Graphical method

At a Glance

• Frequency: 5 sub-parts across 5 of 13 years (2014, 2016, 2017, 2019, 2023)
• Priority tier: T2
• Marks (count): 10 (5)
• Average solve time: ~9 min
• Difficulty mix: easy 3, medium 2
• Section: A | Dominant type: computation

Why This Chapter Matters

This atom appears in 5 of the last 13 years — always as a 10-mark Section A sub-part — making it one of the most predictable linear-programming questions in Paper 2. Two variants have appeared: the standard bounded feasible region where you enumerate corner points, and the trickier unbounded case (2019) where the maximum does not exist. The method is fully mechanical: no calculus, just careful geometry and arithmetic at vertices.

Minimum Theory

Five-step graphical procedure. For a two-variable LPP with decision variables $x_1, x_2 \ge 0$:

1. Plot each constraint as a line in the $x_1$–$x_2$ plane; shade the feasible half-plane (the side satisfying the inequality, including $x_1 \ge 0$, $x_2 \ge 0$).
2. Identify the feasible region $S$ = intersection of all half-planes. If $S = \emptyset$, the problem is infeasible — report that.
3. Locate corner points (vertices) = points where two boundary lines (or a boundary line and a coordinate axis) meet and all other constraints are satisfied. Solve each pair of active-constraint equations simultaneously.
4. Evaluate the objective $Z = c_1 x_1 + c_2 x_2$ at every vertex.
5. Report the optimum: for a bounded region the optimum is the vertex giving the largest (maximisation) or smallest (minimisation) value. For an unbounded region see the criterion below.

Fundamental Theorem. If the feasible region $S$ is a non-empty bounded polygon, the linear objective attains its maximum and minimum at (at least one) vertex of $S$. Consequently, evaluating $Z$ at all vertices and taking the extreme value is both necessary and sufficient.

Unboundedness criterion. If $S$ is unbounded and the objective can increase (for maximisation) or decrease (for minimisation) indefinitely along a feasible recession direction $\mathbf{d}$ (i.e.\ $Z(\mathbf{x}+t\mathbf{d}) \to \pm\infty$ as $t \to +\infty$ for all feasible $\mathbf{x}$), then the problem has no finite optimum. To confirm: exhibit the direction and show the objective is unbounded along it.

Infeasibility. If the constraints are contradictory — no point satisfies all of them simultaneously — the feasible region is empty and no solution exists. A typical sign is that a ”$\le$ region” and a ”$\ge$ region” of the same constraint strip do not overlap.

Question Archetypes

graphical-method (4 questions; 2014, 2016, 2017, 2023)

Solve a two-variable LPP by the graphical (vertex-enumeration) method.

Recognition Cues

• The instruction says “graphical method” or “geometrically”.
• There are 2–5 linear inequality constraints plus $x_1, x_2 \ge 0$.
• A mix of $\le$ and $\ge$ constraints may appear; the feasible region is bounded (it is a polygon with a finite number of vertices).
• The objective is to maximise or minimise a linear function.

Solution Template

1. Write each constraint as an equation; find its two intercepts; draw the line.
2. Determine which side of each line is feasible (test the origin, or use the inequality direction).
3. Shade/identify the feasible polygon; list all vertices.
4. For each vertex, compute $Z$; tabulate.
5. State the optimum value and the achieving vertex. If the question asks for a real-world interpretation (jars, units of goods), translate back.

Worked Example 1

2014 Paper 2, 2014-P2-Q1e (10 marks)

Maximise $Z = 6x_1 + 5x_2$ subject to $2x_1+x_2 \le 16$, $x_1+x_2 \le 11$, $x_1+2x_2 \ge 6$, $5x_1+6x_2 \le 90$, $x_1,x_2 \ge 0$.

Step 1 — Boundary lines and intercepts.

Step 2 — Feasible half-planes. $L_1, L_2, L_4$: origin-side ($\le$). $L_3$: away from origin ($\ge$, origin gives $0 \not\ge 6$).

Step 3 — Vertices. Solve intersections that satisfy all constraints:

(Check each vertex satisfies all five constraints before accepting it.)

Step 4 — Objective values.

Step 5 — Optimum.

$\boxed{Z_{\max} = 60 \text{ at } (x_1, x_2) = (5,6).}$

Worked Example 2

2016 Paper 2, 2016-P2-Q1e (10 marks)

By graphical method, find the maximum of $Z = 5x+2y$ subject to $x+2y \ge 1$, $2x+y \le 1$, $x,y \ge 0$.

Step 1 — Boundary lines.

• $L_1$: $x+2y=1$; intercepts $(1,0)$, $(0,\tfrac{1}{2})$.
• $L_2$: $2x+y=1$; intercepts $(\tfrac{1}{2},0)$, $(0,1)$.

Step 2 — Feasible half-planes. $L_1$: away from origin ($\ge$); $L_2$: origin-side ($\le$, origin gives $0 \le 1$ ✓).

Step 3 — Vertices.

• Axis corner $(0, \tfrac{1}{2})$: on $L_1$ boundary, satisfies $L_2$? $2(0)+\tfrac{1}{2}=\tfrac{1}{2} \le 1$ ✓. Feasible.
• Axis corner $(0, 1)$: on $L_2$ boundary, satisfies $L_1$? $0+2(1)=2 \ge 1$ ✓. Feasible.
• Intersection $L_1 \cap L_2$: solve $x+2y=1$, $2x+y=1$ $\Rightarrow$ subtract: $x-y=0 \Rightarrow x=y$; then $3x=1 \Rightarrow x = y = \tfrac{1}{3}$. Check non-negativity ✓.

Note: The axis point $(\tfrac{1}{2},0)$ violates $L_1$: $\tfrac{1}{2}+0=\tfrac{1}{2} \not\ge 1$. Infeasible.

Step 4 — Objective values.

Step 5 — Optimum.

$\boxed{Z_{\max} = \tfrac{7}{3} \text{ at } (x,y) = \bigl(\tfrac{1}{3},\tfrac{1}{3}\bigr).}$

Worked Example 3

2017 Paper 2, 2017-P2-Q1e (10 marks)

Using the graphical method, maximise $Z = 2x+y$ subject to $4x+3y \le 12$, $4x+y \le 8$, $4x-y \le 8$, $x,y \ge 0$.

Step 1 — Boundary lines.

• $L_1$: $4x+3y=12$; intercepts $(3,0)$, $(0,4)$.
• $L_2$: $4x+y=8$; intercepts $(2,0)$, $(0,8)$.
• $L_3$: $4x-y=8$; intercepts $(2,0)$, $(0,-8)$.

Step 2 — Feasible half-planes. All three constraints are $\le$; origin satisfies all ✓.

Step 3 — Vertices.

• $O = (0,0)$: both axes; satisfies all three ✓.
• $A = (2,0)$: $x_2=0$, $L_2$ active; check $L_1$: $8 \le 12$ ✓; $L_3$: $8-0=8 \le 8$ ✓.
• $B$: $L_1 \cap L_2$: solve $4x+3y=12$, $4x+y=8$ $\Rightarrow 2y=4 \Rightarrow y=2$, $x=\tfrac{3}{2}$. Check $L_3$: $4(\tfrac{3}{2})-2=4 \le 8$ ✓.
• $C = (0,4)$: $x_1=0$, $L_1$ active; check $L_2$: $0+4=4 \le 8$ ✓; $L_3$: $0-4=-4 \le 8$ ✓.

Step 4 — Objective values.

Step 5 — Optimum.

$\boxed{Z_{\max} = 5 \text{ at } (x,y) = \bigl(\tfrac{3}{2},2\bigr).}$

$L_3$ ($4x-y \le 8$) is redundant at the optimum: it is satisfied with slack at every vertex.

Worked Example 4 (Minimisation)

2023 Paper 2, 2023-P2-Q1e (10 marks)

A gardener uses two chemicals (products $P$ and $Q$). Each kilogram of product $P$ contains 1 unit of chemical A, 4 units of B, and 1 unit of C. Each kilogram of $Q$ contains 1 unit of A, 1 unit of B, and 5 units of C. Minimum requirements: 12 units of A, 24 units of B, 20 units of C. Product $P$ costs ₹30/kg, product $Q$ costs ₹50/kg. Find the amounts that minimise cost.

Formulation. Let $x_1$ = kg of $P$, $x_2$ = kg of $Q$.
Minimise $Z = 30x_1 + 50x_2$
subject to:
$x_1 + x_2 \ge 12 \quad (A), \qquad 4x_1 + x_2 \ge 24 \quad (B), \qquad x_1 + 5x_2 \ge 20 \quad (C), \qquad x_1, x_2 \ge 0.$

Step 1 — Boundary lines and intercepts.

Step 2 — Feasible half-planes. All three are $\ge$; feasible region is above/right of all lines.

Step 3 — Vertices.

• $L_A \cap L_B$: $x_1+x_2=12$, $4x_1+x_2=24$ $\Rightarrow 3x_1=12 \Rightarrow x_1=4, x_2=8$. Check $L_C$: $4+40=44 \ge 20$ ✓.
• $L_A \cap L_C$: $x_1+x_2=12$, $x_1+5x_2=20$ $\Rightarrow 4x_2=8 \Rightarrow x_2=2, x_1=10$. Check $L_B$: $40+2=42 \ge 24$ ✓.
• Axis vertex $(0,24)$: on $L_B$; check $L_A$: $24 \ge 12$ ✓; $L_C$: $120 \ge 20$ ✓.
• Axis vertex $(20,0)$: on $L_C$; check $L_A$: $20 \ge 12$ ✓; $L_B$: $80 \ge 24$ ✓.

Step 4 — Objective values.

Step 5 — Optimum (minimisation: take the smallest value).

$\boxed{Z_{\min} = 400 \text{ at } (x_1,x_2) = (10,2).}$

The gardener should buy 10 kg of $P$ and 2 kg of $Q$ at a minimum cost of ₹400.

Common Traps — graphical-method

• Not checking all constraints at each candidate vertex. Two boundary lines intersect at a unique point, but that point need not satisfy every other constraint. Always verify the full set before including a vertex in the table.
• Missing a vertex because of an off-axis intersection. Draw the lines carefully; intersections between two non-axis lines (e.g. $L_1 \cap L_2$ in examples 1 and 3) are easy to overlook if you only pick up axis intercepts.
• Assuming the origin is always feasible. When a constraint is $\ge$, the origin is on the wrong side. In the 2016 example the feasible region does not contain the origin; the vertex $(\tfrac{1}{2},0)$ looks attractive but violates $x+2y \ge 1$.
• Minimisation: choosing the maximum. The objective table looks the same; for a minimisation problem take the row with the smallest value. (2023 answer is 400, not 1200.)
• Non-binding constraints. $L_3$ in the 2017 problem plays no role at the optimum. Sketch all constraints regardless; the region is the intersection of all half-planes.

graphical-unbounded (1 question; 2019)

Detect when the linear objective has no finite optimum over an unbounded feasible region.

Recognition Cues

• The feasible region has no upper (for maximisation) or lower (for minimisation) bound — it extends to infinity in the direction of increasing $Z$.
• Sometimes a “superfluous” variable appears in only one constraint but not in the objective; this variable can be chosen freely to satisfy that constraint, effectively removing it.
• The question may say “determine whether the LPP has an optimal solution” or simply ask for the maximum.

Solution Template

1. Identify all variables in the problem; note which appear in the objective and which appear only in constraints.
2. Eliminate variables that appear only in non-negativity constraints (or constraints trivially satisfied) by choosing them appropriately.
3. Reduce to the effective two-variable (or one-variable) problem.
4. Plot the reduced feasible region. If it is unbounded in the direction of increasing $Z$, demonstrate a feasible ray along which $Z \to +\infty$.
5. Conclude: “The LPP has no finite maximum; $Z$ is unbounded.”

Worked Example

2019 Paper 2, 2019-P2-Q1e (10 marks)

Maximise $Z = 3x_1 + 2x_2$ subject to $x_1 - x_2 \ge 1$, $x_1 + x_3 \ge 3$, $x_1,x_2,x_3 \ge 0$.

Trap: $x_3$ appears in the second constraint but not in the objective.

For any $x_1 \ge 0$, we can always choose $x_3 = \max(0,\, 3-x_1) \ge 0$ to satisfy $x_1+x_3 \ge 3$. The second constraint therefore imposes no restriction on $(x_1,x_2)$. The effective problem reduces to:

$\text{Maximise } Z = 3x_1+2x_2 \quad \text{subject to} \quad x_1-x_2 \ge 1,\; x_1,x_2 \ge 0.$

Reduced feasible region. The constraint $x_1 - x_2 \ge 1$ (i.e. $x_2 \le x_1-1$) combined with $x_1,x_2 \ge 0$ defines an unbounded wedge: the boundary line $x_2 = x_1-1$ passes through $(1,0)$ with slope 1; the feasible region lies on and below this line in the first quadrant (with $x_1 \ge 1$).

Show unboundedness. Along the boundary ray $x_2 = x_1 - 1$ for $x_1 \ge 1$: $Z = 3x_1 + 2(x_1-1) = 5x_1 - 2 \;\longrightarrow\; +\infty \quad \text{as } x_1 \to +\infty.$

Every point on this ray is feasible (satisfies $x_1-x_2 \ge 1$ with equality, $x_2 = x_1-1 \ge 0$ for $x_1 \ge 1$, and $x_1 \ge 0$), so $Z$ takes arbitrarily large values over the feasible region.

$\boxed{\text{The LPP is unbounded: } Z \text{ has no finite maximum.}}$

Common Traps — graphical-unbounded

• Treating $x_3$ as if it forces $x_1 \ge 3$. The constraint is $x_1+x_3 \ge 3$, not $x_1 \ge 3$. Since $x_3 \ge 0$ is free, it can absorb any deficit. Mistakenly writing $x_1 \ge 3$ introduces a false lower bound on $x_1$ and may lead to a spurious finite answer.
• Reporting a “boundary vertex” as the optimum. The unbounded wedge does have a finite vertex at $(1,0)$ with $Z = 3$. This is the minimum value of $Z$ over the feasible region, not the maximum. For maximisation, never stop at a vertex without checking whether the region is bounded in the direction of increasing $Z$.
• Forgetting to state the conclusion explicitly. In UPSC marking, writing ”$Z$ is unbounded” (or “no finite maximum exists”) earns the conclusion mark. Simply leaving the table blank or writing $\infty$ without justification does not.

Marks-Aware Writing

What a full-marks 10-mark answer must show:

1. Formulation (if the question gives a word problem): define variables, write objective and constraints clearly — 2 marks.
2. Graph/vertex identification: either sketch the region or solve each pair of active constraints — 4 marks. Show the algebra; do not “read off” approximate coordinates from a rough sketch.
3. Objective table: evaluate $Z$ at all vertices and display the values — 2 marks.
4. Conclusion: state the optimal value and the optimal point in a box or highlighted sentence — 2 marks.

For the unbounded case (2019): replace the objective table with a proof-of-unboundedness (exhibit the feasible ray and compute $Z \to +\infty$). The conclusion “unbounded” is mandatory.

General tips:

• Solve vertex coordinates exactly (fractions), not as decimal approximations.
• Check each candidate vertex against every constraint (not just the two active ones).
• For minimisation, explicitly say “minimum” to avoid losing the conclusion mark.

Practice Set

• 2024-P2-Q3c (15 m) — — larger LPP; use the same five-step template; watch for redundant constraints
• 2018-P2-Q1e (10 m) — — standard bounded maximisation; good warm-up before attempting the 2014 variant