Unit 8 — Linear Programming Problems (LPP) — carries 8 marks in the CBSE Class 12 Applied Maths board examination, making it one of the higher-weightage units in the course.
The unit covers four key areas: formulation of LPPs by identifying decision variables, the objective function and constraints from real-world word problems; the graphical solution method for plotting constraint lines and shading the feasible region; identification of bounded and unbounded feasible regions; and the Corner Point Method for locating the optimal solution by evaluating the objective function at every vertex of the feasible region.
This page includes 15 MCQs with full solutions, 6 short-answer questions with step-by-step workings, 6 long-answer problems, and 3 case studies — covering manufacturing problems, diet problems, and resource allocation scenarios.
All questions and solutions are aligned to the latest CBSE Applied Maths syllabus 2026–27.
The corner points of the feasible region for an LPP are \((0, 2)\), \((3, 0)\), \((6, 0)\), \((6, 8)\) and \((0, 5)\). If the objective function is \(Z = 4x + 6y\), then the maximum value of \(Z\) occurs at:
A\((0, 2)\)
B\((3, 0)\)
C\((6, 8)\)
D\((0, 5)\)
✓ Correct Answer: (C) \((6, 8)\)
Evaluate \(Z = 4x + 6y\) at each corner point:
At \((0, 2)\): \(Z = 4(0) + 6(2) = 12\)
At \((3, 0)\): \(Z = 4(3) + 6(0) = 12\)
At \((6, 0)\): \(Z = 4(6) + 6(0) = 24\)
At \((6, 8)\): \(Z = 4(6) + 6(8) = 24 + 48 = \mathbf{72}\) ← Maximum
At \((0, 5)\): \(Z = 4(0) + 6(5) = 30\)
The highest value is 72, which occurs at \((6, 8)\).
Question 2
The objective function is \(Z = 3x - 4y\). To find the minimum value of \(Z\) over the feasible region, you should:
APick the corner point with the highest \(y\)-coordinate
BPick the corner point with the highest \(x\)-coordinate
CPick the corner point with the lowest \(x\)-coordinate
DEvaluate \(Z\) at all corner points and compare
✓ Correct Answer: (D) Evaluate \(Z\) at all corner points and compare
The fundamental theorem of LPP guarantees that the optimal value (if it exists) occurs at a corner point. However, you cannot determine which corner point gives the minimum just by looking at coordinates — you must substitute each corner point into \(Z\) and compare the results.
Question 3
The constraints \(x \geq 0\), \(y \geq 0\), \(x + y \leq 5\), \(x + 2y \leq 8\) define:
AUnbounded feasible region
BBounded feasible region
CInfeasible region
DNo feasible region
✓ Correct Answer: (B) Bounded feasible region
Both constraints are of the \(\leq\) type. Combined with \(x \geq 0,\ y \geq 0\), the feasible region is enclosed within the first quadrant. It forms a closed polygon (a bounded region) — you cannot move infinitely far in any direction and remain in the feasible region.
Question 4
If the constraints are \(x + y \geq 4\), \(x - y \geq 0\), \(x \geq 0\), \(y \geq 0\), then the feasible region is:
ABounded
BUnbounded
CDoes not exist
DA single point
✓ Correct Answer: (B) Unbounded
Both constraints use \(\geq\), meaning the feasible region lies above both lines. The region satisfying \(x + y \geq 4\) and \(x \geq y\) in the first quadrant extends infinitely (there is no upper boundary), making the feasible region unbounded.
Question 5
A manufacturer makes two products A and B. Product A requires 2 hours of labour and product B requires 3 hours. Maximum 12 hours of labour are available. Which constraint represents this?
A\(2x + 3y = 12\)
B\(2x + 3y \leq 12\)
C\(2x + 3y \geq 12\)
D\(x + y \leq 12\)
✓ Correct Answer: (B) \(2x + 3y \leq 12\)
Total labour used = \(2x + 3y\) (2 hours per unit of A, 3 hours per unit of B). Since at most 12 hours are available, the constraint is \(2x + 3y \leq 12\). We use \(\leq\) because we cannot exceed the maximum available hours.
Question 6
The objective function of a linear programming problem is:
AA constraint to be satisfied
BA function to be optimized (maximised or minimised)
CA set of feasible values
DA feasible solution itself
✓ Correct Answer: (B) A function to be optimized
In an LPP, the objective function \(Z = c_1x + c_2y\) is the quantity we want to either maximise (e.g., profit) or minimise (e.g., cost). The constraints are the restrictions that must be satisfied. These are two distinct components of every LPP.
Question 7
If the corner points of the feasible region are \((0, 0)\), \((4, 0)\), \((4, 3)\), \((2, 4)\) and \((0, 4)\), and \(Z = 5x + 3y\), then the minimum value of \(Z\) is:
A0
B12
C20
D29
✓ Correct Answer: (A) 0
Evaluate \(Z = 5x + 3y\) at each corner:
At \((0, 0)\): \(Z = 0\) ← Minimum
At \((4, 0)\): \(Z = 20\)
At \((4, 3)\): \(Z = 20 + 9 = 29\)
At \((2, 4)\): \(Z = 10 + 12 = 22\)
At \((0, 4)\): \(Z = 0 + 12 = 12\)
The minimum is 0, occurring at the origin \((0, 0)\).
Question 8
Which of the following is NOT a requirement for a linear programming problem?
AObjective function must be linear
BConstraints must be linear
CDecision variables must be non-negative
DObjective function must be quadratic
✓ Correct Answer: (D) Objective function must be quadratic
Linear Programming requires: (a) a linear objective function, (b) linear constraints, and (c) non-negative decision variables. A quadratic objective function would make it a nonlinear programming problem — that is an entirely different type of optimisation problem.
Question 9
The region represented by \(x \geq 0\), \(y \geq 0\) is:
AFirst quadrant
BSecond quadrant
CThird quadrant
DFourth quadrant
✓ Correct Answer: (A) First quadrant
\(x \geq 0\) means we are on or to the right of the \(y\)-axis. \(y \geq 0\) means we are on or above the \(x\)-axis. Together, these restrict us to the first quadrant (where both coordinates are non-negative). This is why all LPPs in Class 12 are solved in the first quadrant.
Question 10
In a diet problem where we want to minimise cost, the objective function represents:
AMaximum nutrients required
BTotal cost of the diet
CNutritional constraints
DQuantity of food items
✓ Correct Answer: (B) Total cost of the diet
In a diet problem, if \(x\) and \(y\) are the quantities of two foods with costs \(c_1\) and \(c_2\) per unit, the objective function is \(Z = c_1 x + c_2 y\) (total cost to minimise). The nutritional requirements (e.g., minimum vitamins, minerals) form the constraints, not the objective function.
Question 11
If the feasible region for an LPP is unbounded, then:
AMaximum value of objective function may not exist
BMinimum value of objective function may not exist
CBoth maximum and minimum will never exist
DExistence of optimal value depends on the objective function
✓ Correct Answer: (D) Depends on the objective function
An unbounded feasible region does not automatically mean no optimal solution exists. For example, a minimisation problem with an unbounded region may still have a minimum — you just need to verify by checking whether any point in the open half-plane (formed by the objective function at the candidate optimum) lies inside the feasible region. If it does, no finite optimum exists; if it doesn't, the candidate value is optimal.
Question 12
The point \((2, 3)\) lies in the feasible region of \(x + y \leq 6\) because:
A\(2 + 3 < 6\)
B\(2 + 3 = 6\)
C\(2 + 3 \leq 6\)
DNone of these
✓ Correct Answer: (C) \(2 + 3 \leq 6\)
Substituting \(x = 2,\ y = 3\) into the constraint: \(2 + 3 = 5 \leq 6\). Since \(5 \leq 6\) is true, the point satisfies the constraint and lies in the feasible region. Option (C) is the most precise because the constraint uses \(\leq\) (not strict \(<\)), so points on the boundary line also belong to the feasible region.
Question 13
A factory produces two products X and Y. Product X gives a profit of ₹40 per unit and Y gives ₹30 per unit. The objective function for maximum profit is:
AMaximise \(Z = 40x + 30y\)
BMinimise \(Z = 40x + 30y\)
CMaximise \(Z = 30x + 40y\)
DMaximise \(Z = x + y\)
✓ Correct Answer: (A) Maximise \(Z = 40x + 30y\)
Total profit = (profit per unit of X) × (units of X) + (profit per unit of Y) × (units of Y) = \(40x + 30y\). Since we want maximum profit, the objective is to Maximise \(Z = 40x + 30y\).
Question 14
The constraints \(x \geq 0\), \(y \geq 0\) are called:
AStructural constraints
BNon-negativity constraints
CFunctional constraints
DBinding constraints
✓ Correct Answer: (B) Non-negativity constraints
\(x \geq 0\) and \(y \geq 0\) are specifically called non-negativity constraints. They ensure that decision variables (like number of units produced) cannot take negative values, which would be physically meaningless in real-world problems.
Question 15
In the Corner Point Method, the optimal solution:
AAlways occurs at the origin
BOccurs at the centre of the feasible region
COccurs at one of the corner (vertex) points
DCan be anywhere inside the region
✓ Correct Answer: (C) Occurs at one of the corner (vertex) points
This is the fundamental theorem of linear programming: if an optimal solution exists for a bounded feasible region, it must occur at one (or more) of the corner points (vertices) of the feasible polygon. This is why the Corner Point Method works — we only need to evaluate \(Z\) at the vertices, not at infinitely many interior points.
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Line 1: \(x + 3y = 3\) → \((3, 0)\) and \((0, 1)\). Line 2: \(x + y = 2\) → \((2, 0)\) and \((0, 2)\).
Find intersection: Subtract Line 1 from Line 2: \((x+y) - (x+3y) = 2-3 \Rightarrow -2y = -1 \Rightarrow y = 0.5,\ x = 1.5\). Corner point: \((1.5, 0.5)\).
The feasible region is unbounded (all \(\geq\) constraints), but a minimum may still exist. Corner points: \((3, 0)\), \((1.5, 0.5)\), \((0, 2)\).
Evaluate \(Z\): At \((3, 0)\): \(Z = 9\). At \((1.5, 0.5)\): \(Z = 4.5 + 2.5 = 7\). At \((0, 2)\): \(Z = 0 + 10 = 10\).
✓ Minimum \(Z = \mathbf{7}\) at \(x = 1.5,\ y = 0.5\).
Question 4
Minimise and Maximise \(Z = x + 2y\) subject to: \(x + 2y \geq 100\), \(2x - y \leq 0\), \(2x + y \leq 200\), \(x \geq 0\), \(y \geq 0\).
Rewrite \(2x - y \leq 0\) as \(y \geq 2x\). The three boundary lines are \(L_1: x + 2y = 100\), \(L_2: y = 2x\), \(L_3: 2x + y = 200\).
Intersection of \(L_1\) and \(L_2\): Substitute \(y = 2x\) into \(x + 2(2x) = 100 \Rightarrow 5x = 100 \Rightarrow x = 20,\ y = 40\). Point: \((20, 40)\).
Intersection of \(L_2\) and \(L_3\): Substitute \(y = 2x\) into \(2x + 2x = 200 \Rightarrow x = 50,\ y = 100\). Point: \((50, 100)\).
Also check \((0, 100)\) (on \(L_1\) with \(x=0\)) and \((0, 200)\) (on \(L_3\) with \(x=0\))... Verify feasibility of each candidate corner. Valid corners: \((20, 40)\) and \((50, 100)\).
✓ Minimum \(Z = \mathbf{100}\) at \((20, 40)\). Maximum \(Z = \mathbf{250}\) at \((50, 100)\).
Question 5
Minimise \(Z = x + 2y\) subject to: \(2x + y \geq 3\), \(x + 2y \geq 6\), \(x \geq 0\), \(y \geq 0\).
Line 1: \(2x + y = 3\) → \((1.5, 0)\) and \((0, 3)\). Line 2: \(x + 2y = 6\) → \((6, 0)\) and \((0, 3)\).
Both lines pass through \((0, 3)\). Intersection: Subtract \(L_1\) from \(L_2\): \(-x + y = 3 \Rightarrow y = x + 3\). Substituting into \(L_1\): \(2x + (x+3) = 3 \Rightarrow 3x = 0 \Rightarrow x = 0,\ y = 3\). So they meet at \((0, 3)\).
✓ Minimum \(Z = \mathbf{6}\) — occurs at both corner points, meaning it is minimum at every point on the line segment joining \((6, 0)\) and \((0, 3)\).
Question 6
Maximise \(Z = x + 2y\) subject to: \(x - y \leq 0\), \(2y \leq x + 2\), \(x \geq 0\), \(y \geq 0\).
Rewrite: \(x - y \leq 0 \Rightarrow y \geq x\) and \(2y \leq x + 2 \Rightarrow y \leq \frac{x+2}{2}\).
The feasible region is where \(x \leq y \leq \frac{x+2}{2}\), in the first quadrant.
Intersection of \(y = x\) and \(y = \frac{x+2}{2}\): \(x = \frac{x+2}{2} \Rightarrow 2x = x + 2 \Rightarrow x = 2,\ y = 2\). Corner: \((2, 2)\).
At \(x = 0\): \(y = 0\) (from \(y \geq x = 0\)) and \(y \leq 1\). So \((0, 0)\) and \((0, 1)\) are corners.
Evaluate: At \((0, 0)\): \(Z = 0\). At \((0, 1)\): \(Z = 0 + 2 = 2\). At \((2, 2)\): \(Z = 2 + 4 = 6\).
✓ Maximum \(Z = \mathbf{6}\) at \(x = 2,\ y = 2\).
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Question 1
A furniture manufacturer makes chairs and tables. A chair requires 2 hours on machine A and 6 hours on machine B. A table requires 5 hours on machine A and no time on machine B. Machine A is available for 16 hours per day and machine B for 30 hours. Profit from a chair is ₹2 and from a table is ₹10. Formulate this as an LPP to maximise total profit.
Let \(x\) = number of chairs produced per day, \(y\) = number of tables produced per day.
Objective function (maximise profit): \(Z = 2x + 10y\).
Machine A constraint: \(2x + 5y \leq 16\) (chairs use 2 hrs each, tables use 5 hrs each, total ≤ 16 hrs).
Machine B constraint: \(6x + 0y \leq 30 \Rightarrow x \leq 5\) (only chairs use machine B).
A diet must contain at least 80 units of vitamin A and 100 units of minerals. Food F1 costs ₹4/kg and provides 3 units of vitamin A and 4 units of minerals per kg. Food F2 costs ₹6/kg and provides 6 units of vitamin A and 3 units of minerals per kg. Formulate as LPP to minimise cost.
Let \(x\) = kg of Food F1 purchased, \(y\) = kg of Food F2 purchased.
Objective function (minimise cost): \(Z = 4x + 6y\).
Vitamin A constraint: \(3x + 6y \geq 80\) (must have at least 80 units).
Minerals constraint: \(4x + 3y \geq 100\) (must have at least 100 units).
A firm manufactures gold rings and chains. Combined production per day is at most 24 items. It takes 1 hour to make a ring and ½ hour to make a chain; 16 hours are available per day. Profit is ₹300 per ring and ₹190 per chain. Find the number of rings and chains to maximise profit graphically.
Constraints: Production: \(x + y \leq 24\). Time: \(x + 0.5y \leq 16\). Non-negativity: \(x \geq 0,\ y \geq 0\).
Find corner points. Intersection of \(x + y = 24\) and \(x + 0.5y = 16\): subtract to get \(0.5y = 8 \Rightarrow y = 16,\ x = 8\). Corner: \((8, 16)\).
Evaluate \(Z\): At \((0,0)\): 0. At \((16,0)\): 4800. At \((8,16)\): \(2400 + 3040 = 5440\). At \((0,24)\): 4560.
✓ Maximum profit = ₹5440 when 8 rings and 16 chains are manufactured per day.
Question 4
Reshma wishes to mix two types of food P and Q to meet minimum vitamin requirements of at least 8 units of vitamin A and 11 units of vitamin B. Food P: ₹60/kg, 3 units vitamin A and 5 units vitamin B per kg. Food Q: ₹80/kg, 4 units vitamin A and 2 units vitamin B per kg. Find the minimum cost mix.
Let \(x\) = kg of P, \(y\) = kg of Q. Minimise \(Z = 60x + 80y\).
Intersection of the two lines: From \(3x + 4y = 8\) and \(5x + 2y = 11\). Multiply second by 2: \(10x + 4y = 22\). Subtract first: \(7x = 14 \Rightarrow x = 2,\ y = 0.5\). Corner: \((2, 0.5)\).
Other corners: \((8/3, 0)\) (set \(y=0\) in \(3x+4y=8\)) and \((0, 5.5)\) (set \(x=0\) in \(5x+2y=11\)).
Evaluate \(Z\): At \((8/3, 0)\): \(Z = 60 \times 8/3 = 160\). At \((2, 0.5)\): \(Z = 120 + 40 = 160\). At \((0, 5.5)\): \(Z = 440\).
✓ Minimum cost = ₹160, occurring along the segment between \((8/3, 0)\) and \((2, 0.5)\).
Question 5
A company makes products A and B. Each unit of A needs 3 hrs on M1 and 2 hrs on M2; each unit of B needs 2 hrs on M1 and 4 hrs on M2. M1 is available 18 hrs/day, M2 is available 24 hrs/day. Profit: ₹50/unit A, ₹40/unit B. Find maximum profit graphically.
Let \(x\) = units of A, \(y\) = units of B. Maximise \(Z = 50x + 40y\).
Intersection of \(3x + 2y = 18\) and \(x + 2y = 12\): Subtract: \(2x = 6 \Rightarrow x = 3,\ y = 4.5\). Corner: \((3, 4.5)\).
All corners: \((0,0)\), \((6,0)\), \((3, 4.5)\), \((0,6)\).
Evaluate \(Z\): At \((0,0)\): 0. At \((6,0)\): 300. At \((3,4.5)\): \(150 + 180 = 330\). At \((0,6)\): 240.
✓ Maximum profit = ₹330 per day when 3 units of A and 4.5 units of B are produced.
Question 6
A manufacturer produces bikes: Model X (6 man-hours, ₹2000 handling, ₹1000 profit) and Model Y (10 man-hours, ₹3000 handling, ₹1500 profit). Total man-hours: 450/week. Total funds: ₹80,000/week. Formulate and solve graphically to maximise profit.
Let \(x\) = units of Model X, \(y\) = units of Model Y. Maximise \(Z = 1000x + 1500y\).
Find corners: Intersection of \(3x + 5y = 225\) and \(2x + 3y = 80\): Multiply 2nd by 5 and 1st by 3: \(6x + 10y = 450\) and \(6x + 9y = 240\) → \(y = 210\)... check consistency. Better: from \(2x + 3y = 80\) → \(x = (80-3y)/2\). Substitute into \(3x+5y = 225\): \(3(80-3y)/2 + 5y = 225 \Rightarrow 240 - 9y + 10y = 450 \Rightarrow y = 210\)... Re-verify: this gives infeasible \(y\). Check \(x\)-intercepts: \(L_1: (75,0)\), \(L_2: (40,0)\). \(y\)-intercepts: \(L_1: (0,45)\), \(L_2: (0,80/3)\approx(0,26.7)\). Binding constraint is \(L_2\) for \(y\)-intercept.
The two lines may be parallel or intersect outside feasible region. Corners to check: \((0,0)\), \((40,0)\), \((0, 80/3)\). Evaluate \(Z\): At \((0,0)\): 0. At \((40,0)\): 40000. At \((0, 26.7)\): \(\approx 40050\).
✓ Formulation: Maximise \(Z = 1000x + 1500y\) subject to \(3x + 5y \leq 225\), \(2x + 3y \leq 80\), \(x \geq 0\), \(y \geq 0\). Solve graphically by plotting both constraints and evaluating \(Z\) at all corner points of the feasible polygon.
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Case Study Based Questions
Real-world application problems
Case Study 1
Manufacturing Optimisation Problem
A company manufactures two types of products P1 and P2. The company has two machines M1 and M2. Product P1 requires 4 hours on M1 and 2 hours on M2. Product P2 requires 3 hours on M1 and 3 hours on M2. Machine M1 is available for 60 hours and M2 for 48 hours per week. The profit per unit of P1 is ₹100 and P2 is ₹120.
Profit from P1 = ₹100 per unit, from P2 = ₹120 per unit. Total profit \(Z = 100x + 120y\). We want to maximise this.
Q2
What is the constraint for Machine M1?
A\(4x + 3y \leq 60\)
B\(2x + 3y \leq 48\)
C\(4x + 2y \leq 60\)
D\(3x + 3y \leq 60\)
✓ Correct Answer: (A) \(4x + 3y \leq 60\)
P1 requires 4 hours on M1, P2 requires 3 hours on M1. Total hours on M1 = \(4x + 3y\). Since M1 is available for 60 hours: \(4x + 3y \leq 60\).
Q3
Corner points are \((0,0)\), \((15,0)\), \((0,16)\), \((12,4)\). What is the maximum profit?
A₹1500
B₹1680
C₹1920
D₹2000
✓ Correct Answer: (C) ₹1920
Evaluate \(Z = 100x + 120y\):
At \((0,0)\): \(Z = 0\)
At \((15,0)\): \(Z = 1500\)
At \((0,16)\): \(Z = 1920\) ← Maximum
At \((12,4)\): \(Z = 1200 + 480 = 1680\)
Q4
How many units of each product should be manufactured for maximum profit?
A15 units of P1, 0 units of P2
B0 units of P1, 16 units of P2
C12 units of P1, 4 units of P2
D10 units of P1, 10 units of P2
✓ Correct Answer: (B) 0 units of P1, 16 units of P2
Maximum profit of ₹1920 occurs at the corner point \((0, 16)\): produce 0 units of P1 and 16 units of P2.
Case Study 2
Diet Problem
A dietician wishes to mix two foods such that the vitamin contents contain at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg vitamin A and 1 unit/kg vitamin C. Food II contains 1 unit/kg vitamin A and 2 units/kg vitamin C. Food I costs ₹50/kg and Food II costs ₹70/kg.
Q1
The objective function is:
AMinimise \(Z = 50x + 70y\)
BMaximise \(Z = 50x + 70y\)
CMinimise \(Z = 2x + y\)
DMaximise \(Z = 8x + 10y\)
✓ Correct Answer: (A) Minimise \(Z = 50x + 70y\)
We want to find the cheapest mix. Total cost = ₹50 per kg of Food I + ₹70 per kg of Food II = \(50x + 70y\). Minimise this.
Q2
The constraint for vitamin A is:
A\(2x + y \geq 8\)
B\(x + 2y \geq 10\)
C\(2x + y \leq 8\)
D\(x + y \geq 8\)
✓ Correct Answer: (A) \(2x + y \geq 8\)
Food I has 2 units/kg of vitamin A, Food II has 1 unit/kg. Requirement: at least 8 units. So \(2x + y \geq 8\).
Q3
The feasible region for this problem is:
ABounded
BUnbounded
CEmpty
DA single point
✓ Correct Answer: (B) Unbounded
Both constraints are of the \(\geq\) type (minimum requirements). The feasible region lies above both constraint lines — it extends infinitely, making it unbounded.
Q4
If corner points are \((0,10)\), \((2,4)\), and \((8,0)\), what is the minimum cost?
A₹380
B₹400
C₹700
D₹280
✓ Correct Answer: (A) ₹380
Evaluate \(Z = 50x + 70y\):
At \((0,10)\): \(Z = 0 + 700 = 700\)
At \((2,4)\): \(Z = 100 + 280 = 380\) ← Minimum
At \((8,0)\): \(Z = 400 + 0 = 400\)
Case Study 3
Resource Allocation Problem
A company produces leather belts A (superior quality, ₹40 profit) and B (lower quality, ₹30 profit). Belt A takes twice as much time to produce as Belt B. If all production were Belt B, 1000 belts could be made per day. Leather supply is sufficient for 800 belts per day. Belt A requires a fancy buckle; only 400 fancy buckles are available per day.
Q1
If \(x\) and \(y\) are the number of belts of type A and B, the objective function is:
AMaximise \(Z = 40x + 30y\)
BMinimise \(Z = 40x + 30y\)
CMaximise \(Z = 30x + 40y\)
DMaximise \(Z = x + y\)
✓ Correct Answer: (A) Maximise \(Z = 40x + 30y\)
Profit from type A = ₹40 per belt, from type B = ₹30 per belt. Total profit \(Z = 40x + 30y\). Maximise this.
Q2
The time constraint can be expressed as:
A\(2x + y \leq 1000\)
B\(x + 2y \leq 1000\)
C\(x + y \leq 500\)
D\(2x + y \leq 500\)
✓ Correct Answer: (A) \(2x + y \leq 1000\)
Belt A takes twice the time of belt B. Assigning 1 time unit to B and 2 time units to A, total capacity = 1000 B-units. Constraint: \(2x + y \leq 1000\).
Q3
Which constraint represents the leather supply limitation?
A\(x + y \leq 800\)
B\(x \leq 400\)
C\(2x + y \leq 1000\)
D\(y \leq 800\)
✓ Correct Answer: (A) \(x + y \leq 800\)
Leather is sufficient for at most 800 belts per day, and both types A and B use leather. Combined: \(x + y \leq 800\).
Q4
The buckle constraint is:
A\(x \leq 400\)
B\(y \leq 400\)
C\(x + y \leq 400\)
D\(2x \leq 400\)
✓ Correct Answer: (A) \(x \leq 400\)
Only Belt A requires a fancy buckle (one per belt). With only 400 fancy buckles available: \(x \leq 400\).
💡 Exam Tips for Unit 8 — Linear Programming
Common mistakes examiners flag every year
✅ Tip 1 — Always verify corner points by solving simultaneous equations, not by reading the graph.
When two constraint lines intersect, find the exact coordinates algebraically. A graphical estimate like "approximately (3, 4)" is not acceptable in board exams — examiners expect the exact values. Errors here directly affect your objective function evaluation and cost you marks.
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A Linear Programming Problem is a mathematical optimisation technique where we maximise or minimise a linear objective function (e.g., profit or cost) subject to a set of linear constraints (inequalities) and non-negativity restrictions on the decision variables. It is widely used in manufacturing, diet planning, transportation, and resource allocation problems.
What is the Corner Point Method and why does it work?
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The Corner Point Method is based on the fundamental theorem of linear programming: if the optimal value of a linear objective function exists over a feasible region, it must occur at one (or more) of the corner points (vertices) of that region. This is because a linear function over a convex polygon attains its maximum and minimum at the extreme points. You simply evaluate \(Z\) at every vertex and compare.
What is the difference between a bounded and an unbounded feasible region?
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A bounded feasible region is a closed polygon with finite area — it happens when all constraints are of the \(\leq\) type (along with non-negativity). Both a maximum and minimum of the objective function always exist for a bounded region. An unbounded feasible region extends infinitely in at least one direction (typically when constraints use \(\geq\)). For unbounded regions, you must verify whether the optimal value actually exists using a half-plane test.
How do I formulate an LPP from a word problem?
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Follow these steps: (1) Identify the decision variables — what quantities are you deciding? (2) Write the objective function — what are you maximising (profit, revenue) or minimising (cost, time)? (3) Write the structural constraints — these come from resource limitations, minimum requirements, etc. (4) Add non-negativity constraints: \(x \geq 0,\ y \geq 0\). Always check that the inequalities use the correct direction (\(\leq\) for maximum available resources, \(\geq\) for minimum requirements).
Can the optimal solution occur along an edge rather than at a single corner?
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Yes! If the objective function line is parallel to one of the constraint edges of the feasible region, then the optimal value is the same at two adjacent corner points — and at every point on the segment between them. In such cases, there are infinitely many optimal solutions. When this happens in board exams, you should mention that the minimum (or maximum) occurs at all points on the line segment joining the two corners.
How many marks does Unit 8 carry in the board exam?
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Unit 8 — Linear Programming Problems — carries 8 marks in the CBSE Class 12 Applied Maths board examination. Questions typically appear as 1-mark MCQs, 2-mark short-answer questions, and a case study. It is one of the highest-weightage units in the course.
