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All questions of Chapter 12 - Linear Programming for Commerce Exam

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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?

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.

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?

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.

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?

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.

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?

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.

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?

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.
Corner points Z =  - x + 2y

Here , the open half plane has points in common with the feasible region .
Therefore , Z has no maximum value.

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?

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.

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?

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)

For chapter Notes on Linear Programming click on the link given below:
Linear Programming

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?

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.

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.

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.

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?

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,

[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?

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

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?

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 .

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?

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 .

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?

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 .

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?

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.

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?

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.

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?

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 .

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?

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 .

A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
  • a)
    830 dolls of type A and 430 dolls of type B; Maximum profit = Rs 16300
  • b)
    800 dolls of type A and 400 dolls of type B; Maximum profit = Rs 16000
  • c)
    840 dolls of type A and 404 dolls of type B; Maximum profit = Rs 16500
  • d)
    820 dolls of type A and 420 dolls of type B; Maximum profit = Rs 16200
Correct answer is option 'B'. Can you explain this answer?

Manisha Choudhary answered
Here , Maximise Z = 12x + 16y , subject to constraints : : x + y ≤ 1200, x - 2y ≥ 0, x - 3y ≤ 600 , x,y ≥ 0

Here Z = 16000 is maximum.i.e. 800 dolls of type A and 400 dolls of type B; Maximum profit = Rs 16000 .

Chapter doubts & questions for Chapter 12 - Linear Programming - Mathematics CUET Preparation 2024 is part of Commerce exam preparation. The chapters have been prepared according to the Commerce exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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