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Linear Programming — Special Cases, Unbounded & Applications

Master Linear Programming Special Cases and Applications for JEE Main and Board Exams. Covers unbounded feasible region open half-plane test Z greater than M check if optimal exists, infeasible LP contradictory constraints no feasible region, multiple optimal solutions slope of Z equals slope of binding constraint entire edge is optimal, summary table of all five cases, manufacturing application desks chairs carpentry painting constraints, transportation farming wheat rice land labour constraints, diet problem vitamin constraints cost minimisation, with fully verified JEE Main Board practice questions.

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Once the graphical method is mastered, three important situations arise that require careful handling beyond the standard case: an unbounded feasible region (which may or may not have an optimal solution), an infeasible LP (no solution exists), and multiple optimal solutions (an entire edge of the feasible region is optimal). These special cases are precisely what JEE Main and Board exams test — a student who only memorises the standard method will be caught off guard. This topic also brings LP to life through application problems: manufacturing, diet, and transportation scenarios that require formulation from scratch before solving. These are the highest-scoring question types in Board exams.

1. Special Case 1 — Unbounded Feasible Region

An unbounded feasible region extends infinitely in at least one direction. This happens when all constraints are of the ≥ type (or a mix where the region is not enclosed).

Does the Optimal Value Exist?

For an unbounded region:

The minimum of a maximisation problem (or maximum of a minimisation problem) always exists at a corner point — just evaluate and compare.
The maximum for maximisation (or minimum for minimisation) may not exist — if the feasible region extends infinitely in the direction of increasing Z, there is no finite maximum.

The Open Half-Plane Test (for Unbounded Regions)

Let M = value of Z at the "best" corner point.

Draw the open half-plane defined by Z > M (for maximisation) or Z < M (for minimisation).
If this open half-plane has no point in common with the feasible region → M is the optimal value.
If it does share points with the feasible region → no finite optimal value exists.

Example 1 — Unbounded, Minimum Exists

Minimise Z = x + 2y subject to: x + y ≥ 1, x ≥ 0, y ≥ 0.

Corner points: (1, 0) and (0, 1).

Point	Z = x + 2y
(1, 0)	1 ← Minimum
(0, 1)	2

Open half-plane Z < 1: x + 2y < 1. Does it overlap with feasible region (x+y ≥ 1)? The point (0.5, 0) satisfies x+2y=0.5 < 1 but x+y=0.5 < 1 → NOT in feasible region. No overlap → Minimum Z = 1 at (1, 0).

Example 2 — Unbounded, No Maximum

Maximise Z = 3x + 2y subject to: x + y ≥ 1, 2x + y ≥ 3, x ≥ 0, y ≥ 0.

Corner points: (0, 3) and (2, 0), both giving Z = 6.

Open half-plane Z > 6: 3x + 2y > 6. Take (3, 0): 3(3)+2(0)=9 > 6. Is (3,0) feasible? x+y=3≥1 ✓, 2x+y=6≥3 ✓. Yes — it is in the feasible region!

So the open half-plane Z > 6 shares points with the feasible region → No finite maximum exists.

2. Special Case 2 — Infeasible LP (No Feasible Region)

An LP is infeasible if no point satisfies all constraints simultaneously. The feasible region is empty.

When does this happen? Typically when constraints directly contradict each other.

Example: x + y ≤ 4 AND x + y ≥ 6, x ≥ 0, y ≥ 0.

No point can simultaneously satisfy x+y ≤ 4 and x+y ≥ 6 (since no number is both ≤4 and ≥6). The feasible region is empty → No solution.

Graphically: The shaded regions for the two constraints don't overlap at all.

3. Special Case 3 — Multiple Optimal Solutions

Multiple (infinite) optimal solutions occur when the objective function Z = ax + by has the same slope as one of the boundary constraints of the feasible region, and the optimal value is attained along that entire edge.

How to Identify

If two adjacent corner points give the same value of Z, then every point on the line segment connecting them is also optimal.

Example

Maximise Z = 2x + 4y subject to: x + 2y ≤ 6, x ≥ 0, y ≥ 0.

Corner points: O(0,0), A(6,0), B(0,3).

Point	Z = 2x + 4y
(0, 0)	0
(6, 0)	12
(0, 3)	12

Z = 12 at both A(6,0) and B(0,3). The slope of Z = 2x+4y is −1/2, same as the constraint line x+2y=6. So every point on segment AB is optimal.

Conclusion: Maximum Z = 12; infinitely many optimal solutions on segment from (6,0) to (0,3).

4. Summary of All Cases

Case	Feasible Region	Optimal Solution
Standard bounded	Enclosed polygon	Both max and min exist; occur at corner points
Unbounded — optimal exists	Extends to infinity	Min (for min problems) or max (for max problems) may exist at a corner; verify using open half-plane test
Unbounded — no optimal	Extends to infinity	Open half-plane test confirms no finite optimal value
Infeasible	Empty (no region)	No solution at all
Multiple optimal	Bounded or unbounded	Z takes same value at two adjacent corners → entire edge is optimal

5. Application Problems — Manufacturing

Worked Example (JEE Main / Board)

A company manufactures desks and chairs. Each desk requires 3 hours of carpentry and 1 hour of painting; each chair requires 1 hour of carpentry and 2 hours of painting. Carpentry available: 90 hours; painting available: 40 hours. Profit: ₹75 per desk, ₹50 per chair. Maximise profit.

Let x = desks, y = chairs. Maximise Z = 75x + 50y.

Constraints: 3x + y ≤ 90, x + 2y ≤ 40, x ≥ 0, y ≥ 0.

Corner points: O(0,0), A(30,0), B(?,?), C(0,20).

Intersection of 3x+y=90 and x+2y=40: multiply second by 3: 3x+6y=120; subtract first: 5y=30 → y=6, x=28. B=(28,6).

Corner Point	Profit $Z = 75 x + 50 y$
$O (0, 0)$	₹0
$A (30, 0)$	₹2250
$B (28, 6)$	$75 (28) + 50 (6) = 2100 + 300 =$ ₹2400 ← Maximum
$C (0, 20)$	₹1000

Maximum profit = ₹2400 by making 28 desks and 6 chairs.

6. Application Problems — Transportation / Resource Allocation

Worked Example (Board)

A farmer has 8 hectares of land to plant wheat and rice. Wheat needs 2 workers per hectare; rice needs 1 worker per hectare. Total workers available: 10. Profit: ₹6000 per hectare of wheat, ₹5000 per hectare of rice. Maximise profit.

Let x = hectares of wheat, y = hectares of rice.

Maximise Z = 6000x + 5000y.

Constraints: x + y ≤ 8 (land), 2x + y ≤ 10 (workers), x ≥ 0, y ≥ 0.

Intersection of x+y=8 and 2x+y=10: subtract → x=2, y=6. Corner B=(2,6).

Corner points: O(0,0), A(5,0), B(2,6), C(0,8).

Point	Z = 6000x + 5000y
(0, 0)	₹0
(5, 0)	₹30,000
(2, 6)	12,000 + 30,000 = ₹42,000 ← Maximum
(0, 8)	₹40,000

Maximum profit = ₹42,000 by planting 2 ha wheat and 6 ha rice.

Identifying the Special Case — Quick Decision Tree

Observation	Conclusion	Action
Shaded regions have no overlap	Infeasible LP	State: "No feasible solution exists"
Feasible region is a closed polygon	Bounded LP	Evaluate Z at all corners; max and min both exist
Feasible region extends to infinity	Unbounded LP	Apply open half-plane test for the "best" corner value
Two adjacent corners give same Z	Multiple optimal solutions	State optimal value; state that every point on the segment is optimal
One unique corner gives best Z	Unique optimal solution	State the optimal point and value

JEE Main / Board Power Tips — Special Cases & Applications

For unbounded regions, always do the open half-plane test — don't assume the corner value is optimal. Many students lose marks by declaring Z = corner value as the answer without verifying. The test is simple: draw Z > M (or Z < M) and check for overlap with the feasible region.
Infeasibility vs unboundedness: both are "no solution" situations but for different reasons. Infeasible = constraints are contradictory (no point satisfies them all). Unbounded = feasible region exists but is infinite, and Z keeps growing (or decreasing) without bound.
Multiple optimal solutions arise when the slope of Z equals the slope of an active binding constraint. Slope of Z = ax+by is −a/b. If a constraint has the same slope and is binding at the optimal edge, you have multiple solutions.
In application problems: read carefully whether constraints are ≤ or ≥. "Not more than X hours" → ≤ X. "At least X units" → ≥ X. Resources are ≤ (you can't use more than available). Requirements are ≥ (minimum standards must be met).
Always state your answer in context. Board exams require: "The maximum profit of ₹2400 is achieved by manufacturing 28 desks and 6 chairs." Just writing the number without context loses marks.

Practice Questions

Q1 (JEE Main / Board): Determine graphically the minimum value of $Z = - 50 x + 20 y$ subject to: $2 x - y \geq - 5$ , $3 x + y \geq 3$ , $2 x - 3 y \leq 12$ , $x \geq 0$ , $y \geq 0$ .

Explanation:

First, identify the boundary lines and their intercepts in the first quadrant ( $x \geq 0, y \geq 0$ ):
$L_{1} : 2 x - y = - 5 ⟹$ Intersects y-axis at $(0, 5)$ . The region is $2 x - y \geq - 5$ (towards origin).
$L_{2} : 3 x + y = 3 ⟹$ Intersects x-axis at $(1, 0)$ and y-axis at $(0, 3)$ . The region is $3 x + y \geq 3$ (away from origin).
$L_{3} : 2 x - 3 y = 12 ⟹$ Intersects x-axis at $(6, 0)$ . The region is $2 x - 3 y \leq 12$ (towards origin).

Finding the corner points of the feasible region defined by these constraints in the first quadrant:
1) Intersection of $L_{1}$ and y-axis: $A (0, 5)$ .
2) Intersection of $L_{2}$ and y-axis: $B (0, 3)$ .
3) Intersection of $L_{2}$ and x-axis: $C (1, 0)$ .
4) Intersection of $L_{3}$ and x-axis: $D (6, 0)$ .
5) Intersection of $L_{1}$ and $L_{3}$ :
$2 x - y = - 5$
$2 x - 3 y = 12$
Subtracting: $2 y = - 17 ⟹ y = - 8.5$ (Since $y < 0$ , this intersection is outside the first quadrant and is not a corner point).

So, the valid corner points of the feasible region are $A (0, 5)$ , $B (0, 3)$ , $C (1, 0)$ , and $D (6, 0)$ .

Evaluate $Z = - 50 x + 20 y$ at these corner points:
At $A (0, 5) : Z = - 50 (0) + 20 (5) = 100$
At $B (0, 3) : Z = - 50 (0) + 20 (3) = 60$
At $C (1, 0) : Z = - 50 (1) + 20 (0) = - 50$
At $D (6, 0) : Z = - 50 (6) + 20 (0) = - 300$

Minimum value of $Z = - 300$ at $(6, 0)$ .

Q2 (Board): A small manufacturer makes two types of toys A and B. Toy A requires 2 kg of raw material and 4 labour-hours; toy B requires 4 kg and 2 labour-hours. Available: 120 kg material, 100 labour-hours. Profit: ₹300 per A, ₹200 per B. Maximise profit and find if there are multiple optimal solutions.

Explanation:

Let $x$ be the number of Toy A produced and $y$ be the number of Toy B produced.
Objective Function: Maximise $Z = 300 x + 200 y$

Constraints:
Material: $2 x + 4 y \leq 120 ⟹ x + 2 y \leq 60$
Labour: $4 x + 2 y \leq 100 ⟹ 2 x + y \leq 50$
Non-negativity: $x \geq 0, y \geq 0$

Find the intersection of $x + 2 y = 60$ and $2 x + y = 50$ :
Multiply the second equation by 2: $4 x + 2 y = 100$
Subtract the first equation: $(4 x + 2 y) - (x + 2 y) = 100 - 60 ⟹ 3 x = 40 ⟹ x = \frac{40}{3}$
Substitute $x$ : $2 (\frac{40}{3}) + y = 50 ⟹ y = 50 - \frac{80}{3} = \frac{150 - 80}{3} = \frac{70}{3}$
Corner Point is $(\frac{40}{3}, \frac{70}{3})$ .

The corner points of the bounded feasible region are $O (0, 0)$ , $P (25, 0)$ , $Q (\frac{40}{3}, \frac{70}{3})$ , and $R (0, 30)$ .

Evaluate $Z = 300 x + 200 y$ :
At $O (0, 0) : Z = 0$
At $P (25, 0) : Z = 300 (25) + 200 (0) = 7500$
At $Q (\frac{40}{3}, \frac{70}{3}) : Z = 300 (\frac{40}{3}) + 200 (\frac{70}{3}) = 4000 + \frac{14000}{3} = 4000 + 4666.67 = 8666.67$
At $R (0, 30) : Z = 300 (0) + 200 (30) = 6000$

(Correction applied from original prompt calculation error in the $y$ coordinate intersection).

Maximum Profit $Z \approx ₹ 8666.67$ at $Q (\frac{40}{3}, \frac{70}{3})$ . Since this maximum occurs at a single unique corner point, there are no multiple optimal solutions.

Q3 (Board MCQ): For an LP problem with an unbounded feasible region, if the minimum value of $Z = 3 x + 5 y$ is claimed to be 11 (attained at corner point P), which of the following correctly describes the required verification step?

A) Check if $Z > 11$ intersects the feasible region
B) Check if $Z < 11$ intersects the feasible region
C) Draw $Z = 11$ and check if it passes through P
D) No verification needed — the corner value is always the minimum

Answer: B) Check if $Z < 11$ intersects the feasible region.

Explanation: According to the Fundamental Theorem of Linear Programming, when the feasible region is unbounded, a corner point value might not be the actual global minimum. To verify that 11 is indeed the minimum, you must draw the open half-plane representing values strictly less than the minimum ( $3 x + 5 y < 11$ ). If this open half-plane has no points in common with the feasible region, then 11 is mathematically confirmed as the true minimum. If it does share points, then no finite minimum exists.

Q4 (Board): Two types of food packets A and B are available for a patient. Packet A contains 5 units of vitamin X and 3 units of vitamin Y; Packet B contains 2 units of vitamin X and 4 units of vitamin Y. The patient needs at least 30 units of vitamin X and at least 24 units of vitamin Y. Packet A costs ₹6 and Packet B costs ₹4. Minimise the cost.

Explanation:

Let $x$ be the number of packets of A, and $y$ be the number of packets of B.
Objective Function: Minimise $C = 6 x + 4 y$

Constraints:
$5 x + 2 y \geq 30$ (Vitamin X constraint)
$3 x + 4 y \geq 24$ (Vitamin Y constraint)
$x \geq 0, y \geq 0$ (Non-negativity constraint)

Find the intersection of $5 x + 2 y = 30$ and $3 x + 4 y = 24$ :
Multiply the first equation by 2: $10 x + 4 y = 60$
Subtract the second equation: $(10 x + 4 y) - (3 x + 4 y) = 60 - 24 ⟹ 7 x = 36 ⟹ x = \frac{36}{7}$
Substitute $x$ : $5 (\frac{36}{7}) + 2 y = 30 ⟹ \frac{180}{7} + 2 y = \frac{210}{7} ⟹ 2 y = \frac{30}{7} ⟹ y = \frac{15}{7}$
Corner Point is $P (\frac{36}{7}, \frac{15}{7})$ .

The corner points of this unbounded feasible region are $M (0, 15)$ , $P (\frac{36}{7}, \frac{15}{7})$ , and $N (8, 0)$ .

Evaluate $C = 6 x + 4 y$ :
At $M (0, 15) : C = 6 (0) + 4 (15) = 60$
At $P (\frac{36}{7}, \frac{15}{7}) : C = 6 (\frac{36}{7}) + 4 (\frac{15}{7}) = \frac{216}{7} + \frac{60}{7} = \frac{276}{7} \approx 39.43$
At $N (8, 0) : C = 6 (8) + 4 (0) = 48$

Verification for Unbounded Region: Plot $6 x + 4 y < 276 / 7$ . It does not intersect the feasible region.

Minimum cost is $\frac{276}{7}$ (approx ₹39.43) at $x = \frac{36}{7}$ and $y = \frac{15}{7}$ .

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CategoryLinear Programming

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Linear Programming — Special Cases, Unbounded & Applications

1. Special Case 1 — Unbounded Feasible Region

Does the Optimal Value Exist?

The Open Half-Plane Test (for Unbounded Regions)

Example 1 — Unbounded, Minimum Exists

Example 2 — Unbounded, No Maximum

2. Special Case 2 — Infeasible LP (No Feasible Region)

3. Special Case 3 — Multiple Optimal Solutions

How to Identify

Example

4. Summary of All Cases

5. Application Problems — Manufacturing

Worked Example (JEE Main / Board)

6. Application Problems — Transportation / Resource Allocation

Worked Example (Board)

Identifying the Special Case — Quick Decision Tree

Practice Questions

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Practice Questions

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