 is excluded and 15 is the minimum. If n ≥ 14, then 14 is allowed. Small symbol differences can change the operational answer by one whole unit.
Advanced case: feasibility on a graph
A point can satisfy one inequality but fail another. Graphical feasibility is determined by the overlap of all shaded regions. The learner should test a candidate point in every constraint, especially when boundaries are close.
Advanced case: corner points are not automatically answers
In optimisation problems, corner points can be important, but only if the syllabus task and model make that approach relevant. Do not memorise a rule that every feasible-region problem is solved by choosing a corner without understanding the objective.
Advanced case: a bound can expose impossible precision
If an input is rounded, a final answer with many decimal places may suggest more precision than the data support. Bounds help determine what can genuinely be concluded. Reporting extra digits does not create information.
Advanced case: measurement bounds and feasibility
Suppose a shelf length is 120 cm correct to the nearest centimetre and a unit must fit entirely within it. A calculated item length near the boundary may require using the appropriate lower or upper bound, depending on whether you are proving fit or proving non-fit.
Advanced case: lower bound for capacity
If a container’s capacity is given as 10.0 L correct to the nearest 0.1 L, the smallest possible actual capacity is 9.95 L. A plan requiring exactly 10.0 L may not be guaranteed feasible under worst-case interpretation.
Advanced case: upper bound for clearance
If an object must pass through an opening, compare the object’s largest possible dimension with the opening’s smallest possible dimension when the question asks whether fit is guaranteed. Bound direction follows the decision being tested.
Advanced case: a feasible average can hide infeasible individuals
An average load per vehicle may fall below capacity while one vehicle still exceeds it. Feasibility may need item-level distribution, not just a mean. Do not translate a group average into a guarantee about every member.
Advanced case: total budget versus category caps
A project may satisfy the total budget but violate a category limit, such as no more than $200 on equipment. Multi-constraint decisions require all conditions, not only the headline total.
Advanced case: percentage constraints
A rule may require at least 60% of participants to belong to a target group. If 18 of 32 qualify, the proportion is 56.25%, so the condition fails even though 18 is a substantial count. Percentage feasibility depends on the denominator.
Advanced case: threshold changes when denominator changes
If 18 target participants remain but total participation falls from 32 to 28, the proportion rises above 60%. The count is unchanged; feasibility changes because the denominator changed. This links constraints to careful percentage reasoning.
Advanced case: simultaneous time and capacity
A shuttle holds 12 people and can make at most 4 trips before closing. Its maximum service capacity is 48 people. If 52 require transport, adding one more passenger per trip would still be insufficient unless the model permits that capacity change.
Advanced case: infer the bottleneck
When several constraints exist, ask which one is binding at the chosen solution. A binding constraint is active at the boundary. Identifying it helps explain why small changes to that condition can alter the feasible decision.
Advanced case: slack shows room to move
If the budget is $500 and a feasible plan costs $430, there is $70 of slack in the budget constraint. Slack does not automatically mean the whole plan is robust, because another condition may already be at its limit.
Advanced case: robustness of feasibility
A plan barely inside every constraint is more fragile than one with comfortable margins. In modelling discussion, compare how close the solution sits to important boundaries. Do not invent uncertainty unless the task asks for it, but recognise margin when evaluating robustness.
Advanced case: impossible root plus feasible root
A quadratic context may produce two algebraic roots, one negative and one positive. Reject the negative only if the variable definition makes it impossible. State that reason so the solution remains mathematically complete.
Advanced case: both roots can be feasible
In some geometry or timing contexts, two positive roots may both satisfy the domain. Do not discard one simply because many school examples have a single practical answer. Test both against every condition.
Advanced case: neither root may be feasible
A quadratic can yield two real roots that both lie outside the stated domain. The correct contextual conclusion can be no feasible solution even though the equation itself has solutions.
Advanced case: constraints before calculator precision
A calculator may return 19.999999 due to numerical representation. If the exact algebra establishes 20 and the domain includes it, use the exact reasoning. Do not let display artefacts override the model.
Advanced case: constraints after approximation
If an approximate numerical method gives a value near a boundary, follow the accuracy and interpretation required by the task. A rounded display can cross a strict inequality if the underlying value is not checked carefully.
Workshop drill: write the feasible set first
For each context, write the allowed values before solving the main equation. This habit often reveals impossible roots immediately and makes later interpretation easier.
Workshop drill: classify the variable
Mark the variable as continuous, integer count, probability, length, time or another type. Different variable types carry different implied constraints. This one label can prevent automatic rounding errors.
Workshop drill: test one candidate against every condition
After finding a potential solution, substitute it into the original constraints. This catches values that solve an equation but violate a capacity, time or domain condition.
Workshop drill: explain rejection explicitly
Do not write only “reject x = −3.” Write “reject x = −3 because x represents a physical length and must be positive.” The explanation demonstrates contextual reasoning.
Workshop drill: create an empty feasible set
Take two compatible constraints and change one until they no longer overlap. Explain the exact point at which feasibility disappears. This builds intuition for conflicting conditions.
Workshop drill: add a redundant constraint
Add a condition that does not change the feasible set. Explain why it is redundant. This helps distinguish information that truly narrows the problem from information that merely repeats an existing boundary.
Workshop drill: identify the binding constraint
For a chosen feasible point, calculate how close each constraint is to its limit. Name the one at equality or nearest to violation. This supports sensitivity reasoning without requiring advanced optimisation.
Workshop drill: change one constraint
Increase capacity, tighten the budget or shift a minimum requirement. Predict how the feasible set should expand or shrink before recalculating. Direction-of-change reasoning catches inequality errors.
Feasibility and sensitivity belong together
Vol 0067 asks how thresholds move when assumptions change. This workshop asks whether the resulting values remain allowed. A robust mathematical decision often requires both: where is the boundary, and is the chosen region actually feasible?
Feasibility and source translation belong together
Vol 0069 asks what a source means. Constraints often hide in ordinary wording such as at least, no more than, within, must fit and only if. Translating these phrases into mathematical conditions is the first step toward feasibility.
Feasibility and checking belong together
Final checking should include the original constraints, not only arithmetic. A correct equation can still answer the wrong operational question. Ask whether the result can exist, fit, occur or be selected under every stated condition.
Final feasibility standard
The skill is secure when the learner can define the allowed set before calculation, solve the mathematical relationship, reject or adjust values with explicit reasons, and communicate the final decision in the language of the problem.
