Linear Programming - TCE Mathematics Applications (Tasmania)

Syllabus dot point

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

TCEMathematics ApplicationsUnit 38 min readUpdated 2 Jun 2026

Quick answer

You must translate a worded problem into decision variables, an objective function, and a set of linear inequality constraints, then graph the feasible region and test its corner points to find the maximum or minimum.

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).

The three ingredients

Decision variables: the unknown quantities you choose, often $x$ and $y$ . Objective function: a linear expression such as $P = 8x + 5y$ to be maximised or minimised. Constraints: linear inequalities such as $2x + y \le 40$ that limit the variables.

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.

Setting up constraints

A phrase like "at most 40 hours" means $\le 40$ ; "at least 10 units" means $\ge 10$ ; "no more than" means $\le$ ; "a minimum of" means $\ge$ . Translate each sentence into one inequality before drawing anything.

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.

The corner-point principle

For a linear objective over a polygonal feasible region, the maximum and minimum always occur at a vertex (corner) of the region. You only ever need to test the corners.

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.

Solving by the sliding line

For the workshop above, $P = 60x + 40y$ . Draw the line for a trial value, say $P = 1200$ , which passes through $(20,0)$ and $(0,30)$ .

Slide outward

Move the line parallel to itself away from the origin (increasing $P$ ). The feasible region is the polygon with corners $(0,0)$ , $(20,0)$ , $(10,20)$ and $(0,30)$ .

Read the last corner touched

As the line slides out, the last vertex it meets before leaving the region is $(10,20)$ , giving $P = 60(10) + 40(20) = 1400$ .

This matches the corner-point answer of $\$1400$ , confirming 10 tables and 20 chairs is optimal.

Forgetting non-negativity or shading the wrong side

Two classic errors: leaving out $x \ge 0, y \ge 0$ , which can let a corner fall in negative territory, and shading the wrong half-plane. Always test a clear point such as $(0,0)$ against each inequality, and make sure your final region satisfies every constraint at once.

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 questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice7 marks

A 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

\$30

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.

Practice5 marks

A 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.

Practise this

Sources & how we know this

General Mathematics Level 3 (MTG315123) course document (2023 onwards) — TASC