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

Formulate and solve linear programming problems graphically to optimise an objective function.

Defining variables, writing constraints, graphing feasible regions, and using the corner-point method to maximise or minimise an objective in TCE Mathematics Applications.

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

Linear programming (LP) is a method for making the best decision when resources are limited. Every LP problem has the same skeleton: things you control (variables), a quantity to optimise (objective), and limits you cannot break (constraints).

Formulating the problem

Read the problem and define $x$ and $y$ precisely, including units. Each resource limit becomes one inequality. Almost every LP problem also includes the non-negativity constraints $x \ge 0$ and $y \ge 0$ , because you cannot make a negative number of items.

Graphing the feasible region

Each inequality defines a half-plane. Graph its boundary line (treat the inequality as an equation), then shade the side that satisfies it. Test a point such as $(0,0)$ : if it makes the inequality true, shade that side. The feasible region is where all the shaded half-planes overlap; every point in it satisfies every constraint at once.

Solving by corner points

Find the coordinates of each corner (vertex), often by solving the two boundary lines that meet there simultaneously. Substitute each corner into the objective function and compare the values.

Maximising profit

A workshop makes tables and chairs subject to limited cutting and finishing hours. We formulate the constraints, find the corners of the feasible region, and test each corner to identify the production mix that maximises profit.

Let $x$ be the number of tables and $y$ the number of chairs produced. The objective to maximise is $P = 60x + 40y$ .

Step 1: Write the constraints

Each resource limit becomes one inequality. Tables need 2 cutting hours, chairs need 1 cutting hour, and there are 40 hours of cutting available:

2x + y \le 40

Each table and each chair both use 1 finishing hour, and there are 30 finishing hours available:

x + y \le 30

Because negative production is impossible, we also require $x \ge 0$ and $y \ge 0$ .

Step 2: Find the corner points

The corner where each constraint boundary meets an axis is found by setting the other variable to zero. On the $x$ -axis ( $y = 0$ ), the cutting constraint gives $2x = 40$ , so $x = 20$ : corner $(20, 0)$ . On the $y$ -axis ( $x = 0$ ), the finishing constraint gives $y = 30$ : corner $(0, 30)$ . The origin $(0, 0)$ is always a corner for non-negativity problems.

To find where the two boundary lines intersect, subtract $x + y = 30$ from $2x + y = 40$ :

2x + y - (x + y) = 40 - 30 \implies x = 10

Substituting back into $x + y = 30$ gives $y = 20$ . The interior corner is $(10, 20)$ .

Step 3: Evaluate the objective at each corner

The corner-point principle guarantees the maximum lies at one of these vertices, so we test all four:

Corner	$P = 60x + 40y$
$(0,0)$	$0$
$(20,0)$	$1200$
$(0,30)$	$1200$
$(10,20)$	$60(10) + 40(20) = 600 + 800 = 1400$

Step 4: State the optimal solution

The largest value of $P$ is $1400$ , occurring at $(10, 20)$ .

Final answer: Maximum profit is $\$1400$ , achieved by producing 10 tables and 20 chairs.

The sliding-line method

An alternative is to draw the objective line $P = 60x + 40y$ for any convenient value of $P$ , then slide it parallel to itself in the direction of increasing $P$ . The last corner it touches before leaving the feasible region is the optimum. This gives the same answer as testing corners and is a good visual check.

Finding the corner coordinates

The arithmetic that trips students up is finding where two boundary lines cross. Treat each binding constraint as an equation and solve the pair simultaneously, usually by elimination. For example, $2x + y = 40$ and $x + y = 30$ subtract to give $x = 10$ , and back-substitution gives $y = 20$ . Corners on an axis are easier: set $x = 0$ or $y = 0$ in the relevant boundary line. List every corner before testing, because missing one can cause you to miss the true optimum.

Maximisation versus minimisation

The corner-point method works the same way whether you maximise or minimise; only the selection rule changes. For a maximum, pick the corner giving the largest objective value; for a minimum, pick the smallest. Cost problems are usually minimisations with "at least" ( $\ge$ ) constraints, which open the feasible region upward, while profit problems are usually maximisations with "at most" ( $\le$ ) constraints that close it off. Read whether the question wants the most or the least before you choose your corner.

When you report an answer, give both the optimal variable values and the optimal objective value, and check that integer answers make sense if the items must be whole (you cannot sell half a chair).

Exam-style practice questions

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

TCE 20237 marksA craftworker makes bowls (

x

) and platters (

y

). Each bowl uses

1

kg of clay and

2

hours; each platter uses

2

kg of clay and

2

hours. There are

16

kg of clay and

20

hours available. Profit is

\

per bowl and

40

per platter. a) Write the constraints. b) The corners of the feasible region are

(0,0)

(10,0)

(4,6)

and

(0,8)

. Find the production that maximises profit

P = 30x + 40y

and state that profit.

Show worked answer →

a) (3 marks) Clay: $x + 2y \le 16$ . Time: $2x + 2y \le 20$ (that is $x + y \le 10$ ). Plus $x \ge 0$ , $y \ge 0$ .

b) (4 marks) Test each corner in $P = 30x + 40y$ :
$(0,0): P = 0$ . $(10,0): P = 300$ . $(0,8): P = 320$ . $(4,6): P = 30(4) + 40(6) = 120 + 240 = 360$ .

The maximum profit is $\$360$ , made by producing $4$ bowls and $6$ platters. Markers reward both correct constraints, testing every corner, and stating both the optimal point and the optimal profit.

TCE 20245 marksA feasible region is bounded by

x \ge 0

y \ge 0

x + y \le 12

and

x + 3y \ge 12

. The cost to minimise is

C = 5x + 4y

. a) State which corner you would test first and why. b) The corners are

(12,0)

(0,4)

and

(0,12)

. Find the minimum cost.

Show worked answer →

a) (1 mark) Because the objective is a linear cost over a polygonal feasible region, the minimum must occur at a vertex, so test the corners rather than interior points.

b) (4 marks) Evaluate $C = 5x + 4y$ at each corner:
$(12,0): C = 5(12) + 4(0) = 60$ . $(0,4): C = 5(0) + 4(4) = 16$ . $(0,12): C = 5(0) + 4(12) = 48$ .

The minimum cost is $\$16$ at $(0,4)$ . Markers expect every corner tested and the smallest value selected, with the optimal point stated alongside the cost.