SAGeneral MathematicsSyllabus dot point

How do we find the best decision when several linear constraints limit our choices?

Formulate linear programming problems with constraints and an objective function, identify the feasible region, and find the optimal solution at a vertex.

How to set up constraints and an objective function, graph and shade the feasible region, and use the corner-point principle to maximise or minimise the objective.

Generated by Claude Opus 4.77 min answerUpdated 2026-05-29

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
Setting up the problem
Graphing the feasible region
Interpreting the solution

What this dot point is asking

You must translate a worded problem into inequalities and an objective function, graph the feasible region, and find the vertex that optimises the objective.

Setting up the problem

Define decision variables (often $x$ and $y$ for the number of each product). Then write:

Constraints: linear inequalities from limits on resources (time, material, demand), plus $x \ge 0$ and $y \ge 0$ .
Objective function: the quantity to maximise or minimise, such as $P = 5x + 8y$ .

Graphing the feasible region

For each inequality, graph the boundary line (replace the inequality with $=$ ), then shade the side that satisfies it. The feasible region is where all shadings overlap.

Worked example

Maximising profit

A workshop makes tables ( $x$ ) and chairs ( $y$ ). Each table needs 4 hours, each chair 2 hours, and there are 40 hours available: $4x + 2y \le 40$ . Wood limits production to $x + y \le 14$ . Profit is $P = 30x + 20y$ . Maximise $P$ , with $x, y \ge 0$ .

The vertices of the feasible region are found at the corners. Where the two constraint lines meet, solve $4x + 2y = 40$ and $x + y = 14$ . From the second, $y = 14 - x$ :

4x + 2(14 - x) = 40 \implies 4x + 28 - 2x = 40 \implies 2x = 12 \implies x = 6,\; y = 8

The corners are $(0, 0)$ , $(10, 0)$ , $(6, 8)$ and $(0, 14)$ . Testing $P = 30x + 20y$ :

$(0,0): P = 0$
$(10,0): P = 300$
$(6,8): P = 180 + 160 = 340$
$(0,14): P = 280$

The maximum profit is $340, making 6 tables and 8 chairs.

Interpreting the solution

The optimal vertex gives the best combination. Always state the answer in context: how many of each item and the resulting objective value. Check the solution uses whole numbers if the items are indivisible.