﻿ Linear Programming Worksheet - Page 3 | Problems & Solutions

# Linear Programming Worksheet - Page 3

Linear Programming Worksheet
• Page 3
21.
Suppose you want to buy at least 42 pens and at least 49 pencils from a shop. Which one of the following systems of inequalities models this situation?
 a. $x$ ≤ 42, $y$ ≤ 49 b. $x$ ≥ 42, $y$ ≥ 49 c. $x$ ≤ 42, $y$ ≥ 49 d. $x$ ≥ 42, $y$ ≤ 49

#### Solution:

Let x represent the number of pens purchased and y represent the number of pencils purchased.

The minimum number of pens to be purchased is 42. So, x ≥ 42.

The minimum number of pencils to be purchased is 49. So, y ≥ 49.

Therefore, the system of inequalities that models the given situation is
x ≥ 42, y ≥ 49.

22.
David can afford to buy 25 mangoes and 40 oranges. He wants at least 10 mangoes and at least 10 oranges. If $x$ represent the number of mangoes ,$y$ represent the number of oranges that David buy , then which of the following systems of inequalities models this situation?
 a. $x$ ≤ 25, $y$ ≤ 40; $x$ ≥ 10, $y$ ≥ 10 b. $x$ ≥ 25, $y$ ≥ 40, $x$ ≥ 10, $y$ ≥ 10 c. $x$ ≤ 25, $y$ ≥ 40, $x$ ≥ 10, $y$ ≤ 10 d. $x$ ≥ 25, $y$ ≥ 40, $x$ ≤ 10, $y$ ≤ 10

#### Solution:

x represents the number of mangoes and y represents the number of oranges that David buy.

The maximum number of mangoes that can be bought is 25. So, x ≤ 25.

The maximum number of oranges that can be bought is 40. So, y ≤ 40.

The minimum number of mangoes to be bought is 10. So, x ≥10.

The minimum number of oranges to be bought is 10. So, y ≥ 10.

So, the system of inequalities that models the given situation is:
x ≤ 25, y ≤ 40, x ≥ 10, y ≥ 10.

23.
If the cost of a tape is $11 and the cost of a CD is$14, then write the objective function for the cost of $x$ tapes and $y$ CDs .
 a. C = 11$x$ + 14$y$ b. C = 154$\mathrm{xy}$ c. C = 14$x$ + 11$y$ d. C = 25 + $x$ + $y$

#### Solution:

The cost of each tape is $11 and the cost of each CD is$14.

The cost C of x tapes and y CDs is C = 11x + 14y, which is the objective function for the cost.

24.
Find the values of $x$ and $y$ that maximize the objective function P = 2$x$ + 3$y$ subject to the constraints $x$ ≥ 0, $y$ ≥ 0, $x$ - $y$ ≤ 1 and $x$ + $y$ ≤ 3.
 a. $x$ = 1, $y$ = 0 b. $x$ = 3, $y$ = 0 c. $x$ = 0, $y$ = 3 d. $x$ = 0, $y$ = 1

#### Solution:

Objective function is P = 2x + 3y.

Constraints are x ≥ 0, y ≥ 0, x - y ≤ 1 and x + y ≤ 3.

The feasible region determined by the given constraints is shown.

From the graph, the vertices are (0, 0), (1, 0), (2, 1) and (0, 3).

To find the minimum and maximum values of P, we evaluate P = 2x + 3y at each of the four vertices.

At (0, 0), P = 2(0) + 3(0) = 0
[Substitute x = 0, y = 0 in P.]

At (1, 0), P = 2(1) + 3(0) = 2
[Substitute x = 1, y = 0 in P.]

At (2, 1), P = 2(2) + 3(1) = 7
[Substitute x = 2, y = 1 in P.]

At (0, 3), P = 2(0) + 3(3) = 9
[Substitute x = 0, y = 3 in P.]

So, the maximum value of P is 9, which occurs when x = 0, y = 3.

25.
A Factory produces two types of toys A and B with the help of two machines I and II. The two machines I and II can not be used for not more than $1\frac{1}{2}$ hours each to manufacture the toys. The toy A requires 1 minute of machine I and 2 minutes of II, where as a toy B takes 3 minutes of I and 1 minute of II. The profit for A is $30 and for B$20. How many of each type of toys are to be produced to give the maximum profit?
 a. 36 toys of type A, 20 toys of type B b. 36 toys of type A, 18 toys of type B c. 36 toys of type A, 30 toys of type B d. 18 toys of type A, 36 toys of type B

#### Solution:

Let x and y be the number of toys to be made of types A and B.

The objective function is the profit P = 30x + 20y.

To make x number of toys of type A and y number of toys of type B, the machine I should be used for 1(x) + 3(y) minutes. Since the machine I cannot be used for more than 11 / 2 hours, we have x + 3y ≤ 90.
[ 11 / 2 hour = 90 minutes.]

To make x number of toys of type A and y number of toys of type B, the machine II should be used for 2(x) + 1(y) minutes. Since the machine II cannot be used for more than 11 / 2 hours, we have 2x + y ≤ 90.
[11 / 2 hour = 90 minutes.]

x ≥ 0, y ≥ 0
[x, y are the number of toys of each type.]

The feasible region determined by the above constraints is shown.

From the figure, the vertices are (0, 0), (45, 0), (36, 18) and (0, 30).

At (0, 0), P = 30(0) + 20(0) = 0
[Substitute x = 0, y = 0 in P.]

At (45, 0), P = 30(45) + 20(0) = 1350
[Substitute x = 45, y = 0 in P.]

At (36, 18), P = 30(36) + 20(18) = 1440
[Substitute x = 36, y = 18 in P.]

At (0, 30), P = 30(0) + 20(30) = 600
[Substitute x = 0, y = 30 in P.]

So, the maximum profit can be obtained by making 36 toys of type A and 18 toys of type B.

26.
A furniture dealer deals in two items, tables and chairs. He invests $60,000 in buying tables and chairs.He has a storage capacity for$600 pieces. A table costs him $120 and a chair$40. He can sell a table for a profit of $50 and a chair for a profit of$20. How should he invest his money in order to make maximum profit?
 a. 450 tables, 150 chairs b. 450 tables, 200 chairs c. 450 tables, 300 chairs d. 300 tables, 150 chairs

#### Solution:

Let the number of tables and chairs to be purchased be x and y.

The objective function is the profit P = 50x + 20y.

Since the dealer can invest $60,000, we have 120x + 40y ≤ 60000 or 3x + y ≤ 1500. [Cost of a table is$120, cost of a chair is $40.] Since the storage capacity is for 600 pieces, we have x + y ≤ 600. x ≥ 0, y ≥0 [x, y are the number of tables and chairs.] The feasible region determined by the above constraints is shown. From the graph, the vertices are (0, 0), (500, 0), (450, 150) and (0, 600). At (0, 0), P = 50(0) + 20(0) = 0 [Substitute x = 0, y = 0 in P.] At (500, 0), P = 50(500) + 20(0) = 25000 [Substitute x = 500, y = 0 in P.] At (450, 150), P = 50(450) + 20(150) = 25500 [Substitute x = 450, y = 150 in P.] At (0, 600), P = 50(0) + 20(600) = 12000 [Substitute x = 0, y = 600 in P.] So, the maximum profit can be obtained by purchasing 450 tables and 150 chairs. Correct answer : (1) 27. A person wants to invest$20000 in two types of bonds A and B.Bond A yields 10% return and bond B yields 15% return on the amount invested. After some consideration, he decides to invest at least $5000 in bond A and not more than$8000 in bond B. He also wants to invest at least as much in bond A as in bond B. Which of the following suggests him for maximum returns?
 a. Investing $12,000 in bond A, and$8,000 in bond B b. Investing $8,000 in bond A, and$8,000 in bond B c. Investing $12,000 in bond A, and$5,000 in bond B d. Investing $20,000 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.

Since bond A yields 10% return and bond B yields 15% return on the amount invested, we have to maximize 0.10x + 0.15y, which is the objective function Z = 0.10x + 0.15y.

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 = (0.1)(5000) + (0.15)(0) = 500.
[Substitute x = 5000, y = 0 in Z.]

At (20000, 0), Z = (0.1)(20000) + (0.15)(0) = 2000.
[Substitute x = 20000, y = 0 in Z.]

At (12000, 8000), Z = (0.1)(12000) + (0.15)(8000) = 2400.
[Substitute x = 12000, y = 8000 in Z.]

At (8000, 8000), Z = (0.1)(8000) + (0.15)(8000) = 2000.
[Substitute x = 8000, y = 8000 in Z.]

At (5000, 5000), Z = (0.1)(5000) + (0.15)(5000) = 1250.
[Substitute x = 5000, y = 5000 in Z.]

So, investing $12000 in bond A, and$8000 in bond B gives maximum returns.

28.
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 shop makes a profit of $15 on sweet A and$20 on sweet B per pound. Which of the following packs gives maximum profit?
 a. A pack of 3 pounds of sweet A, 2 pounds of sweet B b. A pack of 4 pounds of sweet A, 4 pounds of sweet B c. A pack of 3 pounds of sweet A, 4 pounds of sweet B d. A pack of 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.

Since the profit on sweet A is $15 per pound and the profit on sweet B is$20 per pound, we have the objective function D = 15x + 20y.

Since the pack should not exceed 7 pounds, we have x + y ≤ 7.

Since the pack should have at least 3 pounds of sweet A, we have x ≥ 3.

Since the pack should not have more than 5 pounds of sweet B, we have y ≤ 5.

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 x = 3, y = 0 in P.]

At (7, 0), P = 15(7) + 20(0) = 105
[Substitute x = 7, y = 0 in P.]

At (3, 4), P = 15(3) + 20(4) = 125
[Substitute x = 3, y = 4 in P.]

Since P is maximum at (3, 4), a pack of 3 pounds of sweet A, 4 pounds of sweet B gives maximum profit.

29.
A carpenter makes tables and chairs. The profit on each table is $20 and on each chair is$10. The carpenter can not work more than 36 hours per week . He needs 6 hours to make a table and 3 hours to make a chair. Each week he makes at least 2 times as many chairs as tables. Each table occupy a space of 4 chairs and there is a room for at most four tables each week. Which of the following gives maximum profit to the carpenter?
 a. Making 4 tables, 4 chairs b. Making 6 tables, 6 chairs c. Making 2 tables, 8 chairs d. Making 6 tables, 4 chairs

#### Solution:

Let x be the number of tables to be made, y be the number of chairs to be made per week.

Since the profit on each table is $20 and the profit on each chair is$10, we have to maximize 20x + 10y, which is the objective function P = 20x + 10y.

Since the carpenter takes 6 hours to make a table, 3 hours to make a chair and he cannot spend more than 36 hours per week, we have 6x + 3y ≤ 36.

He makes 2 times as many chairs as tables. Then, we have y ≥ 2x.

Since the tables take up 4 times as much storage space as chairs and since the storage capacity is for 4 tables, we have x + y4 ≤ 4 or 4x + y ≤ 16.

x ≥ 0, y ≥ 0
[x, y are the number of tables and chairs.]

The feasible region of the above constraints is shown.

From the figure, the vertices are (0, 0), (8 / 3, 16 / 3), (2, 8), (0, 12).

At (0, 0), P = 20(0) + 10(0) = 0
[Substitute x = 0, y = 0 in P.]

At (8 / 3, 16 / 3), P = 20(8 / 3) + 10(16 / 3) = 106.6
[Substitute x = 8 / 3, y = 16 / 3 in P.]

At (2, 8), P = 20(2) + 10(8) = 120
[Substitute x = 2, y = 8 in P.]

At (0, 12), P = 20(0) + 10(12) = 120
[Substitute x = 0, y = 12 in P.]

Since P is maximum at (2, 8) and (0, 12), making either 2 tables and 8 chairs or making only 12 chairs gives maximum profit to the carpenter.

30.
Which of the following is the graph of a linear system of constraints?
 a. the feasible region b. the constraint region c. the linear graph d. the linear region

#### Solution:

The graph of the linear system of constraints is called the feasible region.
[Definition of the feasible region.]