This MCQ module is based on: Bounded, Unbounded Regions and Applications
TOPIC 22 OF 25
Bounded, Unbounded Regions and Applications
🎓 Class 12
Mathematics
CBSE
Theory
Ch 12 — Linear Programming
⏱ ~15 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Bounded, Unbounded Regions and Applications
Targeting Class 12 level in Linear Programming, with Advanced difficulty.
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12.4 Different Types of LPPs
LPPs appear in many real-world contexts. We will examine the main families and how each is formulated.
12.4.1 Manufacturing Problems
Decide quantities of each product to manufacture so that profit is maximized (or cost minimized), subject to machine time, raw material, and labour availability.
12.4.2 Diet Problems
Determine the amount of each food to consume so that nutritional requirements are met at minimum cost.
12.4.3 Transportation Problems
Plan deliveries from factories to retail outlets so the total transportation cost is minimised while meeting supply and demand.
12.5 Bounded and Unbounded Feasible Regions
Fig 12.2: Bounded vs Unbounded feasible regions
Theorem (Existence of Optimum)
Bounded region: both max and min of \(Z\) always exist, attained at corner points.Unbounded region: max or min may not exist. If \(M\) is the largest corner-point value:
- \(M\) is the maximum only if the open half-plane \(ax+by > M\) has no point in common with the feasible region.
- Otherwise \(Z\) has no maximum.
Example 4 — Manufacturing (Bounded)
A toy company makes two types of dolls D₁ and D₂. D₁ needs 2 hrs of machine time and 1 hr of craftsmanship; D₂ needs 1 hr of machine time and 2 hrs of craftsmanship. The company has at most 80 machine-hrs and 70 craft-hrs per day. D₁ profit = ₹20, D₂ profit = ₹30. Maximize profit.
Let \(x,y\) = dolls. Max \(Z = 20x + 30y\). Constraints: \(2x+y\le 80,\; x+2y\le 70,\; x,y\ge 0\).
Corners: (0,0), (40,0), intersection of \(2x+y=80\) and \(x+2y=70\): \(3x=90 ⇒ x=30, y=20\), and (0,35).
Z: 0, 800, 600+600=1200, 1050. Max = ₹1,200 at (30, 20).
Corners: (0,0), (40,0), intersection of \(2x+y=80\) and \(x+2y=70\): \(3x=90 ⇒ x=30, y=20\), and (0,35).
Z: 0, 800, 600+600=1200, 1050. Max = ₹1,200 at (30, 20).
Example 5 — Unbounded Region with Minimum
Minimize \(Z = 3x + 2y\) subject to \(x + y \ge 8,\; 3x + 5y \le 15,\; x \ge 0, y \ge 0\).
Check: \(x+y\ge 8\) means region above line \(x+y=8\). \(3x+5y\le 15\) means below \(3x+5y=15\). At \((0,8)\): \(3(0)+5(8)=40 > 15\) ✗; at \((8,0)\): \(3(8)=24>15\) ✗. The two regions do not intersect in the first quadrant. Feasible region is empty — no solution exists.
Example 6 — Transportation
A company has two warehouses W₁ (stocks 100 units) and W₂ (stocks 150 units). Three dealers A, B, C need 60, 50, 40 units. Cost of shipping (₹/unit): from W₁ → A=6, B=3, C=2.50; from W₂ → A=4, B=2, C=3. How many units should ship from W₁ to A (=\(x\)) and W₁ to B (=\(y\)) to minimize cost? (All demand must be met.)
W₁→C = \(100-x-y\); W₂→A = \(60-x\); W₂→B = \(50-y\); W₂→C = \(40-(100-x-y) = x+y-60\).
Non-negativity: \(0\le x\le 60\), \(0\le y\le 50\), \(x+y\le 100\), \(x+y\ge 60\).
Cost \(Z = 6x + 3y + 2.5(100-x-y) + 4(60-x) + 2(50-y) + 3(x+y-60)\)
\(= 6x+3y+250-2.5x-2.5y+240-4x+100-2y+3x+3y-180\) = \(2.5x + 1.5y + 410\). Minimize.
Corner points: (60,0), (60,40), (50,50), (10,50), (60,0)... Evaluate: (60,0)→560; (10,50)→510; (50,50)→610. Min = ₹510 at (10, 50), with shipments W₁→A=10, W₁→B=50, W₁→C=40, W₂→A=50, W₂→B=0, W₂→C=0.
Non-negativity: \(0\le x\le 60\), \(0\le y\le 50\), \(x+y\le 100\), \(x+y\ge 60\).
Cost \(Z = 6x + 3y + 2.5(100-x-y) + 4(60-x) + 2(50-y) + 3(x+y-60)\)
\(= 6x+3y+250-2.5x-2.5y+240-4x+100-2y+3x+3y-180\) = \(2.5x + 1.5y + 410\). Minimize.
Corner points: (60,0), (60,40), (50,50), (10,50), (60,0)... Evaluate: (60,0)→560; (10,50)→510; (50,50)→610. Min = ₹510 at (10, 50), with shipments W₁→A=10, W₁→B=50, W₁→C=40, W₂→A=50, W₂→B=0, W₂→C=0.
Important Observations
Multiple optima: if \(Z = ax+by\) has the same value at two adjacent corner points, every point on the edge joining them is also optimal — infinitely many optima.Integer LPPs: if \(x, y\) must be integers (e.g., number of dolls), the continuous optimum may not be integer — rounding is needed, or advanced techniques (Branch-and-Bound).
Activity: Bounded vs Unbounded Visualisation
L4 AnalyseMaterials: Graph paper, ruler, two coloured pencils.
- On graph paper, shade the region \(x+y\le 10,\; x\ge 0, y\ge 0\) — it is a bounded triangle.
- Next, shade \(x+y\ge 10,\; x\ge 0, y\ge 0\) — unbounded, extending northeast.
- Pick any linear \(Z=2x+3y\). Compute Z at the corner points of region 1 — max/min both exist.
- In region 2, \(Z\) has a finite minimum but no maximum (it grows indefinitely).
- Draw the line \(2x+3y = m\) (minimum) in region 2 and verify the open half-plane \(2x+3y
Unbounded regions share features with bounded ones — optimum values at vertices — but existence is not guaranteed. The trick is to test adjacent half-planes against the region.
Miscellaneous Exercise — Selected
Q1. A dietician wants to develop a special diet using two foods P and Q. Each packet (30g) of P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol, 6 units of vit A. Same units for Q: 3, 20, 4, 3. The diet requires at least 240 units of calcium, 460 of iron, and at most 300 of cholesterol. How many packets of each minimize vitamin A (\(Z=6x+3y\))?
Constraints: \(12x+3y\ge240,\;4x+20y\ge460,\;6x+4y\le300,\;x,y\ge0\). Simplify first two: \(4x+y\ge80,\;x+5y\ge115\). Corner analysis yields \(x=15, y=20\). Min \(Z=6(15)+3(20)=150\) units.
Q2. A manufacturer makes bolts and nuts. 1 kg bolts takes 3 hr on machine A and 1 hr on machine B; 1 kg nuts takes 1 hr on A and 3 hr on B. Profit: ₹17.50/kg bolts, ₹7/kg nuts. Machines run max 12 hr/day. Maximize profit.
Max \(Z=17.5x+7y\), \(3x+y\le12, x+3y\le12, x,y\ge0\). Corners: (0,0),(4,0),(3,3),(0,4). Z: 0, 70, 73.5, 28. Max = ₹73.50 at (3, 3).
Q3. There are two types of fertilizers F₁ (10% N, 6% P) and F₂ (5% N, 10% P). A farmer needs at least 14 kg N and 14 kg P for his crop. F₁ costs ₹6/kg, F₂ ₹5/kg. Minimize cost.
Min \(Z=6x+5y\), \(0.1x+0.05y\ge14\) (i.e. \(2x+y\ge280\)), \(0.06x+0.1y\ge14\) (i.e. \(3x+5y\ge700\)), \(x,y\ge0\). Intersection: from \(2x+y=280,\;3x+5y=700\): \(x=100,y=80\). Corners: (280,0), (100,80), (0,280). Z: 1680, 600+400=1000, 1400. Min = ₹1,000 at (100, 80).
Competency-Based Questions
Scenario: A hospital cafeteria serves two kinds of nutrition bars — A and B. A has 4 g protein and 2 g fibre; B has 2 g protein and 4 g fibre. Each patient needs at least 16 g protein and 20 g fibre per day. Cost: ₹30 per bar A, ₹40 per bar B. Let \(x, y\) be the number of each.
Q1. Formulate the LPP and find the feasible region's corner points.
L3 ApplyMin \(Z=30x+40y\), \(4x+2y\ge16\) (\(2x+y\ge8\)), \(2x+4y\ge20\) (\(x+2y\ge10\)), \(x,y\ge0\). Corners: (10,0), intersection of \(2x+y=8,\;x+2y=10\): \(3y=12 ⇒ y=4, x=2\), and (0,8).
Q2. Analyse whether the feasible region is bounded or unbounded.
L4 AnalyseBoth constraints are \(\ge\) in the first quadrant — region extends to infinity northeast. Unbounded.
Q3. Evaluate the minimum cost and confirm it is truly a minimum despite the unbounded region.
L5 EvaluateZ-values: (10,0)→300, (2,4)→220, (0,8)→320. Smallest = 220. Check \(30x+40y<220\) has no feasible point — the half-plane lies below all constraint lines in the first quadrant. Min cost = ₹220 at (2, 4) — valid.
Q4. Design a revised LPP: add a budget constraint "\(30x+40y \le 500\)", making the feasible region bounded. Re-compute the minimum.
L6 CreateNow region bounded between \(2x+y\ge8,\;x+2y\ge10,\;30x+40y\le500,\;x,y\ge0\). Corner points include (2,4) still feasible (80+160=240<500). Min Z at (2,4)=220 unchanged — budget constraint is non-binding at the optimum.
Assertion–Reason Questions
A: In an unbounded feasible region, \(Z=ax+by\) always has a minimum if \(a,b>0\) and the region lies in the first quadrant.
R: Both coefficients positive and \(x,y\ge0\) force \(Z\ge0\), so 0 is a lower bound.
R: Both coefficients positive and \(x,y\ge0\) force \(Z\ge0\), so 0 is a lower bound.
(a) Bounded-below + continuous linear ⇒ attains minimum at a corner of the feasible region. R explains A.
A: If two corner points give the same maximum, every point on the edge joining them is also optimal.
R: A linear function is affine, so on a line segment it is either monotone or constant.
R: A linear function is affine, so on a line segment it is either monotone or constant.
(a) Equal endpoints ⇒ constant along segment ⇒ every point is optimal. R explains A.
A: Every LPP can be solved graphically.
R: The graphical method uses 2D plotting of half-planes.
R: The graphical method uses 2D plotting of half-planes.
(d) A false — graphical method works only for 2 (or maybe 3) variables; LPPs with many variables need Simplex. R true.
Frequently Asked Questions — Linear Programming
What is Part 2 — Bounded & Unbounded Regions, Applications | Class 12 Maths Ch 12 | MyAiSchool in NCERT Class 12 Mathematics?
Part 2 — Bounded & Unbounded Regions, Applications | Class 12 Maths Ch 12 | MyAiSchool is a key concept covered in NCERT Class 12 Mathematics, Chapter 12: Linear Programming. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Part 2 — Bounded & Unbounded Regions, Applications | Class 12 Maths Ch 12 | MyAiSchool step by step?
To solve problems on Part 2 — Bounded & Unbounded Regions, Applications | Class 12 Maths Ch 12 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 12: Linear Programming?
The essential formulas of Chapter 12 (Linear Programming) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Part 2 — Bounded & Unbounded Regions, Applications | Class 12 Maths Ch 12 | MyAiSchool important for the Class 12 board exam?
Part 2 — Bounded & Unbounded Regions, Applications | Class 12 Maths Ch 12 | MyAiSchool is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Part 2 — Bounded & Unbounded Regions, Applications | Class 12 Maths Ch 12 | MyAiSchool?
Common mistakes in Part 2 — Bounded & Unbounded Regions, Applications | Class 12 Maths Ch 12 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Part 2 — Bounded & Unbounded Regions, Applications | Class 12 Maths Ch 12 | MyAiSchool?
End-of-chapter NCERT exercises for Part 2 — Bounded & Unbounded Regions, Applications | Class 12 Maths Ch 12 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 12, and solve at least one previous-year board paper to consolidate your understanding.
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