﻿ Linear Programming Worksheet | Problems & Solutions Linear Programming Worksheet

Linear Programming Worksheet
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1.
A person wants to invest $20,000 in two types of bonds A and B. Bond A yields 17% returns and bond B yields 22% returns on the amount invested. After some consideration, he decides to invest at least$5,000 in bond A and not more than $8,000 in bond B. He also wants to invest at least as much in bond A as in bond B. Which of the following investments gives him maximum returns? a.$12,000 in bond A and $8,000 in bond B b.$10,000 in bond A and $10,000 in bond B c.$3,400 in bond A and $4,400 in bond B d.$5,000 in bond A and $8,000 in bond B Solution: Let$x be the amount invested in bond A and $y be the amount invested in bond B. The objective function is, Z = 0.17x + 0.22y. [Bond A yields 17% returns and bond B yields 22% returns, we have to maximize 0.17x + 0.22y.] x + y ≤ 20,000 [Sum of the investments.] x ≥ 5,000 [Constraint on investment in bond A.] y ≤ 8,000 [Constraint on investment in bond B.] xy or x - y ≥ 0 [Relation between the investments.] x ≥ 0, y ≥ 0 [Investment can not be negative.] The feasible region determined by the above constraints is shown. From the figure the vertices are (5000, 0), (20000, 0), (12000, 8000), (8000, 8000) and (5000, 5000) [Read the points from the corresponding red marks.] At (5000, 0), Z = (0.17)(5000) + (0.22)(0) = 850. [Substitute the values.] At (20000, 0), Z = (0.17)(20000) + (0.22)(0) = 3400. [Substitute the values.] At (12000, 8000), Z = (0.17)(12000) + (0.22)(8000) = 3800. [Substitute the values.] At (8000, 8000), Z = (0.17)(8000) + (0.22)(8000) = 3120. [Substitute the values.] At (5000, 5000), Z = (0.17)(5000) + (0.22)(5000) = 1950. [Substitute the values.] So, investing$12000 in bond A, and $8000 in bond B gives maximum returns. Correct answer : (1) 2. A sweet shop makes gift packet of sweets which combines two special types of sweets A and B. The weight of the pack is 7 pounds. At least 3 pounds of A and no more than 5 pounds of B should be used. The profit on sweet A is$15 per pound and on sweet B is $20 per pound. Which of the following combination in the pack gives maximum profit for the shop? a. 3 pounds of sweet A & 4 pounds of sweet B b. 4 pounds of sweet A & 4 pounds of sweet B c. 3 pounds of sweet A & 2 pounds of sweet B d. 3 pounds of sweet A & 3 pounds of sweet B Solution: Let a pack of x pounds of sweet A, y pounds of sweet B gives maximum profit. The objective function D = 15x + 20y [Profit on sweet A is$15 per pound and profit on sweet B is $20 per pound.] x + y ≤ 7 [The pack should not exceed 7 pounds.] x ≥ 3. [The pack should have at least 3 pounds of sweet A.] y ≤ 5 [The pack should not have more than 5 pounds of sweet B.] x ≥ 0, y ≥ 0 [x , y are the number of pounds.] The feasible region determined by the above constraints is shown. From the figure the vertices are (3, 0), (7, 0) and (3, 4). At (3, 0), P = 15(3) + 20(0) = 45 [Substitute the values.] At (7, 0), P = 15(7) + 20(0) = 105 [Substitute the values.] At (3, 4), P = 15(3) + 20(4) = 125 [Substitute the values.] P is maximum at (3, 4). So, a pack of 3 pounds of sweet A and 4 pounds of sweet B gives maximum profit. Correct answer : (1) 3. Ed owns a cement manufacturing plant where two types of cement, namely powdered and granulated are produced. His plant works 8 hours a day and has a maximum capacity to produce 1600 bags per day. He has to produce at least 600 bags of powdered cement per day to meet the demand. The time required to make a bag of powdered cement is 0.24 minutes which is half of the time required to produce granulated cement. He earns a profit of$8 per bag for granulated cement and $7 per bag for powdered cement. Find the maximum profit of Ed per day. a.$11800 b. $9800 c.$4200 d. $11600 Solution: Let x be the number of bags of granulated cement produced per day and y be the number of bags of powdered cement produced per day. x + y ≤ 1600 [The plant capacity is 1600 bags per day.] y ≥ 600 [The plant must produce at least 600 bags of powdered cement per day.] (0.48) x + (0.24) y ≤ 8(60) or 2x + y ≤ 2000 [One bag of powdered cement takes 0.24 minutes, one bag of granulated cement takes 0.48 minutes and the plant works for 8 hours a day.] The objective function is, the profit, P = 8x + 7y [The profit on a bag of granulated cement is$8 and the profit on a bag of powdered cement is $7.] The feasible region determined by the above constraints is shown. From the figure the vertices are (0, 600), (700, 600), (400, 1200) and (0, 1600). At (0, 600), P = 8(0) + 7(600) = 4200 [Substitute the values.] At (700, 600), P = 8(700) + 7(600) = 9800 [Substitute the values.] At (400, 1200), P = 8(400) + 7(1200) = 11600 [Substitute the values.] At (0, 1600), P = 8(0) + 7(1600) = 11200 [Substitute the values.] The maximum value of P is$11600. The maximum profit of Ed per day is $11600. Correct answer : (4) 4. Limits on the variables in the objective function are called________. a. extremes b. restrictions c. vertices d. objectives Solution: Limits on the variables in the objective function are called restrictions. [Definition of the restrictions.] Correct answer : (2) 5. The graphical tool which identifies conditions that make as large as possible or as small as possible is known as_________. a. optimization technique b. curve sketching c. graph theory d. linear programming Solution: The graphical tool which identifies conditions that make as large as possible or as small as possible is known as linear programming. [Definition of linear programming.] Correct answer : (4) 6. The quantity to be maximized or minimized is represented by a linear function called as _________. a. the constraint b. the feasible function c. the objective function d. the restriction Solution: The quantity to be maximized or minimized is represented by a linear function called the objective function. [Definition of the objective function.] Correct answer : (3) 7. Josh wants to buy at least 13 books and at least 19 cakes from a shop. If $y$ represents the number of books purchased and $a$ represents the number of cakes purchased, then which of the following systems of inequalities models this situation? a. $y$ ≥ 13, $a$ ≥ 19 b. $y$ ≤ 13, $a$ ≥ 19 c. $y$ ≥ 19, $y$ ≤ 13 d. $y$ ≥ 13, $a$ ≤ 19 Solution: y represents the number of books purchased and a represents the number of cakes purchased. The minimum number of books to be purchased is 13. So, y ≥ 13. The minimum number of cakes to be purchased is 19. So, a ≥ 19. Therefore the system of inequalities that models the given situation is y ≥ 13, a ≥ 19. Correct answer : (1) 8. If the cost of a tape is$14 and the cost of a CD is $22, then write the objective function for the cost(C) of $z$ tapes and $a$ CDs . a. C = 22$z$ + 14$a$ b. C = 14$z$ + 22$a$ c. C = 308$z$$a$ d. C = 36 + $z$ + $a$ Solution: The cost of each tape is$14 and the cost of each CD is \$22.

The cost C of z tapes and a CDs is C = 14z + 22a, which is the objective function for the cost.

9.
What is the maximum value of the objective function C = 6$x$ + 5$y$ subject to the constraints $x$ ≥ 0, $y$ ≥ 0, $x$ + $y$ ≤ 4? a. 44 b. 22 c. 29 d. 24

Solution:

Objective function is C = 6x + 5y

Constraints are x ≥ 0, y ≥ 0, x + y ≤ 4. The feasible region determined by the given constraints is shown in the graph.
[Draw the constraints.]

From the graph, the three vertices are (0, 0), (4, 0) and (0, 4).

To evaluate the minimum or maximum values of C, we evaluate C = 6x + 5y at each of the three vertices.

At (0, 0) , C = 6(0) + 5(0) = 0
[Substitute the values.]

At (4, 0) , C = 6(4) + 5(0) = 24
[Substitute the values.]

At (0, 4) , C = 6(0) + 5(4) = 20
[Substitute the values.]

So, the maximum value of the objective function, C is 24.

10.
Sunny can afford to buy 14 bananas and 19 apples. He wants at least 12 bananas and at least 16 apples. If $x$ represents the number of bananas, $z$ represents the number of apples that Sunny buys, then which of the following systems of inequalities models this situation? a. $x$ ≥ 14, $z$ ≥ 19, $x$ ≥ 12, $z$ ≥ 16 b. $x$ ≤ 12, $z$ ≤ 16, $x$ ≥ 14, $z$ ≥ 19 c. $x$ ≤ 14, $z$ ≤ 19, $x$ ≤ 12, $z$ ≤ 16 d. $x$ ≤ 14, $z$ ≤ 19, $x$ ≥ 12, $z$ ≥ 16

Solution:

x represents the number of bananas and z represents the number of apples that Sunny buys.

The maximum number of bananas that can be bought is 14. So, x ≤ 14.

The maximum number of apples that can be bought is 19. So, z ≤ 19.

The minimum number of bananas to be bought is 12. So, x ≥ 12.

The minimum number of apples to be bought is 16. So, z ≥ 16.

So, the system of inequalities that models the given situation is:
x ≤ 14, z ≤ 19, x ≥ 12, z ≥ 16.