📊 Unit 8: Linear Programming Problems
LPP Formulation, Graphical Method, Bounded & Unbounded Regions
🎯 10 Marks Weightage1. Formulation of LPP
• Identifying decision variables
• Objective function
• Constraints from word problems
• Non-negativity restrictions
2. Graphical Solution Method
• Plotting constraints
• Finding feasible region
• Bounded and unbounded regions
• Corner points identification
3. Corner Point Method
• Finding optimal solution
• Evaluating objective function
• Maximization problems
• Minimization problems
4. Types of LPP
• Manufacturing problems
• Diet problems
• Resource allocation
• Profit maximization
Multiple Choice Questions
Question 1
CBSE 2024
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:
✓ Correct Answer: (C) (6, 8)
Explanation:
Evaluate Z 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) = 72 (Maximum)
At (0, 5): Z = 4(0) + 6(5) = 30
Evaluate Z 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) = 72 (Maximum)
At (0, 5): Z = 4(0) + 6(5) = 30
Question 2
The feasible region for an LPP is shown shaded in the graph. The objective function is Z = 3x - 4y. The minimum value of Z occurs at:
✓ Correct Answer: (d) Must evaluate at all corner points
Solution:
The optimal value of the objective function (maximum or minimum) in a linear programming problem always occurs at one of the corner points of the feasible region. Therefore, we must evaluate Z at all corner points to find the minimum.
The optimal value of the objective function (maximum or minimum) in a linear programming problem always occurs at one of the corner points of the feasible region. Therefore, we must evaluate Z at all corner points to find the minimum.
Question 3
The constraints x ≥ 0, y ≥ 0, x + y ≤ 5, x + 2y ≤ 8 define:
✓ Correct Answer: (b) Bounded feasible region
Solution:
All constraints are of the form ≤ or ≥ with non-negativity restrictions. The inequalities x + y ≤ 5 and x + 2y ≤ 8 along with x ≥ 0, y ≥ 0 form a closed polygon in the first quadrant, which is a bounded feasible region.
All constraints are of the form ≤ or ≥ with non-negativity restrictions. The inequalities x + y ≤ 5 and x + 2y ≤ 8 along with x ≥ 0, y ≥ 0 form a closed polygon in the first quadrant, which is a bounded feasible region.
Question 4
If the constraints are x + y ≥ 4, x - y ≥ 0, x ≥ 0, y ≥ 0, then the feasible region is:
✓ Correct Answer: (b) Unbounded
Solution:
The constraint x + y ≥ 4 with x - y ≥ 0 in the first quadrant creates an open region extending indefinitely. There is no upper bound on x or y values, making it an unbounded feasible region.
The constraint x + y ≥ 4 with x - y ≥ 0 in the first quadrant creates an open region extending indefinitely. There is no upper bound on x or y values, making it an unbounded feasible region.
Question 5
A manufacturer makes two products A and B. Product A requires 2 hours of labor and product B requires 3 hours. If maximum 12 hours of labor are available, which constraint represents this?
✓ Correct Answer: (b) 2x + 3y ≤ 12
Solution:
If x = number of product A and y = number of product B
Labor required = 2x + 3y
Maximum labor available = 12 hours
Therefore: 2x + 3y ≤ 12
If x = number of product A and y = number of product B
Labor required = 2x + 3y
Maximum labor available = 12 hours
Therefore: 2x + 3y ≤ 12
Question 6
The objective function of a linear programming problem is:
✓ Correct Answer: (b) A function to be optimized
Solution:
The objective function is the function that we want to maximize or minimize (optimize) subject to given constraints. For example, maximize profit Z = 5x + 3y or minimize cost C = 2x + 4y.
The objective function is the function that we want to maximize or minimize (optimize) subject to given constraints. For example, maximize profit Z = 5x + 3y or minimize cost C = 2x + 4y.
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 minimum value of Z is:
✓ Correct Answer: (a) 0
Solution:
At (0, 0): Z = 5(0) + 3(0) = 0 (Minimum)
At (4, 0): Z = 5(4) + 3(0) = 20
At (4, 3): Z = 5(4) + 3(3) = 29
At (2, 4): Z = 5(2) + 3(4) = 22
At (0, 4): Z = 5(0) + 3(4) = 12
At (0, 0): Z = 5(0) + 3(0) = 0 (Minimum)
At (4, 0): Z = 5(4) + 3(0) = 20
At (4, 3): Z = 5(4) + 3(3) = 29
At (2, 4): Z = 5(2) + 3(4) = 22
At (0, 4): Z = 5(0) + 3(4) = 12
Question 8
Which of the following is NOT a requirement for a linear programming problem?
✓ Correct Answer: (d) Objective function must be quadratic
Solution:
In linear programming, both the objective function and constraints must be LINEAR (not quadratic). The variables are typically non-negative. A quadratic objective function would make it a nonlinear programming problem.
In linear programming, both the objective function and constraints must be LINEAR (not quadratic). The variables are typically non-negative. A quadratic objective function would make it a nonlinear programming problem.
Question 9
The region represented by x ≥ 0, y ≥ 0 is:
✓ Correct Answer: (a) First quadrant
Solution:
The constraints x ≥ 0 and y ≥ 0 represent the first quadrant of the coordinate plane, where both x and y values are non-negative.
The constraints x ≥ 0 and y ≥ 0 represent the first quadrant of the coordinate plane, where both x and y values are non-negative.
Question 10
In a diet problem, if we want to minimize cost, the objective function represents:
✓ Correct Answer: (b) Total cost of the diet
Solution:
In a diet problem where we minimize cost, the objective function is Z = c₁x₁ + c₂x₂ + ... where c represents cost per unit and x represents quantity. The nutritional requirements form the constraints, not the objective function.
In a diet problem where we minimize cost, the objective function is Z = c₁x₁ + c₂x₂ + ... where c represents cost per unit and x represents quantity. The nutritional requirements form the constraints, not the objective function.
Question 11
If the feasible region for an LPP is unbounded, then:
✓ Correct Answer: (d) Depends on the objective function
Solution:
In an unbounded feasible region, the existence of optimal solution depends on the objective function. A minimum may exist even if maximum doesn't, or vice versa. We need to check by evaluating at corner points and analyzing the direction of optimization.
In an unbounded feasible region, the existence of optimal solution depends on the objective function. A minimum may exist even if maximum doesn't, or vice versa. We need to check by evaluating at corner points and analyzing the direction of optimization.
Question 12
The point (2, 3) lies in the feasible region of x + y ≤ 6 if:
✓ Correct Answer: (c) 2 + 3 ≤ 6
Solution:
To check if point (2, 3) satisfies x + y ≤ 6:
Substitute: 2 + 3 ≤ 6
5 ≤ 6 (True)
Therefore, the point lies in the feasible region.
To check if point (2, 3) satisfies x + y ≤ 6:
Substitute: 2 + 3 ≤ 6
5 ≤ 6 (True)
Therefore, the point lies in the feasible region.
Question 13
A factory produces two products X and Y. Product X gives profit of ₹40 per unit and Y gives ₹30 per unit. The objective function for maximum profit is:
✓ Correct Answer: (a) Z = 40x + 30y (Maximize)
Solution:
If x = units of product X and y = units of product Y
Profit from X = 40x
Profit from Y = 30y
Total profit Z = 40x + 30y
We want to maximize this profit.
If x = units of product X and y = units of product Y
Profit from X = 40x
Profit from Y = 30y
Total profit Z = 40x + 30y
We want to maximize this profit.
Question 14
The constraints x ≥ 0, y ≥ 0 are called:
✓ Correct Answer: (b) Non-negativity constraints
Solution:
The constraints x ≥ 0 and y ≥ 0 are specifically called non-negativity constraints because they ensure that the decision variables cannot take negative values, which makes practical sense in most real-world problems.
The constraints x ≥ 0 and y ≥ 0 are specifically called non-negativity constraints because they ensure that the decision variables cannot take negative values, which makes practical sense in most real-world problems.
Question 15
In the corner point method, the optimal solution is:
✓ Correct Answer: (c) At one of the corner points
Solution:
The fundamental theorem of linear programming states that if an optimal solution exists, it must occur at one (or more) of the corner points (extreme points) of the feasible region. This is why we evaluate the objective function at all corner points.
The fundamental theorem of linear programming states that if an optimal solution exists, it must occur at one (or more) of the corner points (extreme points) of the feasible region. This is why we evaluate the objective function at all corner points.
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Short Answer Questions
Question 1
CBSE 2024
Maximize Z = x + y, subject to constraints: x ≥ 0, y ≥ 0, x - y ≤ -1, x ≤ y
Solution:
Given constraints: x ≥ 0, y ≥ 0, x - y ≤ -1, x ≤ y
Rewrite: x - y ≤ -1 → y ≥ x + 1
x ≤ y is equivalent to y ≥ x
Plot the lines y = x + 1 and y = x in first quadrant
The feasible region is the area above line y = x + 1
The feasible region is UNBOUNDED. Maximum value of Z does not exist.
Question 2
CBSE 2024
Minimize Z = 200x + 500y, subject to: x + 2y ≥ 10, 3x + 4y ≥ 24, x ≥ 0, y ≥ 0
Solution:
Plot the constraint lines:
x + 2y = 10: Passes through (10, 0) and (0, 5)
3x + 4y = 24: Passes through (8, 0) and (0, 6)
Find corner points by solving simultaneous equations:
Corner points are: (0, 6), (4, 3), (10, 0)
Solving x + 2y = 10 and 3x + 4y = 24:
Multiply first by 2: 2x + 4y = 20
Subtract: (3x + 4y) - (2x + 4y) = 24 - 20
x = 4, then y = 3
Evaluate Z at corner points:
At (0, 6): Z = 200(0) + 500(6) = 3000
At (4, 3): Z = 200(4) + 500(3) = 2300
At (10, 0): Z = 200(10) + 500(0) = 2000
Minimum Z = 2000 at x = 10, y = 0
Question 3
CBSE 2023
Minimize Z = 3x + 5y subject to: x ≥ 0, y ≥ 0, x + 3y ≥ 3, x + y ≥ 2
Solution:
Plot the constraint lines:
x + 3y = 3: Points (3, 0) and (0, 1)
x + y = 2: Points (2, 0) and (0, 2)
Feasible region is UNBOUNDED (all ≥ constraints)
Find corner points:
Solving x + 3y = 3 and x + y = 2:
Subtract: (x + 3y) - (x + y) = 3 - 2
2y = 1 → y = 0.5, x = 1.5
Corner points: (3, 0), (1.5, 0.5), (0, 2)
At (3, 0): Z = 3(3) + 5(0) = 9
At (1.5, 0.5): Z = 3(1.5) + 5(0.5) = 7
At (0, 2): Z = 3(0) + 5(2) = 10
Minimum Z = 7 at x = 1.5, y = 0.5
Question 4
CBSE 2025
Minimize and Maximize Z = x + 2y subject to: x + 2y ≥ 100, 2x - y ≤ 0, 2x + y ≤ 200, x ≥ 0, y ≥ 0
Solution:
Constraints: x + 2y ≥ 100, 2x - y ≤ 0, 2x + y ≤ 200, x ≥ 0, y ≥ 0
From 2x - y ≤ 0: y ≥ 2x
Plot all constraints and find feasible region
Find corner points by solving pairs:
Intersection of x + 2y = 100 and y = 2x:
x + 2(2x) = 100 → 5x = 100 → x = 20, y = 40
Intersection of y = 2x and 2x + y = 200:
2x + 2x = 200 → x = 50, y = 100
Evaluate Z at corner points
At (20, 40): Z = 20 + 2(40) = 100
At (50, 100): Z = 50 + 2(100) = 250
Minimum Z = 100, Maximum Z = 250
Question 5
CBSE 2023
Minimize Z = x + 2y subject to: 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0
Solution:
Plot constraints:
2x + y = 3: Points (1.5, 0) and (0, 3)
x + 2y = 6: Points (6, 0) and (0, 3)
Both lines pass through (0, 3)
Solving 2x + y = 3 and x + 2y = 6:
From first: y = 3 - 2x
Substitute: x + 2(3 - 2x) = 6
x + 6 - 4x = 6 → -3x = 0 → x = 0, y = 3
Corner points: (0, 3), (6, 0)
At (0, 3): Z = 0 + 2(3) = 6
At (6, 0): Z = 6 + 2(0) = 6
Minimum Z = 6 (occurs at multiple points on the line segment)
Question 6
CBSE 2024
Maximize Z = x + 2y subject to: x - y ≤ 0, 2y ≤ x + 2, x ≥ 0, y ≥ 0
Solution:
Rewrite constraints:
x ≤ y (or y ≥ x)
y ≤ (x + 2)/2
Plot these lines in first quadrant
Find corner points:
Origin: (0, 0)
Intersection of y = x and y = (x + 2)/2:
x = (x + 2)/2 → 2x = x + 2 → x = 2, y = 2
Intersection with axes
Evaluate Z at corner points
At (0, 0): Z = 0
At (2, 2): Z = 2 + 2(2) = 6
Maximum Z = 6 at x = 2, y = 2
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Long Answer Questions with Complete Solutions
Practice 4-mark and 6-mark questions
Question 1
CBSE 2024
A furniture manufacturer makes two products: chairs and tables. Processing of these products is done on two machines A and B. 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. There are 16 hours per day available on machine A and 30 hours on machine B. Profit gained by the manufacturer from a chair and a table is ₹2 and ₹10, respectively. Formulate this problem as a linear programming problem to maximize the total profit of the manufacturer.
Complete Solution:
Step 1: Define decision variables
Let x = number of chairs manufactured per day
Let y = number of tables manufactured per day
Step 2: Formulate objective function
Profit from chairs = ₹2x
Profit from tables = ₹10y
Total profit Z = 2x + 10y (to be maximized)
Step 3: Formulate constraints
Machine A constraint: 2x + 5y ≤ 16
(Chair needs 2 hrs, table needs 5 hrs, total 16 hrs available)
Machine B constraint: 6x ≤ 30 → x ≤ 5
(Chair needs 6 hrs, table needs 0 hrs, total 30 hrs available)
Non-negativity: x ≥ 0, y ≥ 0
LPP Formulation:
Maximize Z = 2x + 10y
Subject to: 2x + 5y ≤ 16, x ≤ 5, x ≥ 0, y ≥ 0
Maximize Z = 2x + 10y
Subject to: 2x + 5y ≤ 16, x ≤ 5, x ≥ 0, y ≥ 0
Question 2
CBSE 2024
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs ₹4 per kg and F2 costs ₹6 per kg. One kg of food F1 contains 3 units of vitamin A and 4 units of minerals. One kg of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem to find the minimum cost for diet that meets the minimum nutritional requirements.
Complete Solution:
Step 1: Define decision variables
Let x = quantity of food F1 (in kg)
Let y = quantity of food F2 (in kg)
Step 2: Formulate objective function
Cost of F1 = ₹4x
Cost of F2 = ₹6y
Total cost Z = 4x + 6y (to be minimized)
Step 3: Formulate constraints
Vitamin A constraint: 3x + 6y ≥ 80
(F1 has 3 units/kg, F2 has 6 units/kg, need at least 80 units)
Minerals constraint: 4x + 3y ≥ 100
(F1 has 4 units/kg, F2 has 3 units/kg, need at least 100 units)
Non-negativity: x ≥ 0, y ≥ 0
LPP Formulation:
Minimize Z = 4x + 6y
Subject to: 3x + 6y ≥ 80, 4x + 3y ≥ 100, x ≥ 0, y ≥ 0
Minimize Z = 4x + 6y
Subject to: 3x + 6y ≥ 80, 4x + 3y ≥ 100, x ≥ 0, y ≥ 0
Question 3
CBSE 2023
A small firm manufactures gold rings and chains. The combined number of rings and chains manufactured per day is at most 24. It takes one hour to make a ring and half an hour for a chain. The maximum number of hours available per day is 16. If the profit on a ring is ₹300 and on a chain is ₹190, how many of each should be manufactured daily so as to maximize the profit? Formulate as LPP and solve graphically.
Complete Solution:
Step 1: Formulation
Let x = number of rings, y = number of chains
Maximize Z = 300x + 190y
Subject to: x + y ≤ 24, x + 0.5y ≤ 16, x ≥ 0, y ≥ 0
Step 2: Plot constraints
Line x + y = 24: Points (24, 0) and (0, 24)
Line x + 0.5y = 16: Points (16, 0) and (0, 32)
Feasible region is bounded polygon in first quadrant
Step 3: Find corner points
Corner points: (0, 0), (16, 0), (0, 24)
Intersection of x + y = 24 and x + 0.5y = 16:
From second: x = 16 - 0.5y
Substitute: (16 - 0.5y) + y = 24
16 + 0.5y = 24 → y = 16, x = 8
Step 4: Evaluate Z at corner points
At (0, 0): Z = 0
At (16, 0): Z = 300(16) = 4800
At (8, 16): Z = 300(8) + 190(16) = 5440
At (0, 24): Z = 190(24) = 4560
Maximum profit = ₹5440 when 8 rings and 16 chains are manufactured
Question 4
CBSE 2023
Reshma wishes to mix two types of foods P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs ₹60/kg and food Q costs ₹80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
Complete Solution:
Step 1: Formulation
Let x kg of food P and y kg of food Q be mixed
Minimize Z = 60x + 80y (total cost)
Vitamin A: 3x + 4y ≥ 8
Vitamin B: 5x + 2y ≥ 11
x ≥ 0, y ≥ 0
Step 2: Find corner points
Plot 3x + 4y = 8: Points (8/3, 0) and (0, 2)
Plot 5x + 2y = 11: Points (11/5, 0) and (0, 5.5)
Solve 3x + 4y = 8 and 5x + 2y = 11:
Multiply first by 1, second by 2: 3x + 4y = 8, 10x + 4y = 22
Subtract: 7x = 14 → x = 2
From 3x + 4y = 8: 6 + 4y = 8 → y = 0.5
Step 3: Evaluate Z
At (8/3, 0): Z = 60(8/3) = 160
At (2, 0.5): Z = 60(2) + 80(0.5) = 160
At (0, 5.5): Z = 80(5.5) = 440
Minimum cost = ₹160 (occurs at multiple points)
Question 5
CBSE 2025
A company manufactures two types of products A and B. Each unit of A requires 3 hours on machine M1 and 2 hours on machine M2. Each unit of B requires 2 hours on M1 and 4 hours on M2. The machines M1 and M2 are available for 18 and 24 hours respectively per day. The profit is ₹50 per unit of A and ₹40 per unit of B. Formulate the LPP and find the maximum profit graphically.
Complete Solution:
Formulation:
Let x = units of A, y = units of B
Maximize Z = 50x + 40y
Constraints: 3x + 2y ≤ 18, 2x + 4y ≤ 24, x ≥ 0, y ≥ 0
Graphical Solution:
Corner points: (0, 0), (6, 0), (0, 6), and intersection
Solving 3x + 2y = 18 and 2x + 4y = 24:
From second: x = 12 - 2y
Substitute: 3(12 - 2y) + 2y = 18
36 - 6y + 2y = 18 → y = 4.5, x = 3
At (0, 0): Z = 0
At (6, 0): Z = 300
At (3, 4.5): Z = 150 + 180 = 330
At (0, 6): Z = 240
Maximum profit = ₹330 at x = 3, y = 4.5
Question 6
CBSE 2024
A manufacturer produces two models of bikes - Model X and Model Y. Model X takes 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hours available per week. Handling and packaging costs are ₹2000 and ₹3000 per unit respectively for the two models. The total funds available for this operation are ₹80,000 per week. Profits per unit for models X and Y are ₹1000 and ₹1500 respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Solve graphically.
Complete Solution:
Formulation:
Let x = units of Model X, y = units of Model Y
Maximize Z = 1000x + 1500y
Man-hours: 6x + 10y ≤ 450
Cost: 2000x + 3000y ≤ 80000 → 2x + 3y ≤ 80
x ≥ 0, y ≥ 0
Corner Points:
(0, 0), (40, 0), (0, 26.67), intersection point
Solving 6x + 10y = 450 and 2x + 3y = 80:
Multiply second by 3: 6x + 9y = 240
Subtract: y = 210, which is > 26.67, so corner is (0, 26.67)
Check (30, 15): 6(30) + 10(15) = 330 ≤ 450 ✓
2(30) + 3(15) = 105 > 80 ✗
Actual corners from constraints give optimal solution
Solve graphically to find maximum profit at appropriate corner point
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Case Study Based Questions
Real-world application problems
Case Study 1
CBSE 2024
Manufacturing Optimization 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.
Q1
What is the objective function for this LPP?
✓ Correct Answer: (a) Maximize Z = 100x + 120y
Solution:
The objective is to maximize profit. Profit from P1 = ₹100x and from P2 = ₹120y. Therefore, Z = 100x + 120y to be maximized.
The objective is to maximize profit. Profit from P1 = ₹100x and from P2 = ₹120y. Therefore, Z = 100x + 120y to be maximized.
Q2
What is the constraint for Machine M1?
✓ Correct Answer: (a) 4x + 3y ≤ 60
Solution:
P1 requires 4 hours on M1, P2 requires 3 hours on M1. M1 available for 60 hours. Therefore: 4x + 3y ≤ 60
P1 requires 4 hours on M1, P2 requires 3 hours on M1. M1 available for 60 hours. Therefore: 4x + 3y ≤ 60
Q3
The corner points of the feasible region are (0, 0), (15, 0), (0, 16), and (12, 4). What is the maximum profit?
✓ Correct Answer: (b) ₹1680
Solution:
At (0, 0): Z = 0
At (15, 0): Z = 1500
At (0, 16): Z = 1920
At (12, 4): Z = 100(12) + 120(4) = 1200 + 480 = 1680
At (0, 0): Z = 0
At (15, 0): Z = 1500
At (0, 16): Z = 1920
At (12, 4): Z = 100(12) + 120(4) = 1200 + 480 = 1680
Q4
How many units of each product should be manufactured for maximum profit?
✓ Correct Answer: (b) 0 units of P1, 16 units of P2
Solution:
Maximum profit is ₹1920 at (0, 16), meaning 0 units of P1 and 16 units of P2 should be manufactured.
Maximum profit is ₹1920 at (0, 16), meaning 0 units of P1 and 16 units of P2 should be manufactured.
Case Study 2
CBSE 2023
Diet Problem
A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food II contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. Food I costs ₹50 per kg and Food II costs ₹70 per kg.
Q1
The objective function is:
✓ Correct Answer: (a) Minimize Z = 50x + 70y
Solution:
We want to minimize the total cost. Cost = 50x + 70y where x and y are kg of Food I and II respectively.
We want to minimize the total cost. Cost = 50x + 70y where x and y are kg of Food I and II respectively.
Q2
The constraint for vitamin A is:
✓ Correct Answer: (a) 2x + y ≥ 8
Solution:
Food I has 2 units/kg of vitamin A, Food II has 1 unit/kg. Need at least 8 units. Therefore: 2x + y ≥ 8
Food I has 2 units/kg of vitamin A, Food II has 1 unit/kg. Need at least 8 units. Therefore: 2x + y ≥ 8
Q3
The feasible region for this problem is:
✓ Correct Answer: (b) Unbounded
Solution:
Since both constraints are of the form ≥ (greater than or equal to), the feasible region extends indefinitely and is unbounded.
Since both constraints are of the form ≥ (greater than or equal to), the feasible region extends indefinitely and is unbounded.
Q4
If the corner points are (0, 10), (2, 4), and (8, 0), what is the minimum cost?
✓ Correct Answer: (a) ₹380
Solution:
At (0, 10): Z = 700
At (2, 4): Z = 50(2) + 70(4) = 100 + 280 = 380
At (8, 0): Z = 400
Minimum is ₹380
At (0, 10): Z = 700
At (2, 4): Z = 50(2) + 70(4) = 100 + 280 = 380
At (8, 0): Z = 400
Minimum is ₹380
Case Study 3
CBSE 2024
Resource Allocation Problem
A company produces two types of leather belts A and B. Belt A is superior quality and belt B is of lower quality. The respective profits are ₹40 and ₹30 per belt. Each belt of type A requires twice as much time as required by a belt of type B. If all belts were of type B, the company could produce 1000 belts per day. The supply of leather is sufficient for only 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 buckles are available per day.
Q1
If x and y are the number of belts of type A and B respectively, the objective function is:
✓ Correct Answer: (a) Maximize Z = 40x + 30y
Solution:
Profit from belt A = ₹40x, profit from belt B = ₹30y. Objective is to maximize total profit Z = 40x + 30y
Profit from belt A = ₹40x, profit from belt B = ₹30y. Objective is to maximize total profit Z = 40x + 30y
Q2
The time constraint can be expressed as:
✓ Correct Answer: (a) 2x + y ≤ 1000
Solution:
Belt A takes twice the time of B. If all were B, 1000 can be made. So time for A is 2 units, for B is 1 unit. Constraint: 2x + y ≤ 1000
Belt A takes twice the time of B. If all were B, 1000 can be made. So time for A is 2 units, for B is 1 unit. Constraint: 2x + y ≤ 1000
Q3
Which constraint represents the leather supply limitation?
✓ Correct Answer: (a) x + y ≤ 800
Solution:
Leather is sufficient for only 800 belts per day (both types combined). Therefore: x + y ≤ 800
Leather is sufficient for only 800 belts per day (both types combined). Therefore: x + y ≤ 800
Q4
The buckle constraint is:
✓ Correct Answer: (a) x ≤ 400
Solution:
Belt A requires a fancy buckle and only 400 buckles are available. Therefore: x ≤ 400
Belt A requires a fancy buckle and only 400 buckles are available. Therefore: x ≤ 400
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