Unit 8: Linear Programming Problems | Class 12 Applied Maths | Boundless Maths
Unit 8 of 8 8 Marks CBSE 2026–27 High-Scoring Unit

Linear Programming
Unit 8 — Free Study Resources

CBSE Class 12 Applied Mathematics · MCQs, Solved Examples & Case Studies

Unit 8 carries 8 marks in the CBSE board exam — fully covered through Linear Programming Problems. Everything you need to master it: 15 interactive MCQs, 6 short-answer questions, 6 long-answer questions, and 3 board-pattern case studies. Covering LPP Formulation, Graphical Solution Method, Bounded and Unbounded Feasible Regions, and the Corner Point Method. Fully aligned to CBSE 2026–27. Once you’ve attempted the questions and self-assessed your answers, tap ✨ My Report to see your personalised performance breakdown.

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Short Answers
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Long Answers
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Case Studies
Unit 8 · 8 Marks

Topics & Key Formulas

Four topics examined across MCQs, short answers, long answers, and case studies. Master LPP formulation and the Corner Point Method for full marks.

1. Formulation of LPP

Identify decision variables, write objective function and structural constraints from word problems

Objective: \(Z = c_1x + c_2y\) (maximise or minimise) Constraints: \(a_i x + b_i y \leq (\geq)\ k_i\) Non-negativity: \(x \geq 0,\ y \geq 0\)

2. Graphical Solution Method

Plot constraint lines, shade feasible half-planes, identify feasible region

Step 1: Plot line \(ax + by = k\) Step 2: Test origin — if \(0 \leq k\), shade origin side Feasible region = intersection of all half-planes

3. Corner Point Method

Evaluate \(Z\) at every vertex of the feasible region; optimal is the extreme found

Fundamental Theorem: optimal value occurs at a corner vertex Evaluate \(Z\) at all corners → compare → select max or min

4. Types of Feasible Regions

Manufacturing (maximise profit), diet (minimise cost), resource allocation

Bounded (\(\leq\) constraints): max & min always exist Unbounded (\(\geq\) constraints): use open half-plane test Infeasible: no common feasible region exists
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Linear Programming — Complete One-Shot (1 hr 4 min)

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Interactive Practice

Practice MCQs — Unit 8

Select your answer, then click Show Answer to check and reveal the full explanation.

Q1Bounded LPP
A feasible region has corner points \((0,2)\), \((3,0)\), \((6,0)\), \((6,8)\), and \((0,5)\), with objective function \(Z=4x+6y\). At which point does the maximum value of \(Z\) occur?
\((0,2)\)
\((3,0)\)
\((6,8)\)
\((0,5)\)
Evaluate \(Z=4x+6y\) at each corner:
  • \((0,2)\): \(Z=12\)
  • \((3,0)\): \(Z=12\)
  • \((6,0)\): \(Z=24\)
  • \((6,8)\): \(Z=24+48=\mathbf{72}\) ← Maximum
  • \((0,5)\): \(Z=30\)
Maximum \(Z=72\) at \((6,8)\).
Q2Bounded LPP
What method should you use to find the minimum value of \(Z=3x-4y\) over the feasible region?
Pick the corner point with the highest \(y\)-coordinate
Pick the corner point with the highest \(x\)-coordinate
Pick the corner point with the lowest \(x\)-coordinate
Evaluate \(Z\) at all corner points and compare
The fundamental theorem of LPP guarantees the optimal value occurs at a corner point — but you cannot determine which corner by inspection. You must substitute each corner into \(Z\) and compare all values.
Q3Bounded LPP
Given the constraints \(x\geq0\), \(y\geq0\), \(x+y\leq5\), and \(x+2y\leq8\), what is the feasible region?
Unbounded
Bounded
Infeasible
No feasible region
Both constraints are \(\leq\) type. Combined with \(x\geq0,\ y\geq0\), the feasible region is enclosed within the first quadrant — a closed polygon. The region is bounded.
Q4Unbounded LPP
Given the constraints \(x+y\geq4\), \(x-y\geq0\), \(x\geq0\), and \(y\geq0\), what is the feasible region?
Bounded
Unbounded
Does not exist
A single point
Both constraints use \(\geq\), so the feasible region lies above both lines. In the first quadrant satisfying \(x+y\geq4\) and \(x\geq y\), the region extends infinitely. It is unbounded.
Q5Formulation
Product A needs 2 hours of labour and product B needs 3 hours, with a maximum of 12 hours available. What is the labour constraint?
\(2x+3y=12\)
\(2x+3y\leq12\)
\(2x+3y\geq12\)
\(x+y\leq12\)
Total labour \(= 2x+3y\). Since at most 12 hours are available: \(2x+3y\leq12\). Use \(\leq\) for a maximum resource limit.
Q6Formulation
What is the objective function of an LPP?
A constraint to be satisfied
A linear function to be maximised or minimised
A set of feasible values
A feasible solution itself
The objective function \(Z=c_1x+c_2y\) is the quantity we optimise — either maximise (profit) or minimise (cost). Constraints are separate restrictions.
Q7Bounded LPP
A feasible region has corner points \((0,0)\), \((4,0)\), \((4,3)\), \((2,4)\), and \((0,4)\), with objective function \(Z=5x+3y\). What is the minimum value of \(Z\)?
0
12
20
29
Evaluate \(Z=5x+3y\):
  • \((0,0)\): \(Z=\mathbf{0}\) ← Minimum
  • \((4,0)\): 20
  • \((4,3)\): 29
  • \((2,4)\): 22
  • \((0,4)\): 12
Q8Formulation
Which is NOT a requirement for a linear programming problem?
Objective function must be linear
Constraints must be linear
Decision variables must be non-negative
Objective function must be quadratic
LPP requires: (a) linear objective function, (b) linear constraints, (c) non-negative decision variables. A quadratic objective makes it a nonlinear programming problem — entirely different.
Q9Formulation
What region is represented by \(x\geq0\) and \(y\geq0\)?
First quadrant
Second quadrant
Third quadrant
Fourth quadrant
\(x\geq0\): on or right of \(y\)-axis. \(y\geq0\): on or above \(x\)-axis. Together → first quadrant where both coordinates are non-negative.
Q10Formulation
In a diet problem where we minimise cost, what does the objective function represent?
Maximum nutrients required
Total cost of the diet
Nutritional constraints
Quantity of food items
If \(x,y\) are quantities of two foods with costs \(c_1,c_2\), the objective is \(Z=c_1x+c_2y\) — the total cost. Nutritional requirements form the constraints.
Q11Unbounded LPP
If the feasible region for an LPP is unbounded, what does that imply?
Maximum value may not exist
Minimum value may not exist
Both max and min will never exist
Existence of optimal value depends on the objective function
Unbounded region does not mean no solution exists. A minimisation problem with an unbounded region may still have a minimum. Verify using the open half-plane test: if no point in the open half-plane \(Z<Z_0\) lies in the feasible region, then \(Z_0\) is the minimum.
Q12Formulation
The point \((2,3)\) lies in the feasible region of \(x+y\leq6\) because:
\(2+3 < 6\)
\(2+3 = 6\)
\(2+3 \leq 6\)
None of these
Substituting: \(2+3=5\leq6\) ✓. Since the constraint uses \(\leq\) (not strict \(<\)), boundary points also belong to the feasible region. Option (c) is the most precise statement.
Q13Formulation
A factory produces product X (₹40 profit/unit) and product Y (₹30 profit/unit). What is the objective function for maximum profit?
Maximise \(Z=40x+30y\)
Minimise \(Z=40x+30y\)
Maximise \(Z=30x+40y\)
Maximise \(Z=x+y\)
Total profit \(=40x+30y\). We want maximum profit: Maximise \(Z=40x+30y\).
Q14Formulation
What are the constraints \(x\geq0\) and \(y\geq0\) in an LPP called?
Structural constraints
Non-negativity constraints
Functional constraints
Binding constraints
\(x\geq0\) and \(y\geq0\) are non-negativity constraints. They ensure decision variables cannot be negative — which would be physically meaningless in real problems.
Q15Bounded LPP
In the Corner Point Method, where is the optimal solution found?
Always occurs at the origin
Occurs at the centre of the feasible region
Occurs at one of the corner (vertex) points of the feasible region
Can be anywhere inside the feasible region
Fundamental theorem: if an optimal solution exists, it occurs at one or more corner points (vertices) of the feasible polygon. This is why we only evaluate \(Z\) at vertices, not at infinitely many interior points.
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2–3 Mark Questions

Short Answer Questions

Step-by-step solutions. Click Show Solution to reveal each answer.

📝Self-assess your answer after revealing each solution below — your responses build your free Unit 8 AI Performance Report. Tap ✨ My Report (bottom-right) once you're done.
Q1Unbounded LPP
Maximise \(Z=x+y\), subject to: \(x\geq0,\ y\geq0,\ x-y\leq-1,\ x\leq y\).
Rewrite: \(x-y\leq-1 \Rightarrow y\geq x+1\) and \(x\leq y \Rightarrow y\geq x\).
In the first quadrant, the feasible region lies above \(y=x+1\). Both constraints are \(\geq\) type — the region extends indefinitely upward.
The feasible region is unbounded. As we move upward, \(Z=x+y\) increases without bound.
Feasible region is unbounded. Maximum value of \(Z\) does not exist — it can be made arbitrarily large.
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Q2Unbounded LPP
Minimise \(Z=200x+500y\), subject to: \(x+2y\geq10\), \(3x+4y\geq24\), \(x\geq0\), \(y\geq0\).
Line 1: \(x+2y=10\) → passes through \((10,0)\) and \((0,5)\).
Line 2: \(3x+4y=24\) → passes through \((8,0)\) and \((0,6)\).
Intersection: Substitute \(x=10-2y\) into Line 2: \(3(10-2y)+4y=24 \Rightarrow 30-2y=24 \Rightarrow y=3,\ x=4\). Corner: \((4,3)\).
Corner points: \((10,0)\), \((4,3)\), \((0,6)\). Evaluate \(Z\):
\((10,0)\): 2000  ·  \((4,3)\): \(800+1500=2300\)  ·  \((0,6)\): 3000
Minimum \(Z=\mathbf{2000}\) at \(x=10,\ y=0\).
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Q3Unbounded LPP
Minimise \(Z=3x+5y\) subject to: \(x+3y\geq3\), \(x+y\geq2\), \(x\geq0\), \(y\geq0\).
Intersection: Subtract \((x+3y=3)\) from \((x+y=2)\): \(-2y=-1 \Rightarrow y=0.5,\ x=1.5\). Corner: \((1.5,0.5)\).
Corner points: \((3,0)\), \((1.5,0.5)\), \((0,2)\). Evaluate \(Z\):
\((3,0)\): 9  ·  \((1.5,0.5)\): \(4.5+2.5=7\)  ·  \((0,2)\): 10
Minimum \(Z=\mathbf{7}\) at \(x=1.5,\ y=0.5\).
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Q4Bounded LPP
Minimise and Maximise \(Z=x+2y\) subject to: \(x+2y\geq100\), \(2x-y\leq0\), \(2x+y\leq200\), \(x\geq0\), \(y\geq0\).
Rewrite \(2x-y\leq0\) as \(y\geq2x\). Lines: \(L_1: x+2y=100\), \(L_2: y=2x\), \(L_3: 2x+y=200\).
\(L_1\cap L_2\): \(x+2(2x)=100 \Rightarrow 5x=100 \Rightarrow x=20,\ y=40\). Point: \((20,40)\).
\(L_2\cap L_3\): \(2x+2x=200 \Rightarrow x=50,\ y=100\). Point: \((50,100)\).
Evaluate: At \((20,40)\): \(Z=20+80=100\). At \((50,100)\): \(Z=50+200=250\).
Minimum \(Z=\mathbf{100}\) at \((20,40)\). Maximum \(Z=\mathbf{250}\) at \((50,100)\).
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Q5Unbounded LPP
Minimise \(Z=x+2y\) subject to: \(2x+y\geq3\), \(x+2y\geq6\), \(x\geq0\), \(y\geq0\).
Corner points: \((6,0)\) and \((0,3)\).
Evaluate: At \((6,0)\): \(Z=6\). At \((0,3)\): \(Z=6\). Both give the same value.
Minimum \(Z=\mathbf{6}\) — occurs at every point on the segment joining \((6,0)\) and \((0,3)\). Infinitely many optimal solutions.
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Q6Bounded LPP
Maximise \(Z=x+2y\) subject to: \(x-y\leq0\), \(2y\leq x+2\), \(x\geq0\), \(y\geq0\).
Rewrite: \(y\geq x\) and \(y\leq\dfrac{x+2}{2}\). Feasible region: \(x\leq y\leq\dfrac{x+2}{2}\) in first quadrant.
Intersection of \(y=x\) and \(y=\dfrac{x+2}{2}\): \(x=\dfrac{x+2}{2} \Rightarrow 2x=x+2 \Rightarrow x=2,\ y=2\). Corner: \((2,2)\).
Evaluate: At \((0,0)\): 0. At \((0,1)\): 2. At \((2,2)\): \(2+4=6\).
Maximum \(Z=\mathbf{6}\) at \(x=2,\ y=2\).
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4–6 Mark Questions

Long Answer Questions

Graphical LPP problems with complete formulation and corner point evaluation. Click Show Solution to reveal full working.

LA Q1Formulation
A furniture manufacturer makes chairs and tables. Each chair requires 2 hours on Machine A and 6 hours on Machine B, while each table requires 5 hours on Machine A and 0 hours on Machine B. Machine A is available for 16 hours a day and Machine B for 30 hours a day. The profit is ₹2 per chair and ₹10 per table. Formulate this as a linear programming problem to maximise profit.
Let \(x\) = chairs/day, \(y\) = tables/day.
Objective: Maximise \(Z=2x+10y\)
Machine A: \(2x+5y\leq16\)
Machine B: \(6x\leq30 \Rightarrow x\leq5\)
Non-negativity: \(x\geq0,\ y\geq0\)
LPP: Maximise \(Z=2x+10y\) subject to \(2x+5y\leq16\), \(x\leq5\), \(x\geq0\), \(y\geq0\).
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LA Q2Formulation
A diet must contain at least 80 units of vitamin A and at least 100 units of minerals. Food F1 costs ₹4/kg and provides 3 units of vitamin A and 4 units of minerals, while Food F2 costs ₹6/kg and provides 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem to minimise cost.
Let \(x\) = kg of F1, \(y\) = kg of F2.
Objective: Minimise \(Z=4x+6y\)
Vitamin A: \(3x+6y\geq80\)
Minerals: \(4x+3y\geq100\)
Non-negativity: \(x\geq0,\ y\geq0\)
LPP: Minimise \(Z=4x+6y\) subject to \(3x+6y\geq80\), \(4x+3y\geq100\), \(x\geq0\), \(y\geq0\).
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LA Q3Bounded LPP
A firm manufactures gold rings (\(x\)) and chains (\(y\)), with combined daily production limited to at most 24 units. Each ring takes 1 hour to make and each chain takes ½ hour, with 16 hours available per day. The profit is ₹300 per ring and ₹190 per chain. Maximise the profit graphically.
Maximise \(Z=300x+190y\). Constraints: \(x+y\leq24\), \(x+0.5y\leq16\), \(x\geq0\), \(y\geq0\).
Intersection of \(x+y=24\) and \(x+0.5y=16\): subtract → \(0.5y=8 \Rightarrow y=16,\ x=8\). Corner: \((8,16)\).
All corners: \((0,0)\), \((16,0)\), \((8,16)\), \((0,24)\). Evaluate \(Z\):
\((0,0)\): 0  ·  \((16,0)\): 4800  ·  \((8,16)\): \(2400+3040=\mathbf{5440}\)  ·  \((0,24)\): 4560
Maximum profit = ₹5,440 — produce 8 rings and 16 chains per day.
Self-assess: ✓ Saved
LA Q4Unbounded LPP
Reshma mixes food P, which costs ₹60/kg and provides 3 units of vitamin A and 5 units of vitamin B, with food Q, which costs ₹80/kg and provides 4 units of vitamin A and 2 units of vitamin B. The mix must provide a minimum of 8 units of vitamin A and 11 units of vitamin B. Find the minimum cost mix.
Minimise \(Z=60x+80y\). Constraints: \(3x+4y\geq8\), \(5x+2y\geq11\), \(x\geq0\), \(y\geq0\).
Intersection: Multiply \(5x+2y=11\) by 2: \(10x+4y=22\). Subtract \(3x+4y=8\): \(7x=14 \Rightarrow x=2,\ y=0.5\). Point: \((2,0.5)\).
Other corners: \((\tfrac{8}{3},0)\) and \((0,5.5)\). Evaluate:
\((\tfrac{8}{3},0)\): ₹160  ·  \((2,0.5)\): \(120+40=\)₹160  ·  \((0,5.5)\): ₹440
Minimum cost = ₹160, achieved at all points on the segment between \((\tfrac{8}{3},0)\) and \((2,0.5)\).
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LA Q5Bounded LPP
A company manufactures products A and B. Each unit of A needs 3 hours on machine M1 and 2 hours on machine M2, while each unit of B needs 2 hours on M1 and 4 hours on M2. M1 is available for 18 hours a day and M2 for 24 hours a day. The profit is ₹50 per unit of A and ₹40 per unit of B. Find the maximum profit graphically.
Maximise \(Z=50x+40y\). M1: \(3x+2y\leq18\). M2: \(x+2y\leq12\). Non-negativity: \(x\geq0,\ y\geq0\).
Intersection: Subtract \(x+2y=12\) from \(3x+2y=18\): \(2x=6 \Rightarrow x=3,\ y=4.5\). Corner: \((3,4.5)\).
All corners: \((0,0)\), \((6,0)\), \((3,4.5)\), \((0,6)\). Evaluate:
\((0,0)\): 0  ·  \((6,0)\): 300  ·  \((3,4.5)\): \(150+180=\mathbf{330}\)  ·  \((0,6)\): 240
Maximum profit = ₹330/day — produce 3 units of A and 4.5 units of B.
Self-assess: ✓ Saved
LA Q6Formulation
A company manufactures two models of bikes. Model X requires 6 man-hours, costs ₹2000 in handling, and earns ₹1000 profit, while Model Y requires 10 man-hours, costs ₹3000 in handling, and earns ₹1500 profit. The company has 450 man-hours and ₹80,000 in funds available per week. Formulate this as a linear programming problem and solve it to maximise profit.
Let \(x\) = units of X, \(y\) = units of Y. Maximise \(Z=1000x+1500y\).
Man-hours: \(6x+10y\leq450 \Rightarrow 3x+5y\leq225\).
Funds: \(2000x+3000y\leq80000 \Rightarrow 2x+3y\leq80\).
Non-negativity: \(x\geq0,\ y\geq0\).
Corner points: \((0,0)\), \((40,0)\), \((0,26.\overline{6})\). Evaluate:
\((0,0)\): 0  ·  \((40,0)\): ₹40,000  ·  \((0,26.\overline{6})\): ₹40,000
LPP: Maximise \(Z=1000x+1500y\) s.t. \(3x+5y\leq225\), \(2x+3y\leq80\), \(x\geq0\), \(y\geq0\).
Maximum profit ≈ ₹40,000/week — verify corner points graphically.
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4-Mark Questions

Case Study Questions

Board-pattern case studies with full narrative. Read the context carefully, then click Show Solution under each part.

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Case Study 1: Manufacturing Optimisation

PrecisionTech Industries manufactures two products, P1 and P2, using two machines. Product P1 requires 4 hours on Machine M1 and 2 hours on Machine M2, earning a profit of ₹100 per unit. Product P2 requires 3 hours on Machine M1 and 3 hours on Machine M2, earning a profit of ₹120 per unit.

Each week, Machine M1 is available for a maximum of 60 hours and Machine M2 for a maximum of 48 hours.

Let \(x\) = units of P1 and \(y\) = units of P2. Answer the following questions:

(i)Formulation

Write the objective function for this problem.

Profit per unit P1 = ₹100, profit per unit P2 = ₹120.
Objective: Maximise \(Z = 100x + 120y\)
Self-assess: ✓ Saved
(ii)Formulation

Write all constraints including non-negativity constraints.

Machine M1: \(4x+3y\leq60\)
Machine M2: \(2x+3y\leq48\)
\(4x+3y\leq60\),   \(2x+3y\leq48\),   \(x\geq0\),   \(y\geq0\)
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(iii)Bounded LPP

Given the corner points \((0,0)\), \((15,0)\), \((0,16)\), and \((12,4)\), find the maximum profit and the optimal production plan.

Evaluate \(Z=100x+120y\):
\((0,0)\): ₹0  ·  \((15,0)\): ₹1,500  ·  \((0,16)\): ₹1,920  ·  \((12,4)\): ₹1,680
Maximum profit = ₹1,920/week — produce 0 units of P1 and 16 units of P2.
Self-assess: ✓ Saved
🌱

Case Study 2: Diet Problem

A nutritionist is designing a diet plan requiring at least 8 units of vitamin A and at least 10 units of vitamin C per day. Food I (₹50/kg): 2 units vitamin A, 1 unit vitamin C. Food II (₹70/kg): 1 unit vitamin A, 2 units vitamin C.

Let \(x\) = kg of Food I and \(y\) = kg of Food II per day.

(i)Formulation

Write the objective function and all constraints.

Minimise \(Z=50x+70y\)   s.t.   \(2x+y\geq8\),   \(x+2y\geq10\),   \(x\geq0\),   \(y\geq0\)
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(ii)Unbounded LPP

Is the feasible region bounded or unbounded?

The feasible region is unbounded — both constraints are \(\geq\) type. However a minimum cost still exists (verified by the open half-plane test).
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(iii)Unbounded LPP

Given the corner points \((0,10)\), \((2,4)\), and \((8,0)\), find the minimum cost and the optimal quantities.

Evaluate \(Z=50x+70y\):
\((0,10)\): ₹700  ·  \((2,4)\): ₹380  ·  \((8,0)\): ₹400
Minimum cost = ₹380/day — use 2 kg of Food I and 4 kg of Food II.
Self-assess: ✓ Saved
🎒

Case Study 3: Resource Allocation — Leather Belt Manufacturing

Craftex Leathers manufactures two varieties: Belt A (superior, ₹40 profit) and Belt B (standard, ₹30 profit). Belt A takes twice as long to make as Belt B, and the factory's daily capacity is equivalent to 1,000 units of Belt B. The daily leather supply is enough for 800 belts, and only 400 fancy buckles (used exclusively for Belt A) are available per day.

Let \(x\) = Belt A, \(y\) = Belt B per day.

(i)Formulation

Write the objective function.

Objective: Maximise \(Z = 40x + 30y\)
Self-assess: ✓ Saved
(ii)Formulation

Write all structural constraints and non-negativity constraints.

\(2x+y\leq1000\),   \(x+y\leq800\),   \(x\leq400\),   \(x\geq0\),   \(y\geq0\)
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(iii)Bounded LPP

Given the corner points \((0,0)\), \((400,0)\), \((400,200)\), \((200,600)\), and \((0,800)\), find the optimal production plan and the maximum profit.

Evaluate \(Z=40x+30y\):
\((0,0)\): 0  ·  \((400,0)\): ₹16,000  ·  \((400,200)\): ₹22,000
\((200,600)\): ₹26,000 ← Maximum  ·  \((0,800)\): ₹24,000
Maximum profit = ₹26,000/day — produce 200 belts A and 600 belts B.
Self-assess: ✓ Saved
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Score Full Marks

Exam Tips — Unit 8

The most common mistakes in LPP board exam questions — and exactly how to avoid them.

✓ Tip 1 — Find Corner Points Algebraically, Not Graphically

Always solve simultaneous equations exactly to find corner points — never estimate from a graph. When two constraint lines intersect, set them equal and solve. A graphical estimate is not acceptable in board exams and will cost you method marks.

🔒  4 more exam tips for Unit 8 — how to present LPP formulation for full marks, the open half-plane test for unbounded regions, how to identify when the optimal solution occurs along an edge, and the most common constraint sign errors — are included in the AI Question Bank.

🤖 Get All Tips in AI Q-Bank →

✕ Common Mistakes to Avoid in Unit 8

  • Reading corner points off a graph instead of solving equations exactly
  • Forgetting to evaluate \(Z\) at all corner points
  • Writing \(=\) instead of \(\leq\) or \(\geq\) in constraint formulation
  • Not writing non-negativity constraints \(x\geq0,\ y\geq0\) explicitly
  • Declaring an optimal value for an unbounded region without the half-plane test
  • Swapping the objective function and constraints in the formulation
  • Not stating clearly “Maximise” or “Minimise” before \(Z\)
Common Questions

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A Linear Programming Problem involves optimising (maximising or minimising) a linear objective function subject to a set of linear constraints and non-negativity restrictions on decision variables.
Based on the fundamental theorem: if an optimal value exists, it must occur at one or more corner (vertex) points. Evaluate \(Z\) at every corner and compare.
A bounded region is a closed polygon (all \(\leq\) constraints). Both max and min always exist. An unbounded region extends infinitely (\(\geq\) constraints). Verify the optimal value using the open half-plane test.
Unit 8 carries 8 marks in the CBSE Class 12 Applied Mathematics board exam.
Step 1: Identify decision variables. Step 2: Write the objective function. Step 3: Write structural constraints. Step 4: Add non-negativity: \(x\geq0,\ y\geq0\).
Yes. If the objective function line is parallel to a constraint edge, the optimal value is the same at two adjacent corners — infinitely many optimal solutions exist along that edge.
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Linear Programming Problems
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