All subjects · Mathematics · WDM11 (D1)
Linear programming: graphical solution
With two variables, a linear programming problem can be solved by drawing the constraints and finding the best point of the feasible region.
Drawing the region
Draw each constraint as a straight line (find two points, such as the axis intercepts). Shade the side you do not want, so the feasible region (R) is left unshaded. Label every line and the region. Use a ruler; accuracy matters.
Objective line method
Draw a line of the objective, for example 30x + 40y = 600. Slide it parallel to itself, using a ruler, across the region. For a maximum, the last point of the region it touches is optimal; for a minimum, the first point it touches is.
Vertex method
The optimum is always at a vertex of the feasible region. Find the coordinates of every vertex (solving simultaneous equations exactly where lines cross) and evaluate the objective at each. Choose the largest (or smallest).
Interpreting
Finish with a sentence in context: 'Make 9 chairs and 14 tables for a maximum profit of £830.' If the region is unbounded, a minimum may exist but a maximum may not.
Worked example
Maximise P = 30x + 40y subject to 2x + 3y ≤ 60, 2x + y ≤ 32, y ≥ 4, x ≥ 0.
- Vertices of R: (0, 4) P = 160; (0, 20) P = 800; (9, 14) P = 830; (14, 4) P = 580
- 2x + 3y = 60 and 2x + y = 32 meet where 2y = 28, so y = 14 and x = 9
Answer: Maximum P = 830 at x = 9, y = 14
Check yourself
- Maximise P = 30x + 40y subject to 2x + 3y ≤ 60, 2x + y ≤ 32, y ≥ 4, x ≥ 0. Find the maximum value of P.
Show answer
830. Vertex values: (0, 4) gives 160; (0, 20) gives 800; (9, 14) gives 830; (14, 4) gives 580.
- In the same problem, at which x-coordinate do 2x + 3y = 60 and 2x + y = 32 intersect?
Show answer
9. Subtract: 2y = 28, so y = 14; then 2x = 32 - 14 = 18, so x = 9.
- How many vertices does the feasible region for 2x + 3y ≤ 60, 2x + y ≤ 32, y ≥ 4, x ≥ 0 have?
Show answer
4. Vertices: (0, 4); (0, 20); (9, 14); (14, 4).
- Minimise C = 4x + 3y subject to x + y ≥ 10, x + 3y ≥ 18, 2x + y ≥ 12, x ≥ 0, y ≥ 0. Find the minimum value of C.
Show answer
32. Vertices: (0, 12) gives 36; (2, 8) gives 32; (6, 4) gives 36; (18, 0) gives 72. Minimum 32.
- In D1, how is the feasible region usually shown on the graph?
- It is shaded in
- It is outlined in red only
- It is not shown; only the vertices are marked
- It is left unshaded; the unwanted regions are shaded
Show answer
It is left unshaded; the unwanted regions are shaded. Edexcel expects you to shade out the regions that break constraints and label the feasible region R.
- Using the objective line method to maximise, which point of the feasible region do you choose?
- The first point the line touches
- The last point the line touches as it moves away from the origin
- The point closest to the origin
- The midpoint of the region
Show answer
The last point the line touches as it moves away from the origin. For a maximum, slide the line in the direction that increases the objective until it is about to leave R.
- The feasible region is unbounded above and to the right, and the objective has positive coefficients. Which statement is correct?
- A maximum exists but no minimum
- There is neither
- The optimum is at the origin
- A minimum may exist but there is no maximum
Show answer
A minimum may exist but there is no maximum. The objective can grow without limit in an unbounded region, but it can still have a smallest value at a vertex.
Sources
- Pearson Edexcel International Advanced Level in Mathematics specification (issue 3)
- Pearson Edexcel IAL Decision Mathematics 1 scheme of work
Last updated 7 November 2026. Spotted a mistake? Tell us.