The Graphical Method
Before the algebra of the simplex tableau, there is the picture — and for two variables the picture is the whole answer. Plot the constraints, shade the region they carve out, and the corner-point theorem does the rest: the optimum is one of a handful of corners. This chapter turns that idea into a reliable procedure, shows the two ways to find the best corner, and uses the same drawings to make the special cases — alternative, unbounded, and infeasible — obvious at a glance.
- When the graphical method applies, and its step-by-step procedure.
- How to plot constraints, find the feasible region, and locate its corners.
- The corner-point method: evaluate the objective at every vertex.
- The iso-profit / iso-cost line method: slide the objective line to its last point of contact.
- How minimization problems and \(\ge\) constraints change the picture.
- Reading alternative, unbounded, and infeasible cases, and binding constraints and slack, straight from the graph.
When to Use the Graphical Method
The graphical method solves any linear program with two decision variables — occasionally three, but the drawing becomes awkward. Its value is not industrial scale; real LPs have thousands of variables and demand the simplex method of Chapter 8. Its value is understanding: the picture makes the corner-point theorem tangible, shows exactly why the optimum sits at a vertex, and reveals the special cases in a way no tableau ever will. Master the geometry here and every later algebraic step has a mental image behind it.
Steps of the Method
The procedure is the same every time. Two variables \(x_1, x_2\) span a plane; each constraint is a line dividing that plane, and its inequality keeps one side.
| Step | Action |
|---|---|
| 1 | Replace each inequality by an equality and plot the boundary line (two intercepts) |
| 2 | Decide the feasible side of each line by testing a point (usually the origin) |
| 3 | Intersect all half-planes to get the feasible region (include \(x_1, x_2 \ge 0\)) |
| 4 | Identify every corner (vertex) of the region |
| 5 | Find the optimum by the corner-point or iso-profit method |
The Corner-Point Method
The most direct approach: since the optimum lies at a vertex, simply evaluate the objective at each corner and pick the best. Consider the classic problem
The maximum \(z^\star = 36\) occurs at \((2, 6)\). Because the corner-point theorem guarantees the optimum is a vertex, this finite table of corners is a complete search — no interior point can beat them.
The Iso-Profit Line Method
The second approach draws the objective itself. A line \(z = 3x_1 + 5x_2 = c\) is an iso-profit line — every point on it yields the same profit \(c\). As \(c\) grows, the line moves parallel to itself in a fixed direction. Slide it as far as the feasible region allows; the last point it touches is the optimum.
Both methods must agree — and they do, at \((2,6)\). The corner-point method is safer for exams (it cannot miss a vertex); the iso-profit method is faster when you only need the answer and gives immediate insight into alternative optima, which appear whenever the sliding line comes to rest along an entire edge rather than at a single corner.
Minimization Problems
Minimization flips two things. The constraints are often "\(\ge\)" (meet-at-least requirements), so the feasible region typically lies away from the origin and is frequently unbounded above. And the objective line is slid toward smaller values — pushed back toward the origin — until its last contact. Consider
The corners are \((0,5),\ (1,3),\ (4,0)\), giving \(z = 15,\ 11,\ 8\). The minimum is \(z^\star = 8\) at \((4,0)\). Even though the region runs off to infinity in the up-right direction, the minimum is well defined because the objective decreases toward the origin, where the region has a firm lower boundary.
Special Cases Graphically
The special cases of Chapter 6 have unmistakable graphical fingerprints. Learning to spot them on the sketch stops you from hunting for an answer that isn't there.
| Case | On the graph |
|---|---|
| Alternative optima | The iso-profit line rests along a whole edge (parallel to a binding constraint) |
| Unbounded | The region is open in the improving direction; the line slides forever |
| Infeasible | The half-planes do not overlap; no region exists |
| Redundant constraint | A line lies entirely outside the region and never forms an edge |
Binding Constraints and Slack
Once the optimum is found, the graph tells you which constraints are doing the work. A constraint is binding (active) if the optimum lies exactly on its line — its slack is zero, the resource is fully used. It is non-binding if the optimum sits strictly inside it — there is leftover slack, spare capacity.
Two constraints are binding (\(x_2 \le 6\) and \(3x_1 + 2x_2 \le 18\)) and their intersection is the optimum. The first constraint has slack \(2\) — that resource is not the bottleneck. This slack reading is exactly the shadow-price story of Chapter 5 seen from the graph, and it is what the simplex tableau computes automatically.
Worked Examples
Problem. Maximize \(z = 3x_1 + 5x_2\) subject to \(x_1 \le 4,\ 2x_2 \le 12,\ 3x_1 + 2x_2 \le 18,\ x \ge 0\).
Solution. The vertices are \((0,0), (4,0), (4,3), (2,6), (0,6)\). Evaluating \(z\): \(0, 12, 27, 36, 30\). The maximum is \(z^\star = 36\) at \((2,6)\), matching the figure in Section 7-3.
Problem. Confirm the result using the objective-line method.
Solution. Draw \(3x_1 + 5x_2 = 15\), then slide it outward keeping its slope. It leaves the feasible region last at the single corner \((2,6)\), where \(z = 36\). Since it stops at a point (not along an edge), the optimum is unique — the two methods agree.
Problem. Minimize \(z = 2x_1 + 3x_2\) subject to \(x_1 + x_2 \ge 4,\ 2x_1 + x_2 \ge 5,\ x \ge 0\).
Solution. Corners \((0,5), (1,3), (4,0)\) give \(z = 15, 11, 8\). The minimum is \(z^\star = 8\) at \((4,0)\). The region is unbounded upward, but the minimum exists because the objective decreases toward the origin.
Problem. Maximize \(z = 2x_1 + 2x_2\) subject to \(x_1 + x_2 \le 6,\ x_1 \le 4,\ x \ge 0\).
Solution. The objective is proportional to \(x_1 + x_2\), so its lines are parallel to the binding edge \(x_1 + x_2 = 6\). The iso-profit line comes to rest along the whole segment from \((4,2)\) to \((0,6)\); every point on it gives \(z = 12\). There are infinitely many optimal solutions.
Problem. Maximize \(z = 2x_1 + x_2\) subject to \(x_1 - x_2 \le 1,\ x \ge 0\).
Solution. The region is open toward the upper right. The iso-profit line can be slid outward without limit, so \(z \to \infty\) — the problem is unbounded and has no finite maximum. Graphically, no last point of contact exists.
Problem. At the optimum \((2,6)\) of Example 1, state which constraints are binding and compute the slacks.
Solution. \(x_1 = 2 < 4\), so \(x_1 \le 4\) is non-binding with slack \(2\). \(2x_2 = 12\), so \(x_2 \le 6\) is binding (slack \(0\)). \(3(2) + 2(6) = 18\), so \(3x_1 + 2x_2 \le 18\) is binding (slack \(0\)). The optimum is the intersection of the two binding lines, and the one non-binding resource has spare capacity — its shadow price would be zero.
Chapter Summary
The graphical method solves two-variable LPs and builds the geometric intuition for all of LP.
Plot lines, pick feasible sides, intersect to a region, find corners, optimize.
Evaluate \(z\) at every vertex; the best one is optimal.
Slide the objective line to its last point of contact with the region.
Edge contact → alternative; open region → unbounded; no overlap → infeasible.
Optimum on a line → binding (slack 0); inside → non-binding (slack > 0).
Problems
Sketch each region carefully, then solve by the corner-point method and check with an iso-profit line. Difficulty rises down the list.
- Maximize \(z = 4x_1 + 3x_2\) s.t. \(2x_1 + x_2 \le 10,\ x_1 + 3x_2 \le 15,\ x \ge 0\).
- Maximize \(z = 5x_1 + 4x_2\) s.t. \(6x_1 + 4x_2 \le 24,\ x_1 + 2x_2 \le 6,\ x \ge 0\); report the optimum and all slacks.
- Minimize \(z = 3x_1 + 2x_2\) s.t. \(x_1 + x_2 \ge 4,\ x_1 + 2x_2 \ge 6,\ x \ge 0\).
- Solve Problem 2 again using only the iso-profit line method and confirm the answer.
- Maximize \(z = x_1 + x_2\) s.t. \(x_1 + x_2 \le 5,\ x_1 \le 3,\ x \ge 0\); identify whether the optimum is unique or an edge.
- Maximize \(z = 3x_1 + 2x_2\) s.t. \(x_1 - x_2 \le 2,\ x \ge 0\); classify the outcome.
- Show graphically that \(x_1 + x_2 \le 2\) with \(x_1 + x_2 \ge 5\) is infeasible.
- In the system \(x_1 \le 6,\ x_1 + x_2 \le 8,\ x_1 \le 10\), identify the redundant constraint.
- For Problem 1, list the binding constraints at the optimum and compute each slack.
- Explain why a two-variable maximization with all "\(\le\)" constraints and \(x \ge 0\) can never be unbounded.
- A diet must supply at least 8 units of protein and 12 of calcium; food A costs ₹4 (2 protein, 2 calcium) and food B costs ₹3 (1 protein, 4 calcium). Minimize cost graphically.
- A factory makes two products with profits ₹6 and ₹5. Product 1 needs 1 h machine and 3 h labour; product 2 needs 2 h machine and 1 h labour; 14 machine-hours and 15 labour-hours are available. Solve graphically, state the optimum and objective value, report the slack on each resource, and say which resource management should expand first and why.