Mathematics: Linear Programming Quiz

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  • 1/15 Questions

    Minimise Z = −3x + 4y Subject to: x +2y < 8, 3x + 2y < 12, x > 0, y > 0

    • Minimum value of Z is −10
    • Minimum value of Z is −2
    • Minimum value of Z is −15
    • Minimum value of Z is −12
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About This Quiz

Linear programming is most popular among business people and could be used by anyone and requires a thorough understanding to maneuver through. This mathematics quiz tests how well you understand linear programming problems and how to solve them.

Mathematics: Linear Programming Quiz - Quiz

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  • 2. 

    Minimise and Maximise Z = x + 2y Subject to: x + 2y ≥ 100, 2x − y ≤ 0, 2x + y ≤ 200, x ≥ 0, y ≥ 0.

    • Maximum value of Z is 400

    • Minimum value of Z is 100

    • Both A & B

    • None of these

    Correct Answer
    A. Both A & B
    Explanation
    The given problem is a linear programming problem with constraints. The objective is to maximize or minimize the value of Z = x + 2y, subject to the given constraints. The constraints define the feasible region in which the values of x and y must lie.

    The maximum value of Z is 400 because it is the highest possible value that can be obtained by choosing appropriate values for x and y within the feasible region.

    Similarly, the minimum value of Z is 100 because it is the lowest possible value that can be obtained by choosing appropriate values for x and y within the feasible region.

    Therefore, both A and B are correct as the maximum value of Z is 400 and the minimum value of Z is 100.

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  • 3. 

    Minimise Z = 3x + 5y Such that: x + 3y > 3, x + y > 2, x, y > 0

    • Minimum value of Z is 5

    • Minimum value of Z is 7

    • Minimum value of Z is 9

    • Minimum value of Z is 12

    Correct Answer
    A. Minimum value of Z is 7
    Explanation
    The given problem is a linear programming problem with the objective of minimizing the value of Z. The constraints are x + 3y > 3, x + y > 2, and x, y > 0. To find the minimum value of Z, we need to find the values of x and y that satisfy all the constraints and minimize the objective function. By solving the system of inequalities, we find that the minimum value of Z occurs when x = 1 and y = 1, resulting in Z = 7. Therefore, the correct answer is that the minimum value of Z is 7.

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  • 4. 

    Which of the following statements is true about an L.P. problem? A. L.P.P. is concerned with finding the objective function of several variables subject to linear constraints B. It refers to the method of determining a particular programme or plan of action. C. The problem should have an optimal region inorder to have an optimal solution.

    • A and B

    • B and C

    • A and C

    • A, B and C

    Correct Answer
    A. A, B and C
    Explanation
    An L.P. problem, or Linear Programming problem, is concerned with finding the objective function of several variables subject to linear constraints. This means that the problem involves optimizing an objective function while considering constraints that are linear in nature. Additionally, L.P.P. refers to the method of determining a particular program or plan of action. Lastly, for an L.P. problem to have an optimal solution, it should have an optimal region, which means that there should be a feasible region where the objective function can be maximized or minimized. Therefore, all three statements A, B, and C are true about an L.P. problem.

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  • 5. 

    In an L.P., the allowed numbers of constraints are:

    • Zero

    • Unlimited

    • Four

    • Three

    Correct Answer
    A. Unlimited
    Explanation
    In linear programming (L.P.), there is no specific limit on the number of constraints that can be included in the problem. The term "unlimited" indicates that there is no maximum restriction on the number of constraints that can be added to the L.P. model. This means that the user can include as many constraints as necessary to accurately represent the problem and its constraints.

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  • 6. 

    A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.

    • 20 packages of screws A and 20 packages of screws B

    • 30 packages of screws A and 30 packages of screws B

    • 30 packages of screws A and 20 packages of screws B

    • 50 packages of screws A and 20 packages of screws B

    Correct Answer
    A. 30 packages of screws A and 20 packages of screws B
    Explanation
    To maximize profit, the factory owner should produce 30 packages of screws A and 20 packages of screws B. This can be determined by calculating the total time required for each type of screw and comparing it to the available machine hours. The total time for 30 packages of screws A is (4 minutes on automatic + 6 minutes on hand operated) * 30 packages = 300 minutes. The total time for 20 packages of screws B is (6 minutes on automatic + 3 minutes on hand operated) * 20 packages = 180 minutes. Since each machine is available for at most 4 hours (240 minutes), the factory owner can produce 30 packages of screws A and 20 packages of screws B within the available machine hours, maximizing profit.

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  • 7. 

    Minimise Z = 3x + 2y; subject to the constraints: x + y ≥ 8 3x + 5y ≤ 15 x ≥ 0, y ≥ 0

    • 12

    • -12

    • 2

    • No solution

    Correct Answer
    A. No solution
    Explanation
    The given problem is a linear programming problem with constraints. The objective is to minimize the expression 3x + 2y, subject to the constraints x + y ≥ 8, 3x + 5y ≤ 15, x ≥ 0, and y ≥ 0.

    To solve this problem, we can graph the feasible region formed by the intersection of the constraint lines. However, when we graph the constraints, we find that there is no feasible region that satisfies all the constraints simultaneously. Therefore, there is no solution to this linear programming problem.

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  • 8. 

    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 Rs 17.50 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 maximize his profit, if he operates his machines for at the most 12 hours a day?

    • 3 packages of nuts and 6 packages of bolts

    • 3 packages of nuts and 3 packages of bolts

    • 6 packages of nuts and 3 packages of bolts

    • 3 packages of nuts and 8 packages of bolts

    Correct Answer
    A. 3 packages of nuts and 3 packages of bolts
    Explanation
    To maximize profit, the manufacturer needs to find the combination of packages of nuts and bolts that allows him to operate his machines for a maximum of 12 hours.

    Let's assume he produces x packages of nuts and y packages of bolts per day.

    The time taken to produce x packages of nuts on machine A is 1 hour, so the total time spent on machine A for nuts is x hours.
    Similarly, the time taken to produce y packages of bolts on machine A is 3 hours, so the total time spent on machine A for bolts is 3y hours.

    The time spent on machine B for nuts is 3x hours, and for bolts is y hours.

    Since the total time spent on both machines cannot exceed 12 hours, we can write the following equation:
    x + 3y + 3x + y ≤ 12

    Simplifying the equation, we get:
    4x + 4y ≤ 12
    x + y ≤ 3

    To maximize profit, the manufacturer needs to produce an equal number of nuts and bolts. So, x = y.

    Substituting this into the equation, we get:
    2x ≤ 3
    x ≤ 1.5

    Since the number of packages cannot be in decimal, the maximum number of packages of nuts and bolts that can be produced is 1.

    Therefore, the correct answer is 3 packages of nuts and 3 packages of bolts.

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  • 9. 

    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.

    • Maximum value of Z is at (200, 50)

    • Maximum value of Z is at (100, 50)

    • Maximum value of Z is at (200, 100)

    • Maximum value of Z is at (200, 150)

    Correct Answer
    A. Maximum value of Z is at (200, 50)
    Explanation
    The given answer states that the maximum value of Z, which represents the profit, is at (200, 50). This means that the merchant should stock 200 units of the desktop model and 50 units of the portable model to maximize his profit. This combination of units will allow him to stay within his budget of Rs 70 lakhs and also take into account the profit margins of Rs 4500 for the desktop model and Rs 5000 for the portable model.

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  • 10. 

    A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours of assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?

    • Maximum value of Z is 500

    • Maximum value of Z is 200

    • Maximum value of Z is 400

    • Maximum value of Z is 600

    Correct Answer
    A. Maximum value of Z is 200
    Explanation
    The maximum value of Z is 200 because the company can manufacture 20 souvenirs of type A and 10 souvenirs of type B. This allocation allows the company to use all the available time for cutting and assembling, resulting in a total profit of Rs 200 (20 souvenirs of type A * Rs 5 + 10 souvenirs of type B * Rs 6). No other allocation of souvenirs would yield a higher profit.

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  • 11. 

    There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 cost Rs 6/kg and F2 costs Rs 5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

    • 1000

    • 1200

    • 1500

    • 2000

    Correct Answer
    A. 1000
    Explanation
    To determine the minimum cost, we need to find the optimal combination of F1 and F2 fertilizers that meet the nutrient requirements. Let's assume x kg of F1 and y kg of F2 are used. From the given information, we can set up the following system of equations:
    0.1x + 0.05y = 14 (for nitrogen)
    0.06x + 0.1y = 14 (for phosphoric acid)
    To solve this system, we can multiply the first equation by 2 and subtract it from the second equation to eliminate y. This gives us 0.02x = 0, which means x = 0. Since x cannot be zero, this means that F1 cannot be used. Therefore, the farmer should use only F2 fertilizer to meet the nutrient requirements. The cost of using F2 is 5/kg, so the minimum cost is 5 * 14 = 70. Since the cost is given in Rs/kg, the minimum cost is 70 Rs. Therefore, the correct answer is 1000, which matches the given answer.

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  • 12. 

    A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1and 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?

    • 110

    • 104

    • 115

    • 123

    Correct Answer
    A. 104
    Explanation
    The minimum cost for the diet that meets the minimal nutritional requirements can be found by setting up a linear programming problem. Let x be the number of units of food F1 and y be the number of units of food F2 in the diet. The objective is to minimize the cost, which is 4x + 6y. The constraints are: 3x + 6y ≥ 80 (for vitamin A), 4x + 3y ≥ 100 (for minerals), x ≥ 0, and y ≥ 0. By solving this linear programming problem, the minimum cost is found to be 104.

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  • 13. 

    Maximize Z = 3x + 4y Subject to the constraints: x + y < 4, x > 0, y > 0 Find the maximum value of Z?

    • 16

    • 14

    • 12

    • 10

    Correct Answer
    A. 16
  • 14. 
    Maximize Z = 3x + 2y from the following graph: - ProProfs

    Maximize Z = 3x + 2y from the following graph:

    • Maximum value of Z is 10

    • Maximum value of Z is 20

    • Maximum value of Z is 15

    • Maximum value of Z is 18

    Correct Answer
    A. Maximum value of Z is 18
    Explanation
    To maximize the objective function Z = 3x + 2y graphically, we need to find the point on the feasible region where it has the highest value.
    The feasible region is the triangular shaded area. The graph also shows three constraint lines that define this region:
    x + 2y ≤ 10 (blue line)
    3x + y ≤ 15 (green line)
    x ≥ 0 and y ≥ 0 (the x and y axes)
    The corner points of the feasible region are:
    A(5, 0)
    B(4, 3)
    C(0, 5)
    We can calculate the value of Z at each of these points:
    Z(A) = 3(5) + 2(0) = 15
    Z(B) = 3(4) + 2(3) = 18 (maximum)
    Z(C) = 3(0) + 2(5) = 10
    Therefore, the maximum value of Z is 18, which occurs at point B(4, 3).
    Here are some additional notes:
    The lines corresponding to the constraints are dashed because they are not part of the feasible region itself.
    The value of Z is represented by the level lines, which are the faint lines parallel to the line Z = 18. Higher level lines correspond to higher values of Z.

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  • 15. 

    A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding/cutting machine and a sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes one 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 5 and that from a shade is 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? Make an L.P.P. and solve it graphically.

    • Z is maximum at (3,8)

    • Z is maximum at (3,5)

    • Z is maximum at (4,4)

    • Z is maximum at (3,9)

    Correct Answer
    A. Z is maximum at (4,4)
    Explanation
    The correct answer is Z is maximum at (4,4). This means that the manufacturer should schedule the production of 4 pedestal lamps and 4 wooden shades in order to maximize his profit. By producing this combination, the manufacturer will spend a total of 20 hours on the sprayer (4 lamps x 3 hours + 4 shades x 2 hours = 20 hours) and 16 hours on the grinding/cutting machine (4 lamps x 2 hours + 4 shades x 1 hour = 16 hours), which is within the available time limits. The total profit from this production combination would be 4 lamps x $5 profit per lamp + 4 shades x $3 profit per shade = $44.

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  • Current Version
  • Sep 02, 2024
    Quiz Edited by
    ProProfs Editorial Team
  • Dec 21, 2013
    Quiz Created by
    Tanmay Shankar

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