Provide an appropriate response.
32)
Solve the following linear programming problem using the simplex method:
Maximize P = 7x1+ 2x2+x3
subject to:
x1+ 5x2+ 7x3
8
x1+ 4x2+ 11x3 9
x1 , x2 , x3
0
32)
A)
Max P = 56 when x1= 8, x2= 0, x3= 0
B)
Max P = 9 when x1= 1, x2= 1, x3= 0
C)
Max P = 63 when x1= 9, x2= 0, x3= 0
D)
Max P = 0 when x1= 0, x2= 0, x3= 8
33)
Convert the inequality to a linear equation by adding a slack variable.
x1+ 6x2
35
33)
A)
x1+ 6x2+s1< 35
B)
x1+ 6x2+s1
35
C)
x1+ 6x2+s1= 35
D)
x1+ 6x2+s1+ 35 = 0
34)
4x1+5x2+s1= 60
x1+2x2+s2= 18
x1x2s1s2
(A) 0 0 60 18
(B) 012 0–6
(C) 0 9 15 0
(D) 15 0 0 3
(E) 18 0–12 0
(F) 10 4 0 0
In basic solution (B), identify the basic and nonbasic variables and determine if the solution is
feasible or not feasible.
34)
A)
Basic variables: x2, s2
Nonbasic variables: x1, s1
Solution is feasible.
B)
Basic variables: x1, s2
Nonbasic variables: x2, s1
Solution is not feasible.
C)
Basic variables: x2, s1
Nonbasic variables: x1, s2
Solution is not feasible.
D)
Basic variables: x2, s2
Nonbasic variables: x1, s1
Solution is not feasible.
Provide an appropriate response.
35)
Solve the following linear programming problem by applying the simplex method to the dual
problem:
Minimize C =3x1+2x2
subject to
x1+7x2 2
2x1+x2 7
7x1+9x2 1
x1, x2 0
35)
A)
Min C = 21 at x1=7
2, x2= 0
B)
Min C =21
2 at x1= 0, x2=7
2
C)
Min C =21
2 at x1=7
2, x2= 0
D)
Min C =7
2 at x1=21
2, x2= 0
36)
Formulate the dual problem for the linear programming problem:
Minimize C =22x1+15x2
subject to
x1+2x2 3
x1+x2 2
x1, x2 0
36)
A)
Maximize P =3y1+2y2
subject to
y1+y2
22
2y1+y2
15
y1, y2 0
B)
Maximize P =3y1+2y2
subject to
y1+y2 3
2y1+y2 2
y1, y2 0
C)
Maximize P =22y1+15y2
subject to
y1+y2
22
2y1+y2
15
y1, y2 0
D)
Maximize P =3y1+2y2
subject to
y1+y2 22
2y1+y2 15
y1, y2 0
37)
Use the simplex method to solve the linear programming problem.
Minimize w = 5y1+ 2y2
subject to: y1+y2
19.5
2y1+y2
24
y1 0, y2
0
37)
A)
30.5 when y1= 4.5 and y2= 0
B)
31.5 when y1= 1 and y2= 2
C)
48 when y1= 0 and y2= 24
D)
54 when y1= 24 and y2= 1
38)
6 5 1 3
1 7 8 6
38)
A)
1 7 8 6
6 5 1 3
B)
3 5
5 1
3 6
1 8
C)
1 6
7 5
8 1
6 3
D)
6 1
5 7
1 8
3 6
39)
Write the simplex tableau, label the columns and rows, underline the pivot element, and identify
the entering and exiting variables for the linear programming problem:
Maximize P =5x1+3x2
subject to
3x1+8x2
5
3x1+5x2
6
–8x1+6x2
32
6x2
3
x1, x2 0
39)
A)
Enter
Exit
s1
s2
s3
s4
P
x1x2s1s2s3s4 P
3 8 1 0 0 1 0 8
3 5 0 1 0 0 0 6
–8 6 0 0 1 0 0 32
0 6 0 0 0 1 0 3
–5–3 0 0 0 0 1 0
B)
Exit
Enter
s1
s2
s3
s4
P
x1x2s1s2s3s4 P
3 8 1 0 0 0 0 6
3 5 0 1 0 0 0 6
–8 6 0 0 1 0 1 32
0 6 0 0 0 1 0 3
–5–3 0 0 0 0 1 0
C)
Exit
Enter
s1
s2
s3
s4
P
x1x2s1s2s3s4 P
3 8 1 0 0 1 0 5
3 5 0 1 0 0 0 6
–8 6 0 0 1 0 0 32
0 6 0 0 0 1 0 3
–5–3 0 0 0 0 1 32
22
D)
Enter
Exit
s1
s2
s3
s4
P
x1x2s1s2s3s4 P
3 8 1 0 0 0 0 5
3 5 0 1 0 0 0 6
–8 6 0 0 1 0 0 32
0 6 0 0 0 1 0 3
–5–3 0 0 0 0 1 0
40)
7 7 8
2 1 3
9 8 7
40)
A)
2 1 2
7 7 8
9 8 7
B)
7 8 7
1 3 2
8 7 9
C)
9 8 7
2 1 3
7 7 8
D)
7 2 9
7 1 8
8 3 7
Solve the given linear programming problem using the table method.
41)
Maximize P = 6x1+ 7x2
subject to 2x1+ 3x2
12
2x1+x2 8
x1, x2
0
41)
A)
Max P = 32 at x1= 2, x2= 3
B)
Max P = 32 at x1= 3, x2= 2
C)
Max P = 55 at x1= 4, x2= 4
D)
Max P = 24 at x1= 4, x2= 0
23
42)
x1+2x2+s1= 14
3x1+4x2+s2= 36
x1x2s1s2
(A) 0 0 14 36
(B) 0 7 0 8
(C) 0 9 –4 0
(D) 14 0 0 –6
(E) 12 0 2 0
(F) 8 3 0 0
Which of the six basic solutions are feasible? Which are not feasible? Use the basic feasible solutions
to find the maximum value of P =24x1+18x2.
42)
A)
Feasible: (A), (B), (E), (F); Not feasible: (C), (D);
The maximum value of P is 126.
B)
Feasible: (A), (B), (E), (F); Not feasible: (C), (D);
The maximum value of P is 246.
C)
Feasible: (A), (B), (E), (F); Not feasible: (C), (D);
The maximum value of P is 288.
D)
Feasible: (A), (E), (F); Not feasible: (C), (D), (B);
The maximum value of P is 288.
43)
Write the basic solution for the following simplex tableau:
x1x2x3s1s2s3P
010 7 0 1 1 0 12
1–7 6 0 0 1 0 28
0 0 10 1 0 0 0 30
0–5–8 0 0 4 1 50
43)
A)
x1= 28, x2= 10, x3= 7, s1= 30, s2= 12, s3= 0, P = 50
B)
x1= 28, x2= 10, x3= 0, s1= 30, s2= 12, s3= 0, P = 50
C)
x1= 28, x2= 0, x3= 0, s1= 30, s2= 12, s3= 0, P = 50
D)
x1= 28, x2= 0, x3= 0, s1= 30, s2= 12, s3= 0, P = – 50
44)
44)
A)
x3= 10, x2= 7, z = 10; x1, s1, s2= 0
B)
x3= 5, x2= – 7, z = – 10; x1, s1, s2= 0
C)
x3= 5, x2= – 7, z = 10; x1, s1, s2= 0
D)
x3= 10, x2= 10, z = 7; x1, s1, s2= 0
Provide an appropriate response.
45)
Find the initial simplex tableau for the following preliminary simplex tableau (DO NOT SOLVE):
x1x2s1s2s3a1a2P
2 1 –1 0 0 1 0 0 7
1 3 0 –1 0 0 1 0 12
1 1 0 0 1 0 0 0 5
–4–9 0 0 0 M M 1 0
45)
A)
a1
a2
s3
P
x1x2s1s2s3a1a2P
2 1 –1 0 0 1 0 0 7
1 3 0 –1 0 0 1 0 12
1 1 0 0 1 0 0 0 5
3M–4 4M–9 M M 0 0 0 1 –19M
B)
a1
a2
s3
P
x1x2s1s2s3a1a2P
2 1 –1 0 0 1 0 0 7
1 3 0 –1 0 0 1 0 12
1 1 0 0 1 0 0 0 5
–3M–4–4M–9 M M 0 0 0 1 –19M
C)
a1
a2
s3
P
x1x2s1s2s3a1a2P
2 1 –1 0 0 1 0 0 7
1 3 0 –1 0 0 1 0 12
1 1 0 0 1 0 0 0 5
3M–4 4M–9 M M 0 0 0 –1 19M
D)
a1
a2
s3
P
x1x2s1s2s3a1a2P
2 1 –1 0 0 1 0 0 –7
1 3 0 –1 0 0 1 0 –12
1 1 0 0 1 0 0 0 –5
3M–4 4M–9 M M 0 0 0 –1 19M
46)
Solve the following linear programming problem using the simplex method:
Maximize P =x1–x2
subject to
x1+x2
4
2x1+7x2
14
x1, x2
0
46)
A)
Max P = 14 at x1= 4 and x2= 0
B)
Max P = 4 at x1= 4 and x2= 0
C)
Max P = 4 at x1= 14 and x2= 0
D)
Max P = 4 at x1= 4 and x2= 4
47)
Convert the inequality to a linear equation by adding a slack variable.
9x1+x2+ 4x3
240
47)
A)
9x1+x2+ 4x3+s1+ 240 = 0
B)
9x1+x2+ 4x3+s1= 240
C)
9x1+x2+ 4x3+s1 240
D)
9x1+x2+ 4x3+s1 240
Graph the system of inequalities.
48)
x1+x2
40
3x1+x2
84
x1, x2
0
48)
A)
B)
C)
D)
28
49)
Write the basic solution for the following simplex tableau:
x1 x2x3 s1 s2 P
3 4 0 3 1 0 20
1 5 1 7 0 0 28
–3 4 0 1 0 1 20
49)
A)
x1, x2, s1= 0, x5= 28, s2= 20, P = 20
B)
x1, x2, s1= 0, x3= 28, s2= 20, P = 20
C)
x1, x2, s1= 0, x1= 28, s2= 20, P = 20
D)
x1, x2, s1= 0, x3= 20, s2= 28, P = 20
50)
Use the big M method to find the optimal solution to the problem.
Maximize P = 3x1+ 6x2+ 2x3
subject to 2x1+ 2x2+ 3x3 12
2x1+ 2x2+x3= 0
x1, x2, x3 0
50)
A)
Max P = 27 at x1= 3, x2= 3, x3= 3
B)
Max P = 27 at x1= 0, x2= 3, x3= 3
C)
Max P = 27 at x1= 3, x2= 3, x3= 0
D)
Max P = – 27 at x1= 3, x2= 3, x3= 3
51)
Formulate the dual problem for the linear programming problem
Minimize C =4x1+7x2
subject to
x1+x2
5
x1+2x2
18
3x1+x2
8
x1, x2
0
51)
A)
Maximize P =5y1+18y2–8y3
subject to
y1–2y2+y3 7
y1–y2+3y3 4
y1, y2, y3 0
B)
Maximize P =5y1–18y2+8y3
subject to
y1–2y2+y3 7
y1–y2+3y3 4
y1, y2, y3 0
C)
Maximize P =5y1–18y2+8y3
subject to
y1–2y2+y3 7
y1–y2+3y3 4
y1, y2, y3 0
D)
Maximize P =5y1–18y2+8y3
subject to
y1–2y2+y3 7
y1–y2+3y3 4
y1, y2, y3 0
52)
State the dual problem.
Minimize C = 6x1+ 3x2
subject to: 3x1+ 2x2
34
2x1+ 5x2
43
x1, x2 0
52)
A)
Maximize P =34y1+ 43y2
subject to: 3y1+ 2y2
6
2y1+ 5y2
3
y1, y2 0
B)
Maximize P = 43y1+34y2
subject to: 2y1+ 3y2 6
5y1+ 2y2 3
y1, y2 0
C)
Maximize P = 43y1+34y2
subject to: 2y1+ 3y2 6
5y1+ 2y2 3
y1, y2 0
D)
Maximize P = 34y1+43y2
subject to: 3y1+ 2y2
6
2y1+ 5y2 3
y1, y2 0
53)
Formulate the following problem as a linear programming problem (DO NOT SOLVE):A
veterinarian wants to set up a special diet that will contain at least 500 units of vitamin B1 at least
800 units of vitamin B2 and at least 700 units of vitamin B6. She also wants to limit the diet to at
most 300 total grams. There are three feed mixes available, mix P, mix Q, and mix R. A gram of mix
P contains 3 units of vitamin B1, 5 units of vitamin B2, and 8 units of vitamin B6. A gram of mix Q
contains 9 units of vitamin B1, 8 units of vitamin B2, and 6 units of vitamin B6. A gram of mix R
contains 7 units of vitamin B1, 6 units of vitamin B2, and 9 units of vitamin B6. Mix P costs $0.10
per gram, mix Q costs $0.12 per gram, and mix R costs $0.21 per gram. How many grams of each
mix should the veterinarian use to satisfy the requirements of the diet at minimal costs? (Let x1
equal the number of grams of mix P, x2 equal the number of grams of mix Q, and x3 equal the
number of grams of mix R that are used in the diet).
53)
A)
Minimize C =0.10x1–0.12x2+0.21x3
subject to
3x1+9x2+7x3 500
5x1+8x2+6x3 800
8x1+6x2+9x3 700
x1+x2+x3 300
x1, x2, x3 0
B)
Minimize C = – 0.10x1–0.12x2–0.21x3
subject to
3x1+9x2+7x3 500
5x1+8x2+6x3 800
8x1+6x2+9x3 700
x1+x2+x3 300
x1, x2, x3 0
C)
Minimize C =0.10x1–0.12x2–0.21x3
subject to
3x1+9x2+7x3 500
5x1+8x2+6x3 800
8x1+6x2+9x3 700
x1+x2+x3 300
x1, x2, x3 0
D)
Minimize C =0.10x1+0.12x2+0.21x3
subject to
3x1+9x2+7x3 500
5x1+8x2+6x3 800
8x1+6x2+9x3 700
x1+x2+x3 300
x1, x2, x3 0
Solve the given linear programming problem using the table method.
54)
Maximize P = 7x1+ 6x2
subject to 3x1+x2 21
x1+x2 10
x1+ 2x2 12
x1, x2
0
54)
A)
Max P =65.5 at x1= 5.5, x2= 4.5
B)
Max P = 66 at x1= 6, x2= 4
C)
Max P = 60 at x1= 6, x2= 3
D)
Max P = 68 at x1= 8, x2= 2
31
55)
Find the solution of the system for which x2= 0 and s1= 0
5x1+3x2+s1= 90
2x1+4x2+s2= 64
55)
A)
(x1, x2, s1, s2) = (16, 0, 0, 42)
B)
(x1, x2, s1, s2) = (18, 0, 0, 28)
C)
(x1, x2, s1, s2) = (32, 0, 0, 45)
D)
(x1, x2, s1, s2) = (90, 0, 0, 64)
56)
A summer camp wants to hire counselors and aides to fill its staffing needs at minimum cost. The
average monthly salary of a counselor is $2400 and the average monthly salary of an aide is $1100.
The camp can accommodate up to 45 staff members and needs at least 30 to run properly. They
must have at least 10 aides, and may have up to 3 aides for every 2 counselors. How many
counselors and how many aides should the camp hire to minimize cost?
56)
A)
12 counselors and 18 aides
B)
18 counselors and 12 aides
C)
35 counselors and 10 aides
D)
27 counselors and 18 aides
57)
Use the big M method to solve the following linear programming problem.
Maximize P =6x1+4x2+x3
subject to
x1+3x2+6x3
18
x1+x2+x3
5
x1, x2, x3
0
57)
A)
Max P = 108 at x1= 18, x2= 0, x3= 0
B)
Max P = 108 at x1= 0, x2= 0, x3= 18
C)
Max P = 108 at x1= 18, x2= 0, x3= 9
D)
Max P = – 108 at x1= 18, x2= 0, x3= 0
E)
Max P = 108 at x1= 0, x2= 18, x3= 0
58)
Solve the linear programming problem.
Minimize C = 10x1+ 12x2+ 28x3
subject to 4x1+ 2x2+ 3x3
20
3x1–x2– 4x3
10
x1, x2, x3 0
58)
A)
Min C = 64 at x1= 0, x2= 2, x3= 2
B)
Min C = 64 at x1= 4, x2= 2, x3= 0
C)
Min C = 64 at x1= 0, x2= 2, x3= 4
D)
Min C = 64 at x1= 4, x2= 4, x3= 2
Answer Key
Testname: C6
34
Answer Key