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 4.77 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 constraints
Finding the optimum at a vertex
The sliding-line method

What this dot point is asking

You must define variables, write the objective function and constraints, shade the feasible region, identify its vertices, and evaluate the objective at each vertex to find the optimum.

Setting up the problem

Define decision variables (for example, $x$ = number of product A, $y$ = number of product B). The objective function is the quantity to maximise or minimise, such as profit $P = 5x + 8y$ . Constraints are linear inequalities from limits on resources, plus the non-negativity constraints $x \ge 0$ and $y \ge 0$ .

Graphing constraints

For each inequality, graph the boundary line (replace the inequality with $=$ ), then shade the side that satisfies the inequality. A quick test is to substitute the origin $(0,0)$ : if it satisfies the inequality, shade the side containing the origin.

Finding the optimum at a vertex

The optimum of a linear objective over a polygonal feasible region always occurs at a vertex (corner point). So you find every vertex, evaluate the objective at each, and pick the best.

Worked example

Maximising profit

A workshop makes chairs ( $x$ ) and tables ( $y$ ). Profit is $P = 40x + 60y$ . Constraints: $x + y \le 8$ (assembly hours), $x + 2y \le 12$ (finishing hours), $x \ge 0$ , $y \ge 0$ . Find the maximum profit.

The vertices of the feasible region are $(0,0)$ , $(8,0)$ , $(0,6)$ and the intersection of $x+y=8$ with $x+2y=12$ . Subtracting the first from the second gives $y = 4$ , so $x = 4$ ; the vertex is $(4,4)$ .

Evaluate $P$ :

$(0,0): P = 0$
$(8,0): P = 320$
$(0,6): P = 360$
$(4,4): P = 40(4) + 60(4) = 160 + 240 = 400$

The maximum profit is $400 at 4 chairs and 4 tables.

The sliding-line method

An alternative to testing every vertex is to draw the objective line $P = c$ for some value of $c$ and slide it (keeping its gradient) in the direction of increase until it last touches the feasible region. The last point it touches is the optimum.

Worked example

Minimising cost

A diet must supply at least 12 units of protein and 6 of fibre. Food X gives 2 protein, 1 fibre per serve at \ $3; food Y gives 3 protein, 1 fibre per serve at \$ 4. Minimise cost $C = 3x + 4y$ subject to $2x + 3y \ge 12$ , $x + y \ge 6$ , $x, y \ge 0$ .

The boundary lines meet where $2x + 3y = 12$ and $x + y = 6$ . From the second, $x = 6 - y$ ; substitute: $2(6 - y) + 3y = 12$ , so $12 + y = 12$ , giving $y = 0$ , $x = 6$ . Vertices include $(6,0)$ and $(0,6)$ .

$(6,0): C = 18$
$(0,6): C = 24$

Minimum cost is \ $18 at 6 serves of food X (which also satisfies$ 2(6) \ge 12 $and$ 6 \ge 6$).