Did you know that a problem with infinitely many possible solutions can sometimes be solved by testing just four points? Suppose a fictional classroom prepares two kinds of learning kits, A and B, with limited materials. There are many possible mixtures, but the linear programming method turns the restrictions into a feasible polygon and shows where to search for a maximum.
This Mathematics in Bukit Timah guide develops decision variables, simultaneous linear inequalities, feasible regions, objective functions, vertices, slack, graphical optimisation, equality cases, unboundedness and integer restrictions. The kit numbers, resources, planning goals and reward points are entirely fictional mathematical inputs, not eduKateSG fees, real school inventory or Bukit Timah operating data. We use exact numbers to explore a method, rather than pretending that an idealised result decides a real organisation’s plans.
Quick answer: Let x and y be nonnegative numbers of two fictional kit types. Assume resource limits 2x+y≤10 and x+2y≤8, and maximise a made-up score Z=3x+4y. The feasible polygon’s vertices are (0,0), (5,0), (4,2) and (0,4). Their scores are 0,15,20 and 16 respectively. The maximum is Z=20 at (x,y)=(4,2). It satisfies both resource constraints exactly.
Article contents
- The first question: what are we choosing?
- Bukit Timah is context, not a verified resource plan
- From words to a resource constraint
- Worked example 1: test a proposed kit mix
- Feasibility is not the same as optimality
- Define an objective function separately
- Worked example 2: a high score can be infeasible
- Nonnegativity creates the first two boundaries
- A linear inequality shades a half-plane
- Worked example 3: find axis intercepts
- The feasible region is an intersection
- Worked example 4: reject one-sided feasibility
- Find the intersection of the two resource lines
- Worked example 5: verify the common corner
- Four corner points define the feasible polygon
- Worked example 6: evaluate every vertex
- Why corners are enough for a linear objective
- A proof idea using weighted averages
- Worked example 7: score along a feasible edge
- The feasible region must be closed and bounded for a routine maximum
- What is a binding constraint?
- Worked example 8: calculate slack
- A resource can be unused at another corner
- Worked example 9: compare two feasible choices
- The objective line has a gradient
- Worked example 10: objective through the optimal point
- A parallel objective can create multiple optima
- Worked example 11: an entire optimal edge
- The equality case depends on the objective
- A minimisation task may choose the origin
- Worked example 12: add a fictional minimum-demand condition
- Some systems are infeasible
- Worked example 13: prove infeasibility with a bound
- An unbounded feasible region is different
- Worked example 14: demonstrate unboundedness
- Strict inequalities can affect attainment
- Worked example 15: no attained maximum on an open interval
- Integer constraints create a different problem
- Worked example 16: show a fractional relaxation gap
- Rounding a fractional optimum is not automatically valid
- Worked example 17: a rounded point can exceed capacity
- Resource sensitivity asks what changes if a limit changes
- Worked example 18: perturb the first resource
- Worked example 19: perturb the second resource
- Shadow values depend on the current optimum
- A dual bound offers another proof of optimality
- Worked example 20: check the multiplier proof
- Why multiplying constraints needs positive weights
- Worked example 21: an invalid negative-weight shortcut
- Constraints must be dimensionally consistent
- Worked example 22: reject a mismatched quantity
- Graphical optimisation requires correct axis scales
- Worked example 23: an axis-intercept verification
- Convexity explains why mixtures remain feasible
- Worked example 24: two feasible points and their midpoint
- A corner method does not solve every optimisation problem
- Worked example 25: a nonlinear counterexample
- Primary foundations: limited resources and arithmetic checks
- Lower Secondary: simultaneous equations and shaded inequalities
- Advanced extension: continuous versus integer optimisation
- A twenty-minute home diagnostic
- Seven days of practice and transfer
- Practice questions with checked answers
- Frequently asked questions
- Sources and related Mathematics in Bukit Timah
The first question: what are we choosing?
A decision variable represents a quantity the model can change. For the kit problem, x counts type-A kits and y counts type-B kits.
We initially let x and y be real and nonnegative so the geometry forms a continuous feasible region. If the quantities must be whole kits, a later integer restriction is necessary.
Before drawing anything, give every variable a clear meaning and a permitted domain. Otherwise the algebra may allow nonsensical negative quantities or fractional objects.
Bukit Timah is context, not a verified resource plan
Families near Sixth Avenue, Beauty World and King Albert Park may recognise decisions about limited time, materials and competing aims.
But nothing in this article identifies actual resources, timetables, fees or allocations used by any Bukit Timah school. The numbers have been chosen to demonstrate linear optimisation neatly.
In real planning, accuracy would depend on measured constraints, stakeholder priorities and assumptions not represented by two simple inequalities.
From words to a resource constraint
Suppose each type-A kit uses two imaginary units of material 1 and one of material 2.
Each type-B kit uses one unit of material 1 and two units of material 2. Total available amounts are ten units of material 1 and eight of material 2.
Material 1 use is 2x+y, so its inequality is 2x+y≤10. Material 2 use is x+2y, so x+2y≤8.
Worked example 1: test a proposed kit mix
Take x=3 and y=1.
Material 1 use is 2(3)+1=7≤10. Material 2 use is 3+2(1)=5≤8. Nonnegativity also holds.
Thus (3,1) is feasible. Its objective score is Z=3(3)+4(1)=13.
Feasibility is not the same as optimality
A feasible solution meets every stated restriction. An optimal solution is feasible and gives the largest or smallest objective value requested.
The mix (3,1) is valid, but it scores only thirteen, below the achievable twenty from (4,2).
A student who finds one combination that fits has solved a feasibility question, not necessarily an optimisation question.
Define an objective function separately
The fictional score is Z=3x+4y. Each type-A kit contributes three points, each type-B kit four points, by assumption.
A maximisation problem asks which feasible x,y make Z as large as possible.
These scores are not prices, educational outcomes or empirically measured values. They merely set the ranking rule for this particular mathematical exercise.
Worked example 2: a high score can be infeasible
Take x=5,y=2. The objective gives Z=15+8=23, which looks better than twenty.
But material 1 use is 2(5)+2=12>10, violating the first constraint. The choice is not feasible.
An objective value is not meaningful as an answer unless the corresponding decision satisfies all required inequalities.
Nonnegativity creates the first two boundaries
Since kit numbers cannot be negative in this model, include x≥0 and y≥0.
These conditions restrict the feasible region to the first quadrant of the coordinate plane.
If either variable represented signed displacement instead of a count, negative values might be meaningful. The domain follows from the variable’s definition, not from an automatic rule that x and y must always be positive.
A linear inequality shades a half-plane
The boundary 2x+y=10 is a straight line. Its inequality 2x+y≤10 selects the side containing points whose resource use does not exceed ten.
The origin (0,0) gives zero on the left, which satisfies ≤10, so the permitted half-plane includes the origin.
Test points are a useful way to choose the correct side of a line when drawing. The equality line itself is included because ≤ allows exact use of all ten units.
Worked example 3: find axis intercepts
For 2x+y=10, set y=0 to get x=5, giving (5,0). Set x=0 to get y=10, giving (0,10).
For x+2y=8, set y=0 to get x=8, giving (8,0). Set x=0 to get y=4, giving (0,4).
The constraint lines have different axis intercepts. Only the overlapping allowed regions matter for the final feasible set.
The feasible region is an intersection
A point must satisfy both resource inequalities and both nonnegativity conditions.
Therefore the permitted set is the intersection of four half-planes. It is not the union of points satisfying any one constraint.
If a point meets material 1’s limit but exceeds material 2’s, it remains infeasible. The diagram should show their overlap, not combine both shaded regions indiscriminately.
Worked example 4: reject one-sided feasibility
Choose x=0,y=5.
Material 1 use is 5≤10, so the first condition holds. But material 2 use is 2(5)=10>8.
Thus (0,5) is excluded despite meeting one resource limit. The largest feasible y on the y-axis is four.
Find the intersection of the two resource lines
Solve the simultaneous equalities
2x+y=10 and x+2y=8.
From the first, y=10−2x. Substitute into the second: x+2(10−2x)=8, so −3x=−12.
Hence x=4, y=2. This point uses both resources exactly and is a candidate corner.
Worked example 5: verify the common corner
At x=4,y=2:
Material 1: 2(4)+2=10. Material 2: 4+2(2)=8.
Both equal their limits. Since x,y are nonnegative, (4,2) is feasible and lies at the crossing of the two active boundary lines.
Four corner points define the feasible polygon
Along the x-axis, the first resource limit allows x≤5, while the second allows x≤8. The tighter limit is x≤5, giving corner (5,0).
Along the y-axis, the first allows y≤10 while the second allows y≤4, giving (0,4).
Together with the origin (0,0) and resource-line crossing (4,2), these are the four vertices of the feasible polygon.
Worked example 6: evaluate every vertex
For Z=3x+4y:
At (0,0): Z=0. At (5,0): Z=15. At (4,2): Z=12+8=20. At (0,4): Z=16.
The largest is 20 at (4,2). This is the exact continuous linear-programming optimum.
Why corners are enough for a linear objective
A bounded polygonal feasible region is convex. On any straight segment between feasible points, a linear objective changes linearly.
A value in the interior of a segment cannot exceed both endpoint objective values unless the objective is constant along that segment.
Consequently a bounded linear programme over a nonempty polygon has an optimal vertex. In special cases, an entire edge may also be optimal, but then its endpoints are optimal too.
A proof idea using weighted averages
Let points A and B be feasible and let C=(1−t)A+tB for 0≤t≤1 lie on the segment joining them.
For a linear objective Z, Z(C)=(1−t)Z(A)+tZ(B).
This is a weighted average of endpoint objective values, so it lies between them. Thus no interior point on that segment can beat both ends.
Worked example 7: score along a feasible edge
Consider the edge joining (5,0) and (4,2). Write a point on it as (5−t,2t), 0≤t≤1.
Objective is Z=3(5−t)+4(2t)=15+5t.
It increases from 15 at t=0 to 20 at t=1. Along this edge, the best endpoint is (4,2).
The feasible region must be closed and bounded for a routine maximum
In our toy problem, x≥0, y≥0 and resource limits restrict x≤5, y≤4. The feasible region is closed and bounded.
Therefore its linear objective attains a maximum. Not every system of inequalities has these properties.
An unbounded feasible region may permit an objective to grow forever. An infeasible model may have no permitted point at all.
What is a binding constraint?
A constraint is binding or active at a solution when it holds as equality.
At optimum (4,2), both resource inequalities are binding: 2x+y=10 and x+2y=8.
At a different feasible point, one or both resources may be unused. Identifying active constraints helps explain why the optimum occurs where it does.
Worked example 8: calculate slack
At (3,1), material 1 used=7 out of 10, leaving slack three.
Material 2 used=5 out of 8, also leaving slack three.
At optimum (4,2), both slacks are zero. Slack means unused capacity in the exact mathematical model, not physical waste inferred from real operations.
A resource can be unused at another corner
At (5,0), material 1 use is 10, so slack zero.
Material 2 use is 5, so slack is 8−5=3.
This corner is feasible but not optimal for the chosen score because allocating some capacity differently can increase Z while respecting both constraints.
Worked example 9: compare two feasible choices
For (5,0), Z=15 and the second resource has unused capacity three.
For (4,2), Z=20 and both resources are fully used.
The higher objective comes from a different mixture, not simply from using the maximum possible number of type-A kits.
The objective line has a gradient
For fixed score Z=c, solve 3x+4y=c for y:
y=c/4−(3/4)x.
Every such line has slope −3/4. Increasing c translates the line parallel upward in the y-intercept direction.
Graphically, slide the objective line toward larger c while it still touches the feasible region. Its last contact occurs at an optimal vertex or edge.
Worked example 10: objective through the optimal point
At (4,2), Z=20.
The objective level line is 3x+4y=20, or y=5−(3/4)x. Substituting (4,2) gives 2=5−3.
This line touches the feasible polygon at the optimal point and no feasible point has a larger Z.
A parallel objective can create multiple optima
Suppose the fictional score changes to W=2x+y, exactly the left-hand expression of material 1’s constraint.
The resource limit gives W≤10 for every feasible point.
Every feasible point along the edge where 2x+y=10 attains W=10, not just one vertex. The endpoints (5,0) and (4,2) are examples.
Worked example 11: an entire optimal edge
On the edge joining (5,0) and (4,2), write (x,y)=(5−t,2t), 0≤t≤1.
Then W=2(5−t)+2t=10 for every t.
Every point of that edge is optimal for maximising W. If x,y must be integers, only the integer points on that edge remain candidates.
The equality case depends on the objective
The original score Z=3x+4y favours (4,2) uniquely among the feasible vertices.
The alternative score W=2x+y has infinitely many continuous optima along one edge.
The constraints did not change. The goal did. An optimisation answer is meaningful only in relation to its stated objective.
A minimisation task may choose the origin
If the objective is to minimise Z=3x+4y with the same constraints and x,y≥0, the minimum is zero at (0,0).
That may be mathematically correct but unhelpful for a real goal requiring a minimum production level.
To model such a goal, one must add an explicit demand constraint. The mathematics cannot infer an unstated requirement.
Worked example 12: add a fictional minimum-demand condition
Suppose we add x+y≥3. Then (0,0) is no longer feasible.
Among the feasible points, minimising a positive cost 3x+4y would favour the cheaper coefficient of x when possible. The point (3,0) is feasible: resource 1 use six, resource 2 use three, demand three, cost nine.
Because every unit counted toward x+y≥3 costs at least three, no feasible choice can have cost below nine. Thus (3,0) is optimal for this modified model.
Some systems are infeasible
Consider adding the demand x+y≥9 to our original resource constraints.
The original feasible polygon has x+y values 0,5,6 and 4 at its corners. Since x+y is linear, its maximum over the polygon is six.
No feasible point can satisfy x+y≥9. The revised problem has no feasible solution, regardless of how strongly the objective rewards production.
Worked example 13: prove infeasibility with a bound
Add the two resource inequalities:
(2x+y)+(x+2y)≤10+8.
This gives 3x+3y≤18, hence x+y≤6.
If another condition requires x+y≥9, it contradicts x+y≤6. The model is infeasible without any detailed graph.
An unbounded feasible region is different
Suppose the only constraints are x≥0, y≥0 and x+y≥2.
Maximise Z=x+y. For any feasible value of Z, choose a larger positive x while keeping y=0; the objective grows without limit.
The model is feasible but has no finite maximum. This is not the same as being infeasible.
Worked example 14: demonstrate unboundedness
Take points (2,0),(10,0),(100,0), all satisfying x+y≥2.
Their objective values x+y are 2,10 and 100.
There is no largest attainable value because the positive x-coordinate can increase indefinitely. The solution is “unbounded above,” not an invented finite number.
Strict inequalities can affect attainment
A condition x<5 excludes the boundary x=5 even if a candidate optimum would otherwise occur there.
If a feasible set is not closed, an objective may approach a bound but never attain it. In that case the bound is a supremum, not an achieved maximum.
Most introductory graphical linear programmes use ≤ or ≥ boundaries, but strict inequalities deserve separate care.
Worked example 15: no attained maximum on an open interval
Let 0≤x<5 and maximise Z=x.
The value can get arbitrarily close to five, such as 4.9,4.99,4.999, but never equal five.
Therefore Z has supremum five but no maximum attained on that domain. Writing x=5 would violate the strict condition.
Integer constraints create a different problem
In a physical interpretation, kit counts should generally be whole numbers.
A continuous linear programme allows fractional x,y, while an integer linear programme restricts variables to integers.
Our original optimum (4,2) is already an integer pair, so it remains valid under that additional condition. But this coincidence is not guaranteed in every model.
Worked example 16: show a fractional relaxation gap
Let x,y≥0 with 2x+2y≤5, and maximise x+y.
For real variables, the maximum is 2.5 along x+y=2.5. For nonnegative integer variables, x+y must be an integer at most 2.5.
Therefore integer maximum is 2, not 2.5. The continuous answer is an upper bound on the integer optimum.
Rounding a fractional optimum is not automatically valid
Suppose a continuous optimum is (1.7,2.3). Rounding it to (2,2) might violate one of the original constraints.
Even if rounding yields a feasible pair, it need not be the best integer choice.
Integer optimisation requires checking feasible lattice points or an appropriate integer algorithm. Rounding is not a theorem.
Worked example 17: a rounded point can exceed capacity
Let x,y≥0 with x+y≤3.5. The continuous boundary point (1.8,1.7) is feasible.
Rounding each coordinate up gives (2,2), whose sum is four, exceeding 3.5.
The rounded pair is infeasible. A student’s final answer must satisfy the original inequalities exactly.
Resource sensitivity asks what changes if a limit changes
Suppose material 1 availability rises slightly from ten to 10+δ while material 2 stays at eight.
If the same two resource lines remain active at the optimum, solve 2x+y=10+δ and x+2y=8.
The intersection shifts. The objective may increase, but the amount depends on which resource is relaxed and whether the active-corner structure remains valid.
Worked example 18: perturb the first resource
Solve the altered equalities:
2x+y=10+δ; x+2y=8.
The solution is x=4+(2/3)δ, y=2−(1/3)δ.
The fictional score becomes Z=3x+4y=20+(2/3)δ. So for sufficiently small changes keeping this corner feasible and optimal, one extra unit of material 1 raises the best score by about 2/3.
Worked example 19: perturb the second resource
Now keep material 1 at ten but raise material 2 to 8+δ.
The active intersection becomes x=4−(1/3)δ, y=2+(2/3)δ.
Objective is Z=20+(5/3)δ under the same local-active-set assumption.
The resource changes have different marginal effects because their constraints and objective coefficients differ.
Shadow values depend on the current optimum
The preceding marginal coefficients 2/3 and 5/3 are sometimes called shadow prices in linear programming duality.
They describe objective changes for small resource-limit perturbations when the relevant optimum structure remains unchanged.
They are not actual market prices or eduKateSG resource costs. Beyond a certain range, another vertex may become optimal and the marginal effect can change.
A dual bound offers another proof of optimality
From the constraints, take nonnegative multipliers 2/3 and 5/3:
(2/3)(2x+y)+(5/3)(x+2y)≤(2/3)10+(5/3)8.
The left side simplifies to 3x+4y, exactly our objective. The right side simplifies to 20.
Therefore every feasible point has Z≤20. Since (4,2) attains twenty, this directly proves global optimality without enumerating vertices.
Worked example 20: check the multiplier proof
Compute the coefficient of x: (2/3)×2+(5/3)×1=4/3+5/3=3.
Compute the coefficient of y: (2/3)×1+(5/3)×2=2/3+10/3=4.
Thus the weighted resource inequalities produce 3x+4y≤20. The bound is exact at (4,2), confirming the optimum.
Why multiplying constraints needs positive weights
If a≤b and λ≥0, then λa≤λb. Adding inequalities with nonnegative weights preserves their direction.
Multiplying by a negative weight reverses the inequality, so an arbitrary weighted sum could be invalid.
The nonnegative multipliers in the preceding dual bound are mathematically significant, not decorative fractions.
Worked example 21: an invalid negative-weight shortcut
From 2x+y≤10, multiplying by −1 gives −2x−y≥−10, not ≤−10.
Any proof that uses −2x−y≤−10 as though it followed from the original inequality is invalid.
A linear-programming bound must preserve inequality direction at every algebraic step.
Constraints must be dimensionally consistent
A resource inequality should add terms measured in the same type of resource unit.
In the toy model, 2x+y represents units of material 1, while x+2y represents units of material 2. Their right sides are different capacities.
An objective point score is not automatically a resource quantity. The coefficients carry stipulated meanings, even when all are represented by bare numbers in classroom algebra.
Worked example 22: reject a mismatched quantity
Suppose x counts kits while y measures minutes. Writing 2x+y≤10 would add two kits to a number of minutes without a conversion or model coefficient providing consistent units.
The inequality is not meaningful as a physical constraint in that form.
A sound word problem assigns coefficients to convert each decision variable into the same resource quantity before addition.
Graphical optimisation requires correct axis scales
A sloped resource line may look steeper or flatter if the x and y axes are drawn at different physical scales.
The algebraic gradient and intercepts come from the equation, not from guessing a line’s apparent angle on an uneven sketch.
Mark the axis labels, units and exact intercept coordinates. For proof of the optimum, use calculated vertices rather than pixel measurements.
Worked example 23: an axis-intercept verification
For 2x+y=10, x-intercept is five and y-intercept ten.
For x+2y=8, x-intercept eight and y-intercept four.
A sketch suggesting that the first line crosses the y-axis at five instead of ten has been drawn incorrectly. Recompute the intercept before shading any feasible region.
Convexity explains why mixtures remain feasible
If points A and B satisfy linear inequalities, every weighted average (1−t)A+tB with 0≤t≤1 also satisfies them.
This makes a linear-inequality feasible region convex: line segments between its points remain inside.
Convexity supports the vertex method and helps explain why arbitrary holes do not appear in a region defined only by simultaneous non-strict linear half-plane constraints.
Worked example 24: two feasible points and their midpoint
Both (0,4) and (4,2) are feasible in our model.
Their midpoint is (2,3). Check material 1: 2(2)+3=7≤10. Material 2: 2+2(3)=8≤8.
It is feasible, consistent with convexity. Its objective score is 3(2)+4(3)=18, between the endpoint scores 16 and 20.
A corner method does not solve every optimisation problem
If the objective is nonlinear, its maximum need not occur at a vertex of a feasible polygon.
For example, over 0≤x≤2, the concave function f(x)=−(x−1)² reaches its maximum at interior x=1, not endpoints.
The corner-point theorem belongs to linear objectives over suitable polyhedral feasible sets, not arbitrary curved objectives.
Worked example 25: a nonlinear counterexample
Maximise f(x)=−(x−1)² on [0,2].
At both endpoints, f(0)=f(2)=−1. At x=1, f(1)=0, which is larger.
Thus the maximum lies inside. Applying the linear-programming vertex rule to this nonlinear function would fail.
Primary foundations: limited resources and arithmetic checks
Younger pupils can begin with a table of hypothetical kits, resource use and remaining amounts.
They can test whether combinations fit without solving graphs. Fractions, multiplication and clear word meanings come first.
The goal is to understand constraints as rules that every valid choice must obey.
Lower Secondary: simultaneous equations and shaded inequalities
Students can solve two linear equations, draw boundary lines and distinguish intersections from unions of feasible regions.
The kit example reinforces that a boundary crossing is a candidate, not automatically the optimum until the objective is evaluated.
Repair sign errors, graph scales and nonnegativity misconceptions before assigning more complex optimisations.
Advanced extension: continuous versus integer optimisation
A-level, university and operations-research learners can study duality, shadow prices and integer programming algorithms.
The SEAB 2027 G3 Mathematics syllabuses and SEAB 2027 A-Level subject syllabuses provide official course context.
This guide’s detailed dual and integer examples are enrichment, not a claim that every advanced technique belongs in a Secondary Mathematics assessment.
A twenty-minute home diagnostic
Minutes 1–5: Define x,y and the conditions 2x+y≤10, x+2y≤8, x,y≥0.
Minutes 6–10: Find line intercepts, then solve the intersection (4,2). Explain why a point satisfying only one resource condition is invalid.
Minutes 11–15: Evaluate Z=3x+4y at the four vertices and obtain maximum twenty.
Minutes 16–20: Change objective to W=2x+y and identify the entire optimal edge. Ask why the optimum ceases to be unique even though constraints stay the same.
Seven days of practice and transfer
Day one translates words into inequalities. Day two plots boundary lines. Day three shades half-plane intersections.
Day four calculates feasible vertices. Day five evaluates linear objectives and compares ties. Day six tests infeasible and unbounded examples.
Day seven adds an integer restriction and asks for a checked feasible optimum, explaining why ordinary rounding of a continuous answer is insufficient.
Practice questions with checked answers
A. Is (3,1) feasible under 2x+y≤10 and x+2y≤8? Answer: yes.
B. Is (5,2) feasible? Answer: no; first resource use is twelve.
C. Intersection of 2x+y=10 and x+2y=8? Answer: (4,2).
D. Objective 3x+4y at (4,2)? Answer: 20.
E. Maximum objective over the original feasible polygon? Answer: 20.
F. Maximise W=2x+y. Answer: 10 along the feasible boundary edge from (5,0) to (4,2).
G. Add x+y≥9. Does a feasible point remain? Answer: no.
H. For integer x,y≥0 under 2x+2y≤5, maximise x+y. Answer: 2.
Frequently asked questions
What is linear programming?
It is optimisation of a linear objective subject to linear equalities or inequalities and stated domains.
What is a feasible region?
The set of all points satisfying every constraint simultaneously.
Why do corner points matter?
A bounded feasible polygon with a linear objective has at least one optimal vertex, even when an entire edge is also optimal.
Can several solutions be equally best?
Yes. An objective line parallel to an active boundary edge can make every point of that edge optimal.
Is a high score enough to make a choice valid?
No. The choice must satisfy all constraints before its score is relevant to the optimisation.
What is slack?
Unused capacity in a ≤ resource constraint at a specified feasible point.
Is the continuous optimum always the integer optimum?
No. Integer constraints can exclude fractional continuous solutions; rounding can violate feasibility.
What if the feasible region is empty?
The problem is infeasible and has no solution satisfying all restrictions.
Do these scores describe eduKateSG revenue or actual resources?
No. Every numerical resource and objective score is a fictional mathematical example.
How can parents recognise mastery?
The learner can define variables, identify the overlapping feasible region, test all vertices, preserve constraints and explain why the result is a global optimum.
Sources and related Mathematics in Bukit Timah
Use the SEAB 2027 G3 syllabuses directory and 2027 A-Level Mathematics syllabuses to distinguish current school topics from advanced operations-research enrichment. The resources, scores and decisions here are wholly invented.
Continue with Mathematics in Bukit Timah: Linear Inequalities and Feasible Regions, Simultaneous Equations, AM–GM Inequality and Area Optimisation and Graph Colouring and Timetable Conflicts for different mathematical optimisation questions.
The immutable Clementi Secondary 1 Mathematics tutoring reference anchors accessible explanations, carefully checked worked examples and diagnosis of the earliest incorrect assumption.
The lasting habit is to state the decision and its domain, satisfy every constraint, locate feasible vertices, evaluate the objective, and check whether integer, strict or additional conditions change the final answer.
Explore the other three Mathematics in Bukit Timah guides in this batch: Eigenvalues, eigenvectors and repeated matrix transformations · Information entropy, binary bits and prefix-code efficiency · Sudoku, constraint satisfaction and backtracking. The four-part series explores constraints, linear transformations, uncertainty and systematic problem-solving, with checked worked examples and clear explanations of mathematical assumptions.