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How do we find the best decision when choices are limited by constraints?

Formulate linear programming problems, graph feasible regions, and locate the optimal solution at a vertex of the feasible region.

How to turn a worded optimisation problem into an objective function and inequality constraints, graph the feasible region, and test corner points to maximise or minimise the objective.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Formulating the problem
Graphing the feasible region
The corner-point method
Interpreting the answer

What this dot point is asking

You must formulate a worded problem into an objective and constraints, graph the feasible region, and test corner points to optimise.

Formulating the problem

Start by defining the decision variables clearly, for example "let $x$ be the number of chairs and $y$ the number of tables".

Each "no more than" limit gives a $\le$ inequality; each "at least" requirement gives a $\ge$ inequality. Reading these correctly from the wording is where most marks are won or lost.

Graphing the feasible region

For each constraint, graph its boundary line (replace the inequality with equals), then shade the side that satisfies the inequality. The feasible region is the area satisfying all constraints simultaneously, usually a shaded polygon bounded by the axes and the constraint lines.

The corner-point method

Because the objective is linear, its value increases steadily across the region, so the maximum or minimum is reached at a corner of the feasible region.

Intersections are found by solving the two boundary equations simultaneously, usually by substitution or elimination.

Worked example

Minimising cost subject to constraints

A farmer mixes two feeds. Feed X costs $\$4$ /kg and supplies $2$ units of protein and $1$ unit of fibre per kg; Feed Y costs $\$3$ /kg and supplies $1$ unit of protein and $1$ unit of fibre per kg. The mix needs at least $8$ units of protein and at least $6$ units of fibre. Minimise cost.

Step 1: Define variables and objective

Naming the decision variables first makes every subsequent expression unambiguous. Let $x$ be kg of Feed X and $y$ kg of Feed Y. Minimise $C = 4x + 3y$ .

Step 2: Write the constraints

Each nutritional requirement becomes a "greater than or equal to" inequality because the farmer needs at least the stated amount. Protein: $2x + y \ge 8$ . Fibre: $x + y \ge 6$ . Plus $x \ge 0$ , $y \ge 0$ .

Step 3: Find the vertices of the feasible region

The feasible region is the set of points satisfying every constraint, so its corners are found where the boundary lines cross the axes or each other. The region is unbounded above, with corners at $(0,8)$ (on the $y$ axis where protein binds), $(6,0)$ (on the $x$ axis where fibre binds), and the intersection of $2x + y = 8$ and $x + y = 6$ . Subtracting: $x = 2$ , so $y = 4$ ; the intersection is $(2,4)$ .

Step 4: Test the corners in the objective function

Because the objective is linear, its minimum must occur at one of the corner points, so substituting each one identifies the optimum. $C(0,8) = 3(8) = 24$ . $C(6,0) = 4(6) = 24$ . $C(2,4) = 4(2) + 3(4) = 8 + 12 = 20$ .

Step 5: State the answer in context

The smallest cost value identifies the optimal mix; always express the answer using the original units and variables. The cheapest mix is $2$ kg of Feed X and $4$ kg of Feed Y, costing $\$20$ .

Interpreting the answer

Always state the solution in context: which quantities, and the optimal objective value. If the variables must be whole numbers (you cannot make half a chair), check whether the optimal vertex has integer coordinates and, if not, test the nearest feasible integer points.

A sliding-line (or "objective-line") method is an alternative to testing every corner: draw the objective function for one value, then slide it parallel across the feasible region. For a maximum, the optimum is the last vertex the line touches as it slides outward; for a minimum, the first it touches. This confirms the corner-point result and is handy when there are many vertices, but you still read the optimum off a vertex.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20228 marksA workshop makes

x

chairs and

y

tables. Each chair needs

2

hours of cutting and

1

hour of finishing; each table needs

1

hour of cutting and

3

hours of finishing. There are

40

cutting hours and

45

finishing hours available. Profit is

\

per chair and

50

per table. (a) Write the constraints. (b) Find the production that maximises profit.

Show worked answer →

Translate each resource limit into an inequality, then test the corner points.

(a) Constraints: $2x + y \le 40$ (cutting), $x + 3y \le 45$ (finishing), $x \ge 0$ , $y \ge 0$ . Objective: maximise $P = 30x + 50y$ . (3 marks)

(b) Vertices of the feasible region: $(0,0)$ , $(20,0)$ , $(0,15)$ , and the intersection of the two lines. Solving $2x + y = 40$ and $x + 3y = 45$ : from the first, $y = 40 - 2x$ ; substitute, $x + 3(40 - 2x) = 45$ , so $x + 120 - 6x = 45$ , $-5x = -75$ , $x = 15$ , $y = 10$ . Profits: $(20,0)\to 600$ , $(0,15)\to 750$ , $(15,10)\to 30(15)+50(10)=450+500=950$ . Maximum profit is $\$950$ at $15$ chairs and $10$ tables. (5 marks)

Markers reward both resource inequalities, finding the intersection vertex, and testing every corner to identify the maximum.

WACE 20244 marksExplain why the optimal solution of a linear programming problem always occurs at a vertex of the feasible region, and describe how you would find it once the region is graphed.

Show worked answer →

The objective is linear, so its value changes steadily across the region.

The objective function $P = ax + by$ has straight, parallel level lines. As the objective increases, these lines sweep across the feasible region, and the last point they touch before leaving the region is always a corner (vertex), so the optimum lies at a vertex. (2 marks)

To find it, identify the coordinates of every vertex (intersections of constraint boundaries and the axes), substitute each into the objective function, and choose the vertex giving the largest value for a maximum (or smallest for a minimum). (2 marks)

Markers reward the parallel-level-lines argument and the corner-point testing method.