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 4.77 min answerUpdated 2026-05-29

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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 ( $x$ ) and chairs ( $y$ ). Each table needs 2 hours of cutting and 1 hour of finishing; each chair needs 1 hour of cutting and 1 hour of finishing. There are 40 cutting hours and 30 finishing hours available. Profit is $\$ 60 $per table and$ \ $40$ per chair. Maximise profit $P = 60x + 40y$ .

Constraints

Cutting: $2x + y \le 40$ . Finishing: $x + y \le 30$ . Plus $x \ge 0$ , $y \ge 0$ .

Find the corners

The corners are $(0,0)$ , $(20,0)$ from $2x+y=40$ on the $x$ -axis, and $(0,30)$ from $x+y=30$ on the $y$ -axis.

Intersection of the two lines: subtract $x+y=30$ from $2x+y=40$ to get $x = 10$ , then $y = 20$ . So $(10,20)$ is a corner.

Test each corner

$(0,0): P = 0$ . $(20,0): P = 1200$ . $(0,30): P = 1200$ . $(10,20): P = 60(10)+40(20) = 600 + 800 = 1400$ .

Maximum profit is $\$ 1400$, made 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.

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).