Graphical Method of Solution: Explained with Examples

The graphical method is the standard way to solve a linear programming problem with two decision variables. You draw each constraint as a line on graph paper, shade the side of the line that satisfies the inequality, and the overlapping shaded part is the feasible region where every valid solution lives.

This method matters because it turns abstract inequalities into a picture you can read: the best value of the objective function is always found at a corner of that shaded region. It works only for two variables, but it is the method examiners expect for every two-variable LPP.

Key Concepts

• Constraint line: the line a₁x + b₁y = c₁ drawn from its intercepts (c₁/a₁, 0) and (0, c₁/b₁).
• Half-plane: the side of a constraint line where the inequality holds; found with a test point such as (0, 0).
• Feasible region: the intersection of all the required half-planes.
• Iso-profit line: a line Z = ax + by = constant; all iso-profit lines are parallel, and sliding the line outward increases Z.
• Optimal corner: the corner point of the feasible region giving the best objective value.

Formulas

• Objective function: Z = ax + by to be maximised or minimised.
• Constraint lines: a₁x + b₁y = c₁, a₂x + b₂y = c₂, drawn using x-intercept (c/a, 0) and y-intercept (0, c/b).
• Test-point check: substitute (0, 0) into the inequality; if it holds, shade the side containing the origin.
• Corner evaluation: compute Z at every corner point and compare the values.

Worked Example

Maximise Z = 4x + 3y subject to x + y ≤ 8, 2x + y ≤ 12, x ≥ 0, y ≥ 0. Draw x + y = 8 through intercepts (8, 0) and (0, 8); testing (0, 0) gives 0 ≤ 8, true, so shade toward the origin. Draw 2x + y = 12 through (6, 0) and (0, 12); (0, 0) satisfies 0 ≤ 12, so shade toward the origin again. The two lines intersect where x + y = 8 and 2x + y = 12; subtracting gives x = 4, and then y = 4, so the intersection is (4, 4). The feasible region has corners O(0, 0), P(6, 0), Q(4, 4), R(0, 8). Evaluating Z = 4x + 3y: at O, Z = 0; at P, Z = 4(6) + 3(0) = 24; at Q, Z = 4(4) + 3(4) = 16 + 12 = 28; at R, Z = 4(0) + 3(8) = 24. The maximum is Z = 28 at (4, 4). Check feasibility of (4, 4): 4 + 4 = 8 ≤ 8 and 2(4) + 4 = 12 ≤ 12, both satisfied, so the solution is valid.

Common Misconceptions

• Any intersection of lines is a corner. Wrong — only intersections that satisfy all the constraints are corners; an intersection lying outside the feasible region must be discarded.
• The interior point gives the best value. No — for a linear objective over a polygon, the optimum always occurs at a corner (or along an entire edge).
• Shading the wrong side of the line. Always test with (0, 0) unless the line passes through the origin, in which case pick another easy point such as (1, 0).

Real-World Applications

• Small businesses decide product mixes, such as how many tables and chairs to make with limited wood and labour.
• Teachers use it to demonstrate optimisation visually before introducing the simplex method.
• Farmers and shopkeepers solve two-product planning problems on paper without software.

The graphical method converts inequalities into a shaded region and reads the optimum from its corners. Draw carefully, test each corner, and the best value of the objective function reveals itself.