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)
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
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:
ACorner point with highest y-coordinate
BCorner point with highest x-coordinate
CCorner point with lowest x-coordinate
DMust evaluate at all corner points
✓ Correct Answer: (D) Must evaluate at all corner points
Solution:
The optimal value of the objective function 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:
AUnbounded feasible region
BBounded feasible region
CInfeasible region
DNo feasible region
✓ Correct Answer: (B) Bounded feasible region
Solution:
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:
ABounded
BUnbounded
CDoes not exist
DA single point
✓ Correct Answer: (B) Unbounded
Solution:
The constraint x + y ≥ 4 with x - y ≥ 0 in the first quadrant creates an open region extending indefinitely, 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?
A2x + 3y = 12
B2x + 3y ≤ 12
C2x + 3y ≥ 12
Dx + y ≤ 12
✓ Correct Answer: (B) 2x + 3y ≤ 12
Solution:
Labor required = 2x + 3y. Maximum available = 12 hours. Therefore: 2x + 3y ≤ 12
Question 6
The objective function of a linear programming problem is:
AA constraint to be satisfied
BA function to be optimized
CA set of values
DA feasible solution
✓ Correct Answer: (B) A function to be optimized
Solution:
The objective function is the function we want to maximize or minimize (optimize) subject to given constraints.
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:
A0
B12
C20
D29
✓ Correct Answer: (A) 0
Solution:
At (0, 0): Z = 0 (Minimum) · At (4, 0): Z = 20 · At (4, 3): Z = 29 · At (2, 4): Z = 22 · At (0, 4): Z = 12
Question 8
Which of the following is NOT a requirement for a linear programming problem?
AObjective function must be linear
BConstraints must be linear
CVariables must be non-negative
DObjective function must be quadratic
✓ Correct Answer: (D) Objective function must be quadratic
Solution:
In linear programming, both the objective function and constraints must be LINEAR. A quadratic objective function makes it a nonlinear programming problem.
Question 9
The region represented by x ≥ 0, y ≥ 0 is:
AFirst quadrant
BSecond quadrant
CThird quadrant
DFourth quadrant
✓ Correct Answer: (A) First quadrant
Solution:
x ≥ 0 and y ≥ 0 represent the first quadrant where both coordinates are non-negative.
Question 10
In a diet problem, if we want to minimize cost, the objective function represents:
AMaximum nutrients required
BTotal cost of the diet
CNutritional constraints
DQuantity of food
✓ Correct Answer: (B) Total cost of the diet
Solution:
The objective function Z = c₁x₁ + c₂x₂ represents the total cost. Nutritional requirements form the constraints.
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 may not exist
DDepends on the objective function
✓ Correct Answer: (D) Depends on the objective function
Solution:
In an unbounded feasible region, the existence of optimal solution depends on the objective function. 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:
A2 + 3 < 6
B2 + 3 = 6
C2 + 3 ≤ 6
DNone of these
✓ Correct Answer: (C) 2 + 3 ≤ 6
Solution:
Substitute: 2 + 3 = 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:
AZ = 40x + 30y (Maximize)
BZ = 40x + 30y (Minimize)
CZ = 30x + 40y (Maximize)
DZ = x + y (Maximize)
✓ Correct Answer: (A) Z = 40x + 30y (Maximize)
Solution:
Total profit Z = 40x + 30y. We maximize this.
Question 14
The constraints x ≥ 0, y ≥ 0 are called:
AStructural constraints
BNon-negativity constraints
CFunctional constraints
DBinding constraints
✓ Correct Answer: (B) Non-negativity constraints
Solution:
x ≥ 0 and y ≥ 0 are non-negativity constraints ensuring decision variables cannot take negative values.
Question 15
In the corner point method, the optimal solution is:
AAlways at the origin
BAt the center of feasible region
CAt one of the corner points
DCan be anywhere in the region
✓ 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 of the feasible region.
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A furniture manufacturer makes two products: 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. There are 16 hours per day available on machine A and 30 hours on machine B. Profit from a chair is ₹2 and from a table is ₹10. Formulate this as an LPP to maximize total profit.
Maximize Z = 2x + 10y subject to: 2x + 5y ≤ 16, x ≤ 5, x ≥ 0, y ≥ 0
Question 2CBSE 2024
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Food F1 costs ₹4/kg (3 units vitamin A, 4 units minerals per kg). Food F2 costs ₹6/kg (6 units vitamin A, 3 units minerals per kg). Formulate as LPP to minimize cost.
Complete Solution:
Let x = kg of F1, y = kg of F2
Minimize Z = 4x + 6y
Vitamin A: 3x + 6y ≥ 80 · Minerals: 4x + 3y ≥ 100
Minimize Z = 4x + 6y subject to: 3x + 6y ≥ 80, 4x + 3y ≥ 100, x ≥ 0, y ≥ 0
Question 3CBSE 2023
A firm manufactures gold rings and chains. Combined production per day is at most 24. It takes 1 hour for a ring and ½ hour for a chain; 16 hours available per day. Profit: ₹300/ring, ₹190/chain. Maximize profit graphically.
Complete Solution:
Let x = rings, y = chains. Maximize Z = 300x + 190y
Constraints: x + y ≤ 24, x + 0.5y ≤ 16, x ≥ 0, y ≥ 0
Intersection of x + y = 24 and x + 0.5y = 16: y = 16, x = 8 → (8, 16)
At (0,0): Z=0 · At (16,0): Z=4800 · At (8,16): Z=5440 · At (0,24): Z=4560
Maximum profit = ₹5440 when 8 rings and 16 chains are manufactured
Question 4CBSE 2023
Reshma wishes to mix foods P and Q. Vitamin contents needed: at least 8 units of vitamin A and 11 units of vitamin B. Food P: ₹60/kg, 3 units vitamin A, 5 units vitamin B. Food Q: ₹80/kg, 4 units vitamin A, 2 units vitamin B. Find minimum cost.
Complete Solution:
Minimize Z = 60x + 80y; constraints: 3x + 4y ≥ 8, 5x + 2y ≥ 11, x ≥ 0, y ≥ 0
Solving 3x + 4y = 8 and 5x + 2y = 11: 7x = 14 → x = 2, y = 0.5
At (8/3, 0): Z = 160 · At (2, 0.5): Z = 160 · At (0, 5.5): Z = 440
Minimum cost = ₹160 (occurs at multiple points)
Question 5CBSE 2025
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 available 18 hrs/day, M2 available 24 hrs/day. Profit: ₹50/unit A, ₹40/unit B. Find maximum profit graphically.
Complete Solution:
Maximize Z = 50x + 40y; constraints: 3x + 2y ≤ 18, 2x + 4y ≤ 24, x ≥ 0, y ≥ 0
Solving: y = 4.5, x = 3 → corner point (3, 4.5)
At (0,0): Z=0 · At (6,0): Z=300 · At (3,4.5): Z=330 · At (0,6): Z=240
Maximum profit = ₹330 at x = 3, y = 4.5
Question 6CBSE 2024
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. Maximize profit graphically.
Complete Solution:
Let x = Model X, y = Model Y. Maximize Z = 1000x + 1500y
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Case Study Based Questions
Real-world application problems
Case Study 1CBSE 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?
AMaximize Z = 100x + 120y
BMinimize Z = 100x + 120y
CMaximize Z = 4x + 3y
DMinimize Z = 60x + 48y
✓ Correct Answer: (A) Maximize Z = 100x + 120y
Profit from P1 = ₹100x, from P2 = ₹120y. Objective: maximize Z = 100x + 120y.
Q2
What is the constraint for Machine M1?
A4x + 3y ≤ 60
B2x + 3y ≤ 48
C4x + 2y ≤ 60
D3x + 3y ≤ 60
✓ Correct Answer: (A) 4x + 3y ≤ 60
P1 requires 4 hours on M1, P2 requires 3 hours; M1 available 60 hours. Therefore: 4x + 3y ≤ 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
At (0,0): Z=0 · At (15,0): Z=1500 · At (0,16): Z=1920 (Maximum) · At (12,4): Z=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 ₹1920 occurs at (0,16): 0 units of P1 and 16 units of P2.
Case Study 2CBSE 2023
Diet Problem
A dietician wishes to mix two foods such that 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:
AMinimize Z = 50x + 70y
BMaximize Z = 50x + 70y
CMinimize Z = 2x + y
DMaximize Z = 8x + 10y
✓ Correct Answer: (A) Minimize Z = 50x + 70y
We want to minimize total cost = 50x + 70y.
Q2
The constraint for vitamin A is:
A2x + y ≥ 8
Bx + 2y ≥ 10
C2x + y ≤ 8
Dx + y ≥ 8
✓ Correct Answer: (A) 2x + y ≥ 8
Food I has 2 units/kg vitamin A, Food II has 1 unit/kg. Need ≥ 8 units: 2x + y ≥ 8.
Q3
The feasible region for this problem is:
ABounded
BUnbounded
CEmpty
DA single point
✓ Correct Answer: (B) Unbounded
Both constraints are ≥ type, so the feasible region extends indefinitely — it is unbounded.
Q4
If the 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
At (0,10): Z=700 · At (2,4): Z=100+280=380 · At (8,0): Z=400. Minimum is ₹380.
Case Study 3CBSE 2024
Resource Allocation Problem
A company produces leather belts A (superior, ₹40 profit) and B (lower quality, ₹30 profit). Belt A takes twice the time of B. If all were B, 1000 belts/day could be produced. Leather supply sufficient for 800 belts/day. Belt A requires a fancy buckle; only 400 buckles available/day.
Q1
If x and y are the number of belts of type A and B, the objective function is:
AMaximize Z = 40x + 30y
BMinimize Z = 40x + 30y
CMaximize Z = 30x + 40y
DMaximize Z = x + y
✓ Correct Answer: (A) Maximize Z = 40x + 30y
Profit from A = ₹40x, from B = ₹30y. Maximize total profit Z = 40x + 30y.
Q2
The time constraint can be expressed as:
A2x + y ≤ 1000
Bx + 2y ≤ 1000
Cx + y ≤ 500
D2x + y ≤ 500
✓ Correct Answer: (A) 2x + y ≤ 1000
Belt A takes twice the time of B (2 units vs 1 unit). Total capacity = 1000 B-units. So 2x + y ≤ 1000.
Q3
Which constraint represents the leather supply limitation?
Ax + y ≤ 800
Bx ≤ 400
C2x + y ≤ 1000
Dy ≤ 800
✓ Correct Answer: (A) x + y ≤ 800
Leather is sufficient for only 800 belts per day (both types combined): x + y ≤ 800.
Q4
The buckle constraint is:
Ax ≤ 400
By ≤ 400
Cx + y ≤ 400
D2x ≤ 400
✓ Correct Answer: (A) x ≤ 400
Only belt A requires a fancy buckle; 400 buckles available. Therefore: x ≤ 400.
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