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All questions of Chapter 12 - Linear Programming for JEE Exam
How many of the following points satisfy the inequality 2x – 3y > -5?
(1, 1), (-1, 1), (1, -1), (-1, -1), (-2, 1), (2, -1), (-1, 2) and (-2, -1)- a)2
- b)4
- c)6
- d)5
Correct answer is option 'D'. Can you explain this answer?
How many of the following points satisfy the inequality 2x – 3y > -5?
(1, 1), (-1, 1), (1, -1), (-1, -1), (-2, 1), (2, -1), (-1, 2) and (-2, -1)
(1, 1), (-1, 1), (1, -1), (-1, -1), (-2, 1), (2, -1), (-1, 2) and (-2, -1)
a)
2
b)
4
c)
6
d)
5
|
Megha Chopra answered |
It is not possible to answer this question without knowing the values of the points. Please provide more information.
Objective function of a LPP is- a)function to be optimized
- b)A constraint
- c)Relation between variables
- d)Equation in a line
Correct answer is option 'A'. Can you explain this answer?
Objective function of a LPP is
a)
function to be optimized
b)
A constraint
c)
Relation between variables
d)
Equation in a line
|
Ishan Choudhury answered |
Linear programming is an optimization technique for a system of linear constraints and a linear objective function. An objective function defines the quantity to be optimized, and the goal of linear programming is to find the values of the variables that maximize or minimize the objective function.
The set of all feasible solutions of a LPP is a ____ set.- a)Concave
- b)Convex
- c)Feasible
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The set of all feasible solutions of a LPP is a ____ set.
a)
Concave
b)
Convex
c)
Feasible
d)
None of these
|
Varun Kapoor answered |
The set of feasible solutions to an LP (feasible region) forms a (possibly unbounded) convex set.
Maximize Z = 100x + 120y , subject to constraints 2x + 3y ≤ 30, 3x + y ≤ 17, x ≥ 0, y ≥ 0.- a)1260
- b)1200
- c)1300
- d)1280
Correct answer is option 'A'. Can you explain this answer?
Maximize Z = 100x + 120y , subject to constraints 2x + 3y ≤ 30, 3x + y ≤ 17, x ≥ 0, y ≥ 0.
a)
1260
b)
1200
c)
1300
d)
1280
|
Amrita Sarkar answered |
We have , Maximize Z = 100x + 120y , subject to constraints 2x + 3y ≤ 30, 3x + y ≤ 17, x ≥ 0, y ≥ 0.
In linear programming problems the function whose maxima or minima are to be found is called- a)Delta function
- b)Selection function
- c)Subjective function
- d)Objective function
Correct answer is option 'D'. Can you explain this answer?
In linear programming problems the function whose maxima or minima are to be found is called
a)
Delta function
b)
Selection function
c)
Subjective function
d)
Objective function
|
Sinjini Datta answered |
In linear programming problems the function whose maxima or minima are to be found is called Objective function .
Which of the following types of problems cannot be solved by linear programming methods- a)Diet problems
- b)Transportation problems
- c)Manufacturing problems
- d)Traffic signal control
Correct answer is option 'D'. Can you explain this answer?
Which of the following types of problems cannot be solved by linear programming methods
a)
Diet problems
b)
Transportation problems
c)
Manufacturing problems
d)
Traffic signal control
|
Nandini Choudhury answered |
Traffic signal control types of problems cannot be solved by linear programming methods .
In a LPP, the objective function is always- a)cubic
- b)quadratic
- c)Linear
- d)constant
Correct answer is option 'C'. Can you explain this answer?
In a LPP, the objective function is always
a)
cubic
b)
quadratic
c)
Linear
d)
constant
|
Pragati Patel answered |
In a LPP, the objective function is always linear.
Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function. Maximum of F – Minimum of F =- a)60
- b)48
- c)42
- d)18
Correct answer is option 'A'. Can you explain this answer?
Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function. Maximum of F – Minimum of F =
a)
60
b)
48
c)
42
d)
18
|
Rhea Joshi answered |
Here the objective function is given by : F = 4x +6y .
Maximum of F – Minimum of F = 72 – 12 = 30 .
Maximum of F – Minimum of F = 72 – 12 = 30 .
A maximum or a minimum may not exist for a linear programming problem if- a)The feasible region is bounded
- b)if the constraints are non linear
- c)if the objective function is continuous
- d)The feasible region is unbounded
Correct answer is option 'D'. Can you explain this answer?
A maximum or a minimum may not exist for a linear programming problem if
a)
The feasible region is bounded
b)
if the constraints are non linear
c)
if the objective function is continuous
d)
The feasible region is unbounded
|
Milan Roy answered |
A maximum or a minimum may not exist for a linear programming problem if The feasible region is unbounded .
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time. What number of rackets and bats must be made if the factory is to work at full capacity?- a)5 tennis rackets and 15 cricket bats
- b)4 tennis rackets and 12 cricket bats
- c)4 tennis rackets and 14 cricket bats
- d)3 tennis rackets and 13 cricket bats
Correct answer is option 'B'. Can you explain this answer?
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time. What number of rackets and bats must be made if the factory is to work at full capacity?
a)
5 tennis rackets and 15 cricket bats
b)
4 tennis rackets and 12 cricket bats
c)
4 tennis rackets and 14 cricket bats
d)
3 tennis rackets and 13 cricket bats
|
Advait Ghoshal answered |
Let number of rackets made = x
And number of bats made = y
Therefore , the above L.P.P. is given as :
Maximise , Z = x +y , subject to the constraints : 1.5x +3y ≤ 42 and. 3x +y ≤ 24, i.e.0.5x + y ≤ 14 i.e. x +2y ≤ 28 and 3x +y ≤ 24 , x, y ≥ 0.
Here Z = 16 is maximum. i.e Maximum number of rackets = 4 and number of bats = 12.
And number of bats made = y
Therefore , the above L.P.P. is given as :
Maximise , Z = x +y , subject to the constraints : 1.5x +3y ≤ 42 and. 3x +y ≤ 24, i.e.0.5x + y ≤ 14 i.e. x +2y ≤ 28 and 3x +y ≤ 24 , x, y ≥ 0.
Here Z = 16 is maximum. i.e Maximum number of rackets = 4 and number of bats = 12.
In linear programming, optimal solution- a)satisfies all the constraints only
- b)is not unique
- c)satisfies all the constraints as well as the objective function
- d)maximizes the objective function only
Correct answer is option 'C'. Can you explain this answer?
In linear programming, optimal solution
a)
satisfies all the constraints only
b)
is not unique
c)
satisfies all the constraints as well as the objective function
d)
maximizes the objective function only
|
Sameer Saha answered |
Linear Programming and Optimal Solution
Introduction to Linear Programming:
Linear programming is a mathematical optimization technique that helps to optimize a linear objective function subject to a set of linear constraints. It is widely used in various fields such as economics, engineering, operations research, and management.
Optimal Solution in Linear Programming:
The optimal solution in linear programming refers to the best possible solution that satisfies both the objective function and the given constraints. It represents the optimal values of the decision variables that maximize or minimize the objective function while satisfying all the constraints.
Satisfying the Constraints:
The optimal solution should satisfy all the constraints given in the linear programming problem. These constraints define the limitations and boundaries within which the decision variables can vary. The optimal solution ensures that all these constraints are met.
Satisfying the Objective Function:
In addition to satisfying the constraints, the optimal solution should also satisfy the objective function. The objective function represents the goal or objective of the linear programming problem. It can be either maximized or minimized based on the problem's requirements. The optimal solution ensures that the objective function is optimized to its maximum or minimum value.
Uniqueness of Optimal Solution:
The uniqueness of the optimal solution depends on the specific linear programming problem. In some cases, the optimal solution may be unique, meaning there is only one solution that satisfies all the constraints and optimizes the objective function. However, in other cases, there may be multiple optimal solutions that yield the same optimal value for the objective function.
Conclusion:
In linear programming, the optimal solution is the solution that satisfies both the constraints and the objective function. It ensures that all the constraints are met and the objective function is optimized. While the optimal solution may or may not be unique, it represents the best possible solution according to the given problem's requirements.
Introduction to Linear Programming:
Linear programming is a mathematical optimization technique that helps to optimize a linear objective function subject to a set of linear constraints. It is widely used in various fields such as economics, engineering, operations research, and management.
Optimal Solution in Linear Programming:
The optimal solution in linear programming refers to the best possible solution that satisfies both the objective function and the given constraints. It represents the optimal values of the decision variables that maximize or minimize the objective function while satisfying all the constraints.
Satisfying the Constraints:
The optimal solution should satisfy all the constraints given in the linear programming problem. These constraints define the limitations and boundaries within which the decision variables can vary. The optimal solution ensures that all these constraints are met.
Satisfying the Objective Function:
In addition to satisfying the constraints, the optimal solution should also satisfy the objective function. The objective function represents the goal or objective of the linear programming problem. It can be either maximized or minimized based on the problem's requirements. The optimal solution ensures that the objective function is optimized to its maximum or minimum value.
Uniqueness of Optimal Solution:
The uniqueness of the optimal solution depends on the specific linear programming problem. In some cases, the optimal solution may be unique, meaning there is only one solution that satisfies all the constraints and optimizes the objective function. However, in other cases, there may be multiple optimal solutions that yield the same optimal value for the objective function.
Conclusion:
In linear programming, the optimal solution is the solution that satisfies both the constraints and the objective function. It ensures that all the constraints are met and the objective function is optimized. While the optimal solution may or may not be unique, it represents the best possible solution according to the given problem's requirements.
A linear programming problem is one that is concerned with- a)finding the upper limits of a linear function of several variables
- b)finding the lower limit of a linear function of several variables
- c)finding the limiting values of a linear function of several variables
- d)finding the optimal value (maximum or minimum) of a linear function of several variables
Correct answer is option 'D'. Can you explain this answer?
A linear programming problem is one that is concerned with
a)
finding the upper limits of a linear function of several variables
b)
finding the lower limit of a linear function of several variables
c)
finding the limiting values of a linear function of several variables
d)
finding the optimal value (maximum or minimum) of a linear function of several variables
|
Manisha Choudhary answered |
A linear programming problem is one that is concerned with finding the optimal value (maximum or minimum) of a linear function of several variables .
In linear programming problems the optimum solution- a)satisfies a set of piecewise – linear inequalities (called constraints)
- b)satisfies a set of linear inequalities (called linear constraints)
- c)satisfies a set of quadratic inequalities (calledconstraints)
- d)satisfies a set of cubic inequalities (calledconstraints)
Correct answer is option 'B'. Can you explain this answer?
In linear programming problems the optimum solution
a)
satisfies a set of piecewise – linear inequalities (called constraints)
b)
satisfies a set of linear inequalities (called linear constraints)
c)
satisfies a set of quadratic inequalities (calledconstraints)
d)
satisfies a set of cubic inequalities (calledconstraints)
|
Jaideep Mukherjee answered |
Explanation:
Linear Programming Problems:
Linear programming involves optimizing a linear objective function subject to a set of linear constraints. The goal is to find the values of the decision variables that maximize or minimize the objective function while satisfying all constraints.
Optimum Solution:
The optimum solution in linear programming problems refers to the values of the decision variables that result in the maximum or minimum value of the objective function while still meeting all the constraints.
Linear Inequalities:
In linear programming, the constraints are typically represented as linear inequalities. These are mathematical statements that involve linear expressions (variables raised to the power of 1) connected by inequality symbols such as ≤, ≥, or =.
Optimum Solution and Linear Inequalities:
The optimum solution in linear programming satisfies a set of linear inequalities, also known as linear constraints. These constraints define the feasible region within which the optimal solution must lie. The optimal solution is the point within this feasible region that maximizes or minimizes the objective function.
Conclusion:
Therefore, in linear programming problems, the optimum solution always satisfies a set of linear inequalities or linear constraints. This is a fundamental aspect of linear programming and is crucial for finding the most efficient solutions to optimization problems.
Linear Programming Problems:
Linear programming involves optimizing a linear objective function subject to a set of linear constraints. The goal is to find the values of the decision variables that maximize or minimize the objective function while satisfying all constraints.
Optimum Solution:
The optimum solution in linear programming problems refers to the values of the decision variables that result in the maximum or minimum value of the objective function while still meeting all the constraints.
Linear Inequalities:
In linear programming, the constraints are typically represented as linear inequalities. These are mathematical statements that involve linear expressions (variables raised to the power of 1) connected by inequality symbols such as ≤, ≥, or =.
Optimum Solution and Linear Inequalities:
The optimum solution in linear programming satisfies a set of linear inequalities, also known as linear constraints. These constraints define the feasible region within which the optimal solution must lie. The optimal solution is the point within this feasible region that maximizes or minimizes the objective function.
Conclusion:
Therefore, in linear programming problems, the optimum solution always satisfies a set of linear inequalities or linear constraints. This is a fundamental aspect of linear programming and is crucial for finding the most efficient solutions to optimization problems.
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 Rs 4 per unit food and F2 costs Rs 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.- a)Minimum cost = Rs 114
- b)Minimum cost = Rs 134
- c)Minimum cost = Rs 124
- d)Minimum cost = Rs 104
Correct answer is option 'D'. Can you explain this answer?
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 Rs 4 per unit food and F2 costs Rs 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
a)
Minimum cost = Rs 114
b)
Minimum cost = Rs 134
c)
Minimum cost = Rs 124
d)
Minimum cost = Rs 104
|
Devansh Joshi answered |
Let number of units of food F1 = x
And number of units of food F2 = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 4x +6y , subject to the constraints : 3 x + 6y ≥ 80, 4x + 3y ≥ 100, x,y ≥ 0.
Corner points Z =4x +6 y B(80/3 , 0) 320/3 D(24,4/3 ) 104…………………(Min.) A(0,100/3) 200 Here Z = 104 is minimum. i.e. Minimum cost = Rs 104.
And number of units of food F2 = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 4x +6y , subject to the constraints : 3 x + 6y ≥ 80, 4x + 3y ≥ 100, x,y ≥ 0.
Corner points Z =4x +6 y B(80/3 , 0) 320/3 D(24,4/3 ) 104…………………(Min.) A(0,100/3) 200 Here Z = 104 is minimum. i.e. Minimum cost = Rs 104.
Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities,- a)optimal value must occur at a corner point (vertex) of the feasible region.
- b)optimal value must occur at the centroid of the feasible region.
- c)optimal value must occur at the midpoints of the corner points (vertices) of the feasible region
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities,
a)
optimal value must occur at a corner point (vertex) of the feasible region.
b)
optimal value must occur at the centroid of the feasible region.
c)
optimal value must occur at the midpoints of the corner points (vertices) of the feasible region
d)
None of these
|
Poulomi Datta answered |
Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities,then , optimal value must occur at a corner point (vertex) of the feasible region.
The optimal value of the objective function Z = ax + by may or may not exist, if the feasible region for a LPP is- a)a circle
- b)a polygon
- c)Bounded
- d)Unbounded
Correct answer is option 'D'. Can you explain this answer?
The optimal value of the objective function Z = ax + by may or may not exist, if the feasible region for a LPP is
a)
a circle
b)
a polygon
c)
Bounded
d)
Unbounded
|
Akshay Kapoor answered |
The optimal value of the objective function Z = ax + by may or may not exist, if the feasible region for a LPP is unbounded.
Explanation:
To understand why the optimal value of the objective function may or may not exist when the feasible region for a Linear Programming Problem (LPP) is unbounded, let's first define what an unbounded feasible region is.
An unbounded feasible region is a region in the graph of the constraints where there are no restrictions on the values of the decision variables. This means that the feasible region extends infinitely in one or more directions.
Optimal Value of the Objective Function:
The optimal value of the objective function refers to the maximum or minimum value that can be achieved for the given objective function within the feasible region. In other words, it represents the best possible value that can be obtained for the objective function.
Possible Scenarios:
When the feasible region is unbounded, there are two possible scenarios in terms of the optimal value of the objective function:
1. Optimal value exists: In some cases, even though the feasible region is unbounded, there may still exist an optimal value for the objective function. This occurs when the objective function is bounded and has a maximum or minimum value within the feasible region. The optimal value can be found at the extreme point(s) of the feasible region.
2. Optimal value does not exist: In other cases, the objective function may not have an optimal value within the unbounded feasible region. This occurs when the objective function is unbounded as well, meaning it can increase or decrease indefinitely without reaching a maximum or minimum value. In such cases, the objective function does not have a finite optimal value.
Conclusion:
Therefore, the optimal value of the objective function Z = ax + by may or may not exist when the feasible region for a LPP is unbounded. It depends on whether the objective function is bounded or unbounded within the feasible region. If the objective function is bounded, an optimal value exists, whereas if the objective function is unbounded, an optimal value does not exist.
Explanation:
To understand why the optimal value of the objective function may or may not exist when the feasible region for a Linear Programming Problem (LPP) is unbounded, let's first define what an unbounded feasible region is.
An unbounded feasible region is a region in the graph of the constraints where there are no restrictions on the values of the decision variables. This means that the feasible region extends infinitely in one or more directions.
Optimal Value of the Objective Function:
The optimal value of the objective function refers to the maximum or minimum value that can be achieved for the given objective function within the feasible region. In other words, it represents the best possible value that can be obtained for the objective function.
Possible Scenarios:
When the feasible region is unbounded, there are two possible scenarios in terms of the optimal value of the objective function:
1. Optimal value exists: In some cases, even though the feasible region is unbounded, there may still exist an optimal value for the objective function. This occurs when the objective function is bounded and has a maximum or minimum value within the feasible region. The optimal value can be found at the extreme point(s) of the feasible region.
2. Optimal value does not exist: In other cases, the objective function may not have an optimal value within the unbounded feasible region. This occurs when the objective function is unbounded as well, meaning it can increase or decrease indefinitely without reaching a maximum or minimum value. In such cases, the objective function does not have a finite optimal value.
Conclusion:
Therefore, the optimal value of the objective function Z = ax + by may or may not exist when the feasible region for a LPP is unbounded. It depends on whether the objective function is bounded or unbounded within the feasible region. If the objective function is bounded, an optimal value exists, whereas if the objective function is unbounded, an optimal value does not exist.
Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px+qy, where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is- a)p = 2q
- b)p = q
- c)p = 3q
- d)p = q/2
Correct answer is 'D'. Can you explain this answer?
Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px+qy, where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is
a)
p = 2q
b)
p = q
c)
p = 3q
d)
p = q/2
|
Rohit Roy answered |
We have Z = px + qy , At (3, 0) Z = 3p ……………………………….(1) At (1 , 1) Z = p + q …………………………(2) Therefore , from (1) and (2) : We have : p = q/2 .
Maximize Z = – x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.- a)Maximum Z = 12 at (2, 6)
- b)Z has no maximum value
- c)Maximum Z = 10 at (2, 6)
- d)Maximum Z = 14 at (2, 6)
Correct answer is option 'B'. Can you explain this answer?
Maximize Z = – x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
a)
Maximum Z = 12 at (2, 6)
b)
Z has no maximum value
c)
Maximum Z = 10 at (2, 6)
d)
Maximum Z = 14 at (2, 6)
|
Sushant Khanna answered |
Objective function is Z = - x + 2 y ……………………(1).
The given constraints are : x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
The given constraints are : x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
Corner points Z = - x + 2y
Here , the open half plane has points in common with the feasible region .
Therefore , Z has no maximum value.
Here , the open half plane has points in common with the feasible region .
Therefore , Z has no maximum value.
In linear programming infeasible solutions- a)fall on the x = 0 plane
- b)fall inside the feasible region
- c)fall outside the feasible region
- d)fall inside the a regular polygon
Correct answer is option 'C'. Can you explain this answer?
In linear programming infeasible solutions
a)
fall on the x = 0 plane
b)
fall inside the feasible region
c)
fall outside the feasible region
d)
fall inside the a regular polygon
|
Lekshmi Bose answered |
In linear programming infeasible solutions fall outside the feasible region .
Maximise Z = 3x + 4y subject to the constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.- a)Maximum Z = 18 at (1, 4)
- b)Maximum Z = 19 at (1, 5)
- c)Maximum Z = 16 at (0, 4)
- d)Maximum Z = 17 at (0, 5)
Correct answer is option 'C'. Can you explain this answer?
Maximise Z = 3x + 4y subject to the constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.
a)
Maximum Z = 18 at (1, 4)
b)
Maximum Z = 19 at (1, 5)
c)
Maximum Z = 16 at (0, 4)
d)
Maximum Z = 17 at (0, 5)
|
Manisha Mehta answered |
Objective function is Z = 3x + 4 y ……(1).
The given constraints are : x + y ≤ 4, x ≥ 0, y ≥ 0.
therefore Z = 16 is maximum at ( 0 , 4 ) .
The given constraints are : x + y ≤ 4, x ≥ 0, y ≥ 0.
therefore Z = 16 is maximum at ( 0 , 4 ) .
Let R be the feasible region for a linear programming problem,and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and a minimum value on R and- a)each of these occurs at a corner point (vertex) of R.
- b)each of these occurs at themidpoints of the edges of R
- c)each of these occurs at the centre of R.
- d)each of these occurs at some points except corner points of R.
Correct answer is option 'A'. Can you explain this answer?
Let R be the feasible region for a linear programming problem,and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and a minimum value on R and
a)
each of these occurs at a corner point (vertex) of R.
b)
each of these occurs at themidpoints of the edges of R
c)
each of these occurs at the centre of R.
d)
each of these occurs at some points except corner points of R.
|
|
Nandini Choudhury answered |
Let R be the feasible region for a linear programming problem,and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and a minimum value on R and each of these occurs at a corner point (vertex) of R.
Shape of the feasible region formed by the following constraints is x + y ≤ 2, x + y ≥ 5, x ≥ 0, y ≥ 0- a)No feasible region
- b)Triangular region
- c)Unbounded solution
- d)Trapezium
Correct answer is option 'A'. Can you explain this answer?
Shape of the feasible region formed by the following constraints is x + y ≤ 2, x + y ≥ 5, x ≥ 0, y ≥ 0
a)
No feasible region
b)
Triangular region
c)
Unbounded solution
d)
Trapezium
|
Prajjawal Sahu answered |
No feasible region simply means that no solution
Since equation 1 and 2 have no common area
Hence Correct Answer is Option (A)
A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?- a)3 packages of nuts and 3 packages of bolts; Maximum profit = Rs 73.50.
- b)4 packages of nuts and 4 packages of bolts; Maximum profit = Rs 76.50.
- c)3 packages of nuts and 4 packages of bolts; Maximum profit = Rs 74.50.
- d)4 packages of nuts and 3 packages of bolts; Maximum profit = Rs 75.50.
Correct answer is option 'A'. Can you explain this answer?
A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?
a)
3 packages of nuts and 3 packages of bolts; Maximum profit = Rs 73.50.
b)
4 packages of nuts and 4 packages of bolts; Maximum profit = Rs 76.50.
c)
3 packages of nuts and 4 packages of bolts; Maximum profit = Rs 74.50.
d)
4 packages of nuts and 3 packages of bolts; Maximum profit = Rs 75.50.
|
Disha Nair answered |
To maximize the profit, we need to determine the number of packages of nuts and bolts that should be produced each day, given the constraints of the machines' operating hours.
Let's assume that x packages of nuts and y packages of bolts are produced each day.
The time constraint can be represented by the following equation:
1x + 3y ≤ 12 (since machine A takes 1 hour for nuts and 3 hours for bolts, and the total operating hours is 12)
Next, we need to consider the profit constraint. The profit for nuts is Rs17.50 per package, and the profit for bolts is Rs7.00 per package.
The profit equation can be represented as:
Profit = 17.50x + 7.00y
To determine the maximum profit, we need to find the maximum value of the profit equation subject to the constraints mentioned above.
Solving the above equations graphically or using linear programming techniques, we find that the maximum profit is obtained when x = 3 and y = 3.
Substituting these values into the profit equation:
Profit = 17.50(3) + 7.00(3) = Rs 52.50 + Rs 21.00 = Rs 73.50
Therefore, the correct answer is option 'A': 3 packages of nuts and 3 packages of bolts, with a maximum profit of Rs 73.50.
Let's assume that x packages of nuts and y packages of bolts are produced each day.
The time constraint can be represented by the following equation:
1x + 3y ≤ 12 (since machine A takes 1 hour for nuts and 3 hours for bolts, and the total operating hours is 12)
Next, we need to consider the profit constraint. The profit for nuts is Rs17.50 per package, and the profit for bolts is Rs7.00 per package.
The profit equation can be represented as:
Profit = 17.50x + 7.00y
To determine the maximum profit, we need to find the maximum value of the profit equation subject to the constraints mentioned above.
Solving the above equations graphically or using linear programming techniques, we find that the maximum profit is obtained when x = 3 and y = 3.
Substituting these values into the profit equation:
Profit = 17.50(3) + 7.00(3) = Rs 52.50 + Rs 21.00 = Rs 73.50
Therefore, the correct answer is option 'A': 3 packages of nuts and 3 packages of bolts, with a maximum profit of Rs 73.50.
The maximum value of Z = 4x + 3y subjected to the constraints 3x + 2y ≤ 160, 5x + 2y ≤ 200, x + 2y ≤ 80 ;
x, y ≥ 0 attains at- a)x = 40, y = 0
- b)x = 40, y = 25
- c)x = 0, y = 25
- d)x = 30. y = 25
Correct answer is option 'D'. Can you explain this answer?
The maximum value of Z = 4x + 3y subjected to the constraints 3x + 2y ≤ 160, 5x + 2y ≤ 200, x + 2y ≤ 80 ;
x, y ≥ 0 attains at
x, y ≥ 0 attains at
a)
x = 40, y = 0
b)
x = 40, y = 25
c)
x = 0, y = 25
d)
x = 30. y = 25
|
Harshad Nair answered |
Given object function is
Z = 4x+3y
Constraints are
3x + 2y ≥ 160
5x + 2y ≥ 200
x + 2y ≥ 80
x ≥ 0
y≥ 0.
Consider, the inequalities as equalities for some time,
3x + 2y = 160 ; 5x + 2y = 200 and x + 2y = 80
If we convert these into intercept line format equations, we get,
From this form of line, we can say that the line 3x + 2y = 160 meets the x-axis at (,0) and y-axis at (0,80).
This shows the inequality 3x + 2y ≥ 160 holds.
Similarly, from the intercept line format, we can say that the line 5x + 2y = 200 meets the x-axis at (40,0) and y-axis at (0,100).
This shows the inequality 5x + 2y ≥ 200 holds .
Similarly from the intercept line format, we can say that the line x + 2y = 80 meets the x-axis at (80,0) and y-axis at (0,40).
This shows the inequality x + 2y ≥ 80 holds .
Now considering the inequalities, x ≥ 0 and y ≥ 0, this clearly shows the region where both x and y are positive. This represents the 1st quadrant of the graph.
So, the solutions of the LPP are in the first quadrant where the inequalities meet.
We can clearly see that, there is no area in the 1st quadrant where all the three inequalities met.
This clearly says that there is no solution for the LPP with the given constraints.
Hence the option D, is the solution to the problem.
Z = 4x+3y
Constraints are
3x + 2y ≥ 160
5x + 2y ≥ 200
x + 2y ≥ 80
x ≥ 0
y≥ 0.
Consider, the inequalities as equalities for some time,
3x + 2y = 160 ; 5x + 2y = 200 and x + 2y = 80
If we convert these into intercept line format equations, we get,
[Dividing the whole equation with the right hand side number of the equation]
From this form of line, we can say that the line 3x + 2y = 160 meets the x-axis at (,0) and y-axis at (0,80).
This shows the inequality 3x + 2y ≥ 160 holds.
Similarly, from the intercept line format, we can say that the line 5x + 2y = 200 meets the x-axis at (40,0) and y-axis at (0,100).
This shows the inequality 5x + 2y ≥ 200 holds .
Similarly from the intercept line format, we can say that the line x + 2y = 80 meets the x-axis at (80,0) and y-axis at (0,40).
This shows the inequality x + 2y ≥ 80 holds .
Now considering the inequalities, x ≥ 0 and y ≥ 0, this clearly shows the region where both x and y are positive. This represents the 1st quadrant of the graph.
So, the solutions of the LPP are in the first quadrant where the inequalities meet.
Now by plotting all the graphs 3x + 2y ≥ 160 , 5x + 2y ≥ 200 and x + 2y ≥ 80 we get the below graph.
We can clearly see that, there is no area in the 1st quadrant where all the three inequalities met.
This clearly says that there is no solution for the LPP with the given constraints.
Hence the option D, is the solution to the problem.
The optimum value of the objective function is attained at the points- a)Corner points of feasible region
- b)Any point of the feasible region
- c)On x-axis
- d)On y-axis
Correct answer is option 'A'. Can you explain this answer?
The optimum value of the objective function is attained at the points
a)
Corner points of feasible region
b)
Any point of the feasible region
c)
On x-axis
d)
On y-axis
|
Anjali Reddy answered |
Given that,
• There is an objective function
• There are optimal values
From the definition of optimal value of a Linear Programming Problem(LPP):
An optimal/ feasible solution is any point in the feasible region that gives a maximum or minimum value if substituted in the objective function.
Here feasible region of an LPP is defined as:
A feasible region is that common region determined by all the constraints including the non-negative constraints of the LPP.
So the Feasible region of a LPP is a convex polygon where, its vertices (or corner points) determine the optimal values (either maximum/minimum) of the objective function.
For Example,
5x + y ≤ 100 ; x + y ≤ 60 ; x ≥ 0 ; y ≥ 0
The feasible solution of the LPP is given by the convex polygon OADC.
Here, points O, A ,D and C will be optimal solutions of the taken LPP
• There is an objective function
• There are optimal values
From the definition of optimal value of a Linear Programming Problem(LPP):
An optimal/ feasible solution is any point in the feasible region that gives a maximum or minimum value if substituted in the objective function.
Here feasible region of an LPP is defined as:
A feasible region is that common region determined by all the constraints including the non-negative constraints of the LPP.
So the Feasible region of a LPP is a convex polygon where, its vertices (or corner points) determine the optimal values (either maximum/minimum) of the objective function.
For Example,
5x + y ≤ 100 ; x + y ≤ 60 ; x ≥ 0 ; y ≥ 0
The feasible solution of the LPP is given by the convex polygon OADC.
Here, points O, A ,D and C will be optimal solutions of the taken LPP
Can you explain the answer of this question below:A ……… of a feasible region is a point in the region, which is the intersection of two boundary lines.
- A:
Section point
- B:
Corner point
- C:
Reasonable point
- D:
Vertex point
The answer is b.
A ……… of a feasible region is a point in the region, which is the intersection of two boundary lines.
Section point
Corner point
Reasonable point
Vertex point
|
Rahul Bansal answered |
Corner Point:
A vertex of the feasible region. Not every intersection of lines is a corner point. The corner points only occur at a vertex of the feasible region. If there is going to be an optimal solution to a linear programming problem, it will occur at one or more corner points, or on a line segment between two corner points.
One kind of cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.- a)Maximum number of cakes = 30 , 20 of kind one and 10 cakes of another kind
- b)Maximum number of cakes = 32 , 20 of kind one and 12 cakes of another kind
- c)Maximum number of cakes = 34 , 27 of kind one and 7 cakes of another kind
- d)Maximum number of cakes = 33 , 22 of kind one and 11 cakes of another kind
Correct answer is option 'A'. Can you explain this answer?
One kind of cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.
a)
Maximum number of cakes = 30 , 20 of kind one and 10 cakes of another kind
b)
Maximum number of cakes = 32 , 20 of kind one and 12 cakes of another kind
c)
Maximum number of cakes = 34 , 27 of kind one and 7 cakes of another kind
d)
Maximum number of cakes = 33 , 22 of kind one and 11 cakes of another kind
|
|
Pragati Patel answered |
Let number of cakes of first type = x
And number of cakes of second type = y
Therefore , the above L.P.P. is given as :
Minimise , Z = x +y , subject to the constraints : 200x +100y ≤ 5000 and. 25x +50y ≤ 1000, i.e. 2x + y ≤ 50 and x +2y ≤ 40 x, y ≥ 0.
i.e Maximum number of cakes = 30 , 20 of kind one and 10 cakes of another kind .
And number of cakes of second type = y
Therefore , the above L.P.P. is given as :
Minimise , Z = x +y , subject to the constraints : 200x +100y ≤ 5000 and. 25x +50y ≤ 1000, i.e. 2x + y ≤ 50 and x +2y ≤ 40 x, y ≥ 0.
i.e Maximum number of cakes = 30 , 20 of kind one and 10 cakes of another kind .
The feasible region for a LPP is shown in Figure. Evaluate Z = 4x + y at each of the corner points of this region. Find the minimum value of Z, if it exists
- a)Minimum value = 2
- b)Minimum value = 5
- c)Minimum value = 3
- d)Minimum value = 4
Correct answer is option 'A'. Can you explain this answer?
The feasible region for a LPP is shown in Figure. Evaluate Z = 4x + y at each of the corner points of this region. Find the minimum value of Z, if it exists
a)
Minimum value = 2
b)
Minimum value = 5
c)
Minimum value = 3
d)
Minimum value = 4
|
Bhavana Das answered |
In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same- a)Minimum value
- b)Mean value
- c)Maximum value
- d)Upper limit value
Correct answer is option 'C'. Can you explain this answer?
In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same
a)
Minimum value
b)
Mean value
c)
Maximum value
d)
Upper limit value
|
|
Rhea Joshi answered |
In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same maximum value .
Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then- a)the objective function Z has only a maximum value on R
- b)the objective function Z has only a minimum value on R
- c)the objective function Z has both a maximum and a minimum value on R
- d)the objective function Z has no minimum value on R
Correct answer is option 'C'. Can you explain this answer?
Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then
a)
the objective function Z has only a maximum value on R
b)
the objective function Z has only a minimum value on R
c)
the objective function Z has both a maximum and a minimum value on R
d)
the objective function Z has no minimum value on R
|
|
Athul Patel answered |
Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and a minimum value on R .
Maximize Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.- a)Maximum Z = 20 at (4, 3)
- b)Maximum Z = 22 at (4, 3)
- c)Maximum Z = 24 at (4, 3)
- d)Maximum Z = 18 at (4, 3)
Correct answer is option 'D'. Can you explain this answer?
Maximize Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
a)
Maximum Z = 20 at (4, 3)
b)
Maximum Z = 22 at (4, 3)
c)
Maximum Z = 24 at (4, 3)
d)
Maximum Z = 18 at (4, 3)
|
Dipika Choudhury answered |
Objective function is Z = 3x + 2 y ……………………(1).
The given constraints are : x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
The given constraints are : x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
Maximize Z = 3x + 4y, subject to the constraints: x + y ≤ 1, x ≥ 0, y ≥ 0.- a)2
- b)4
- c)6
- d)5
Correct answer is option 'B'. Can you explain this answer?
Maximize Z = 3x + 4y, subject to the constraints: x + y ≤ 1, x ≥ 0, y ≥ 0.
a)
2
b)
4
c)
6
d)
5
|
Rounak Desai answered |
Here , maximize , Z = 3x + 4y, subject to the constraints: x + y ≤ 1, x ≥ 0, y ≥ 0.
A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.
If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What isthe maximum amount of nitrogen added?- a)150 bags of brand P and 50 bags of brand Q; Maximum amount of nitrogen = 625 kg
- b)160 bags of brand P and 52 bags of brand Q; Maximum amount of nitrogen = 635 kg
- c)140 bags of brand P and 50 bags of brand Q; Maximum amount of nitrogen = 595 kg
- d)145 bags of brand P and 55 bags of brand Q; Maximum amount of nitrogen = 555 kg
Correct answer is option 'C'. Can you explain this answer?
A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.
If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What isthe maximum amount of nitrogen added?
a)
150 bags of brand P and 50 bags of brand Q; Maximum amount of nitrogen = 625 kg
b)
160 bags of brand P and 52 bags of brand Q; Maximum amount of nitrogen = 635 kg
c)
140 bags of brand P and 50 bags of brand Q; Maximum amount of nitrogen = 595 kg
d)
145 bags of brand P and 55 bags of brand Q; Maximum amount of nitrogen = 555 kg
|
Akash Shah answered |
Let the number of bags used for fertilizer of brand P = x And the number of bags used for fertilizer of brand Q = y . Here , Z = 3x + 3.5y subject to constraints : :1.5 x +2 y ≤ 310, x + 2y ≥ 240, 3x + 1.5y ≥ 270 , x,y ≥ 0
Here Z = 595 is maximum i.e. 140 bags of brand P and 50 bags of brand Q; Maximum amount of nitrogen = 595 kg .
Here Z = 595 is maximum i.e. 140 bags of brand P and 50 bags of brand Q; Maximum amount of nitrogen = 595 kg .
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximize his profit?- a)5 Pedestal lamps and 4 wooden shades; Maximum profit = Rs 35
- b)5 Pedestal lamps and 5 wooden shades; Maximum profit = Rs 38
- c)4 Pedestal lamps and 4 wooden shades; Maximum profit = Rs 32
- d)4 Pedestal lamps and 5 wooden shades; Maximum profit = Rs 36
Correct answer is option 'C'. Can you explain this answer?
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximize his profit?
a)
5 Pedestal lamps and 4 wooden shades; Maximum profit = Rs 35
b)
5 Pedestal lamps and 5 wooden shades; Maximum profit = Rs 38
c)
4 Pedestal lamps and 4 wooden shades; Maximum profit = Rs 32
d)
4 Pedestal lamps and 5 wooden shades; Maximum profit = Rs 36
|
|
Bhavana Das answered |
Let number of pedestal lamps manufactured = x
And number of wooden shades manufactured = y
Therefore , the above L.P.P. is given as :
Maximise , Z = 5x +3y , subject to the constraints : 2x +y ≤ 12 and. 3x +2y ≤ 20 , x, y ≥ 0.
Here Z = 32 is maximum.
i.e 30 packages of screws A and 20 packages of screws B; Maximum profit = Rs 410.
i.e. 4 Pedestal lamps and 4 wooden shades; Maximum profit = Rs 32 .
And number of wooden shades manufactured = y
Therefore , the above L.P.P. is given as :
Maximise , Z = 5x +3y , subject to the constraints : 2x +y ≤ 12 and. 3x +2y ≤ 20 , x, y ≥ 0.
Here Z = 32 is maximum.
i.e 30 packages of screws A and 20 packages of screws B; Maximum profit = Rs 410.
i.e. 4 Pedestal lamps and 4 wooden shades; Maximum profit = Rs 32 .
In Corner point method for solving a linear programming problem the first step is to- a)Find the feasible region of the linear programming problem and determine its center points (vertices).
- b)Find the infeasible region of the linear programming problem and determine its complement
- c)Find the infeasible regions of the linear programming problem and determine theunion of the infeasible regions
- d)Find the feasible region of the linear programming problem and determine its corner points (vertices).
Correct answer is option 'D'. Can you explain this answer?
In Corner point method for solving a linear programming problem the first step is to
a)
Find the feasible region of the linear programming problem and determine its center points (vertices).
b)
Find the infeasible region of the linear programming problem and determine its complement
c)
Find the infeasible regions of the linear programming problem and determine theunion of the infeasible regions
d)
Find the feasible region of the linear programming problem and determine its corner points (vertices).
|
Mira Roy answered |
In Corner point method for solving a linear programming problem the first step is : To find the feasible region of the linear programming problem and determine its corner points (vertices)
Which of the following sets are convex ?- a){(x, y) : x2 + y2 ≥ 1}
- b){(x,y) : y≥2 , y≤4}
- c){(x, y) : 3x2 + 4y2 ≥ 5}
- d){(x, y) : y2 ≥ x}
Correct answer is option 'B'. Can you explain this answer?
Which of the following sets are convex ?
a)
{(x, y) : x2 + y2 ≥ 1}
b)
{(x,y) : y≥2 , y≤4}
c)
{(x, y) : 3x2 + 4y2 ≥ 5}
d)
{(x, y) : y2 ≥ x}
|
Gowri Iyer answered |
{(x,y) : y≥2 , y≤4} is the region between two parallel lines, so any line segment joining any two points in it lies in it. Hence, it is the convex set.
Problems which seek to maximise or, minimise profit or, cost form a general class of problems called ………- a)Simple problems
- b)Difficult problems
- c)Non-linear problems
- d)Optimisation problems
Correct answer is option 'C'. Can you explain this answer?
Problems which seek to maximise or, minimise profit or, cost form a general class of problems called ………
a)
Simple problems
b)
Difficult problems
c)
Non-linear problems
d)
Optimisation problems
|
Ritika Kulkarni answered |
In a non linear problem, the constant of variation changes from on data point to another. An example of a non-linear proble
A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden? should the delivery be scheduled in order that the transportation cost is minimum? 
- a)45 bags of brand P and 105 bags of brand Q; Minimum amount of nitrogen = 480 kg.
- b)40 bags of brand P and 100 bags of brand Q; Minimum amount of nitrogen = 470 kg.
- c)50 bags of brand P and 100 bags of brand Q; Minimum amount of nitrogen = 490 kg.
- d)47 bags of brand P and 107 bags of brand Q; Minimum amount of nitrogen = 499 kg.
Correct answer is option 'B'. Can you explain this answer?
A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden? should the delivery be scheduled in order that the transportation cost is minimum?
a)
45 bags of brand P and 105 bags of brand Q; Minimum amount of nitrogen = 480 kg.
b)
40 bags of brand P and 100 bags of brand Q; Minimum amount of nitrogen = 470 kg.
c)
50 bags of brand P and 100 bags of brand Q; Minimum amount of nitrogen = 490 kg.
d)
47 bags of brand P and 107 bags of brand Q; Minimum amount of nitrogen = 499 kg.
|
|
Nandini Choudhury answered |
Let the number of bags used for fertilizer of brand P = x And the number of bags used for fertilizer of brand Q = y . Here , Z = 3x + 3.5y subject to constraints : :1.5 x +2 y ≤ 310, x + 2y ≥ 240, 3x + 1.5y ≥ 270 , x,y ≥ 0
Here Z = 470 is minimum i.e. 40 bags of brand P and 100 bags of brand Q; Minimum amount of nitrogen = 470 kg.
Here Z = 470 is minimum i.e. 40 bags of brand P and 100 bags of brand Q; Minimum amount of nitrogen = 470 kg.
An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:
Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum?What is the minimum cost?- a)From A: 550, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4800
- b)From A: 540, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4700
- c)From A: 500, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4400
- d)From A: 520, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4600
Correct answer is option 'C'. Can you explain this answer?
An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:
Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum?What is the minimum cost?
a)
From A: 550, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4800
b)
From A: 540, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4700
c)
From A: 500, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4400
d)
From A: 520, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4600
|
Prisha Yadav answered |
Here objective function is : Z = 3x + y + 39500 , subject to constraints : : x + y ≤ 7000, x ≤ 4500, x + y ≥ 3500, , y ≤ 3000x , x,y ≥ 0
Here Z = 4400 is minimum.i.e. . From A: 500, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4400 .
Here Z = 4400 is minimum.i.e. . From A: 500, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F respectively; Minimum cost = Rs 4400 .
A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.- a)200 units of desktop model and 50 units of portable model; Maximum profit = Rs 1150000
- b)260 units of desktop model and 50 units of portable model; Maximum profit = Rs 1150400
- c)220 units of desktop model and 50 units of portable model; Maximum profit = Rs 1150020
- d)240 units of desktop model and 50 units of portable model; Maximum profit = Rs 1150040
Correct answer is option 'A'. Can you explain this answer?
A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.
a)
200 units of desktop model and 50 units of portable model; Maximum profit = Rs 1150000
b)
260 units of desktop model and 50 units of portable model; Maximum profit = Rs 1150400
c)
220 units of desktop model and 50 units of portable model; Maximum profit = Rs 1150020
d)
240 units of desktop model and 50 units of portable model; Maximum profit = Rs 1150040
|
Sanchita Iyer answered |
Let number of desktop model computers = x
And number of portable model computers = y
Therefore , the above L.P.P. is given as :
Maximise , Z = 4500x +5000y , subject to the constraints : x +y ≤ 250 and 25000x +40000y ≤ 700000 .i.e. x +y ≤ 250 and 5x +8y ≤ 1400 , x, y ≥ 0.
Here Z = 1150000 is maximum.
i.e. 200 units of desktop model and 50 units of portable model; Maximum profit = Rs 1150000 .
And number of portable model computers = y
Therefore , the above L.P.P. is given as :
Maximise , Z = 4500x +5000y , subject to the constraints : x +y ≤ 250 and 25000x +40000y ≤ 700000 .i.e. x +y ≤ 250 and 5x +8y ≤ 1400 , x, y ≥ 0.
Here Z = 1150000 is maximum.
i.e. 200 units of desktop model and 50 units of portable model; Maximum profit = Rs 1150000 .
Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.- a)Minimum Z = 7 at all the points on the line segment joining the points (6, 0) and (0, 3).
- b)Minimum Z = 6 at all the points on the line segment joining the points (6, 0) and (0, 3).
- c)Minimum Z = 8 at all the points on the line segment joining the points (6, 0) and (0, 3).
- d)Minimum Z = 9 at all the points (6, 0) and (0, 3).
Correct answer is option 'B'. Can you explain this answer?
Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
a)
Minimum Z = 7 at all the points on the line segment joining the points (6, 0) and (0, 3).
b)
Minimum Z = 6 at all the points on the line segment joining the points (6, 0) and (0, 3).
c)
Minimum Z = 8 at all the points on the line segment joining the points (6, 0) and (0, 3).
d)
Minimum Z = 9 at all the points (6, 0) and (0, 3).
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Avantika Saha answered |
Objective function is Z = x + 2y ……………………(1).
The given constraints are : 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0 .
Here , Z = 18 is minimum at (0, 3) and (6 , 0) .
Minimum Z = 6 at all the points on the line segment joining the points (6, 0) and (0, 3).
The given constraints are : 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0 .
Here , Z = 18 is minimum at (0, 3) and (6 , 0) .
Minimum Z = 6 at all the points on the line segment joining the points (6, 0) and (0, 3).
Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.- a)Minimum Z = 7 at 3/2,1/2
- b)Minimum Z = 8 at 3/2,1/2
- c)Minimum Z = 9 at 3/2,1/2 110 V 60 Hz
- d)Minimum Z = 10 at 3/2,1/2
Correct answer is option 'A'. Can you explain this answer?
Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
a)
Minimum Z = 7 at 3/2,1/2
b)
Minimum Z = 8 at 3/2,1/2
c)
Minimum Z = 9 at 3/2,1/2 110 V 60 Hz
d)
Minimum Z = 10 at 3/2,1/2
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Mira Roy answered |
Objective function is Z = 3x + 5 y ……………………(1).
The given constraints are : x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0 .
The given constraints are : x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0 .
The linear inequalities or equations or restrictions on the variables of a linear programming problem are called …… The conditions x ≥ 0 , y ≥ 0 are called …….- a)Objective functions, optimal value
- b)Constraints, non-negative restrictions
- c)Objective functions, non-negative restrictions
- d)Constraints, negative restrictions
Correct answer is option 'B'. Can you explain this answer?
The linear inequalities or equations or restrictions on the variables of a linear programming problem are called …… The conditions x ≥ 0 , y ≥ 0 are called …….
a)
Objective functions, optimal value
b)
Constraints, non-negative restrictions
c)
Objective functions, non-negative restrictions
d)
Constraints, negative restrictions
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Abhay Chauhan answered |
Understanding Linear Programming Terms
Linear programming is a mathematical method used to maximize or minimize a linear objective function, subject to a set of linear inequalities or equations. In this context, it’s essential to differentiate between various components of the problem.
What are Constraints?
- Constraints are the linear inequalities or equations that limit the values of the variables in a linear programming problem.
- They define the feasible region, representing all possible solutions that satisfy these restrictions.
- For example, if we have constraints like x + y ≤ 10, this means that the sum of x and y cannot exceed 10.
What are Non-Negative Restrictions?
- The conditions x ≥ 0 and y ≥ 0 are referred to as non-negative restrictions.
- These ensure that the values of the variables are not negative, which is crucial in many real-world scenarios, such as production quantities or resource allocations.
- Non-negativity is integral to ensuring realistic and practical solutions in linear programming problems.
Why Option B is Correct?
- The correct answer is option 'B' because it appropriately labels the linear inequalities as "constraints" and the conditions x ≥ 0, y ≥ 0 as "non-negative restrictions."
- This terminology aligns with standard definitions in linear programming, making it crucial for understanding the problem structure and finding optimal solutions.
In summary, recognizing the terminology in linear programming helps clarify the problem and guides the approach to finding feasible and optimal solutions.
Linear programming is a mathematical method used to maximize or minimize a linear objective function, subject to a set of linear inequalities or equations. In this context, it’s essential to differentiate between various components of the problem.
What are Constraints?
- Constraints are the linear inequalities or equations that limit the values of the variables in a linear programming problem.
- They define the feasible region, representing all possible solutions that satisfy these restrictions.
- For example, if we have constraints like x + y ≤ 10, this means that the sum of x and y cannot exceed 10.
What are Non-Negative Restrictions?
- The conditions x ≥ 0 and y ≥ 0 are referred to as non-negative restrictions.
- These ensure that the values of the variables are not negative, which is crucial in many real-world scenarios, such as production quantities or resource allocations.
- Non-negativity is integral to ensuring realistic and practical solutions in linear programming problems.
Why Option B is Correct?
- The correct answer is option 'B' because it appropriately labels the linear inequalities as "constraints" and the conditions x ≥ 0, y ≥ 0 as "non-negative restrictions."
- This terminology aligns with standard definitions in linear programming, making it crucial for understanding the problem structure and finding optimal solutions.
In summary, recognizing the terminology in linear programming helps clarify the problem and guides the approach to finding feasible and optimal solutions.
Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.- a)Minimum Z = – 14 at (5, 0)
- b)Minimum Z = – 12 at (4, 0)
- c)Minimum Z = – 15 at (5, 1)
- d)Minimum Z = – 13 at (4, 1)
Correct answer is option 'B'. Can you explain this answer?
Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
a)
Minimum Z = – 14 at (5, 0)
b)
Minimum Z = – 12 at (4, 0)
c)
Minimum Z = – 15 at (5, 1)
d)
Minimum Z = – 13 at (4, 1)
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Devansh Joshi answered |
Objective function is Z = - 3x + 4 y ……………………(1).
The given constraints are : x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
Here , Z = -12 is minimum at C (4 , 0) .
The given constraints are : x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
Here , Z = -12 is minimum at C (4 , 0) .
Chapter doubts & questions for Chapter 12 - Linear Programming - Mathematics (Maths) Class 12 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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