Math 201 - Exam 3 Solutions

Exam Version A

Problem 1A For the following system of equations:

3x1 − 6x2 = 0

x1 − 4x2 = −6

(a) Write the system as a matrix equation:

Let A =

[3 −61 −4

]Let X =

[x1

x2

]and let B =

[0−6

]So AX = B is:[3 −61 −4

] [x1

x2

]=

[0−6

]

(b) A−1 =

[23−1

16−1

2

]Which can easily be verified with matrix multiplication:

AA−1 =

[3 −61 −4

] [23−1

16−1

2

]=

[1 00 1

]

1

(c) Solve the system of linear equations by using the inverse from part (b):

AX = B

A−1AX = A−1B

X = A−1B

So:[x1

x2

]=

[23−1

16−1

2

] [0−6

]=

[63

]

Problem 2A State the linear inequality whose graph is given in the figure.Write your answers in the form Ax + by = C, with A,B and C integers.

The graph has a solid line so cannot have strict inequality. The points (0,−4)and (−7, 0) are on the line. Therefore:

m = ∆y∆x

= y1−y2x1−x2

= y2−y1x2−x1

= −47

Applying point-slope form {y − y1 = m(x− x1)} yields:

7y + 4x = −28

Testing some points reveals that:

7y + 4x ≤ −28

Which is equivalent to

−7y − 4x ≥ 28

2

Problem 3A The graph has a dotted line, so the inequality is strict. y > 1is clearly the solution.

Problem 4A How many 4-letter code words are possible from the first 6letters of the alphabet if no letter is repeated?

Using the Multiplication Principle for Counting states that if n operationsO1, O2, O3, ..., On are performed in order, with possible number of outcomesN1, N2, ..., Nn, respectively, then there are N1 ∗N2 ∗ ...∗Nn possible outcomesof the operations performed.

Therefore, since the set to consider is {A,B,C,D,E, F}, the number of out-comes is 6 ∗ 5 ∗ 4 ∗ 3 = 360

3

Problem 5A

The yellow region is the intersection of the three equations. The region isunbounded.

Problem 6A The equation for Addition Principle is either:

n(A ∪B) = n(A) + n(B)− n(A ∩B)

Orn(A OR B) = n(A) + n(B)− n(A AND B)

Or

n(A UNION B) = n(A) + n(B)− n(A INTERSECTION B)

4

Problem 7A

(a) The feasible region is:

(b) Corner points:

The corner points are: (0, 6), (6, 0), and (2, 4).

(c) A corner point table is:Corner Point Z

(0,6) 120(6,0) 60(2,2) 60

Therefore Z is minimized at (6,0) and (2,2) - so the entire line seg-ment between these points is a minimum. The region is unbounded, sothere is NO maximum.

5

Problem 8A Simplex Method:

The initial tableau is:

The pivot column and pivot row are circled below:

The pivot is already 1, so the rest of the elements in the pivot column needto be made zero. This can be accomplished with the following row operations:

r2 − r1 → r2, r3 − r1 → r3, and r4 + 8r1 → r4.

Which produces:

The pivot column and pivot row are circled below:

6

The pivot needs to be 1.

Then the following row operations produce a pivot column with all zerosother than the pivot:

r1 + r3 → r1, r2 + r3 → r2 and r4 + 10r3 = r4.Which produces:

There are no more negative indicators, so we stop. The solution is:

x1 = 3, x2 = 2, and Z = 28.

7

Problem 9A Survey...

1 Venn Diagram:

2 Using the Venn Diagram...

(a) How many businesses offer either health insurance or dental in-surance?

n(H ∪D) = n(H) + n(D)− n(H ∩D)

Orn(H ∪D) = 95 + 30− 15 = 110

(b) How many businesses offer neither health insurance nor dental in-surance?

There are 150 businesses, and 100 offer either. So 150 - 110 =40 offer neither.

(c) How many businesses offer health insurance but not dental insur-ance?

From the Venn Diagram, there are 80 businesses that offer healthinsurance but not dental insurance.

8

Problem 10A Application Problem:

(a) The data is as follows:

Double Tent 5-Person Tent Available Weekly Hours

Assembly 3 8 1200Packaging 1 2 400

Profit per Tent $45 $100

(b) Define variables:

Let x1 = number of double tents.

Let x2 = number of 5-person tents.

(c) The objective function is P = 45x1 + 100x2. The problem states thatyou should maximize P.

(d) Constraints are as follows:

3x1 + 8x2 ≤ 1200

x1 + 2x2 ≤ 400

x1, x2 ≥ 0

9

Exam Version B

Problem 1B

(a) Write the system as a matrix equation:

Let A =

[2 −41 −4

]Let X =

[x1

x2

]and let B =

[0−6

]So AX = B is:[2 −41 −4

] [x1

x2

]=

[0−6

]

(b) A−1 =

[1 −114−1

2

]Which can easily be verified with matrix multiplication:

AA−1 =

[2 −41 −4

] [1 −114−1

2

]=

[1 00 1

](c) Solve the system of linear equations by using the inverse from part (b):

AX = B

A−1AX = A−1B

X = A−1B

So:[x1

x2

]=

[1 −114−1

2

] [0−6

]=

[63

]

10

Problem 2B State the linear inequality whose graph is given in the figure.Write your answers in the form Ax + by = C, with A,B and C integers.

The graph has a solid line so cannot have strict inequality. The points (0,−4)and (−7, 0) are on the line. Therefore:

m = ∆y∆x

= y1−y2x1−x2

= y2−y1x2−x1

= −47

Applying point-slope form {y − y1 = m(x− x1)} yields:

7y + 4x = −28

Testing some points reveals that:

7y + 4x ≤ −28

Which is equivalent to

−7y − 4x ≥ 28

Problem 3B The graph has a dotted line, so the inequality is strict. y > 1is clearly the solution.

Problem 4B How many 4-letter code words are possible from the first 7letters of the alphabet if no letter is repeated?

Using the Multiplication Principle for Counting states that if n operationsO1, O2, O3, ..., On are performed in order, with possible number of outcomesN1, N2, ..., Nn, respectively, then there are N1 ∗N2 ∗ ...∗Nn possible outcomesof the operations performed.

Therefore, since the set to consider is {A,B,C,D,E, F,G}, the number ofoutcomes is 7 ∗ 6 ∗ 5 ∗ 4 = 840

11

Problem 5B

The yellow region is the intersection of the three equations. The region isbounded.

Problem 6B The equation for Addition Principle is either:

n(A ∪B) = n(A) + n(B)− n(A ∩B)

Orn(A OR B) = n(A) + n(B)− n(A AND B)

Or

n(A UNION B) = n(A) + n(B)− n(A INTERSECTION B)

12

Problem 7B

(a) The feasible region is:

(b) Corner points:

The corner points are: (0, 6), (6, 0), and (2, 4).

(c) A corner point table is:Corner Point Z

(0,6) 30(6,0) 60(2,2) 30

Therefore Z is minimized at (0,6) and (2,2) - so the entire line seg-ment between these points is a minimum. The region is unbounded, sothere is NO maximum.

13

Problem 8B Simplex Method:

The initial tableau is:

The pivot column and pivot row are circled below:

The pivot is already 1, so the rest of the elements in the pivot column needto be made zero. This can be accomplished with the following row operations:

r2 − r1 → r2, r3 − r1 → r3, and r4 + 8r1 → r4.

Which produces:

The pivot column and pivot row are circled below:

14

The pivot needs to be 1.

Then the following row operations produce a pivot column with all zerosother than the pivot:

r1 + r3 → r1, r2 + r3 → r2 and r4 + 10r3 = r4.Which produces:

There are no more negative indicators, so we stop. The solution is:

x1 = 3, x2 = 2, and Z = 28.

15

Problem 9B Survey...

1 Venn Diagram:

2 Using the Venn Diagram...

(a) How many businesses offer either health insurance or dental in-surance?

n(H ∪D) = n(H) + n(D)− n(H ∩D)

Orn(H ∪D) = 60 + 25− 15 = 70

(b) How many businesses offer neither health insurance nor dental in-surance?

There are 130 businesses, and 70 offer either. So 130 - 70 = 60offer neither.

(c) How many businesses offer health insurance but not dental insur-ance?

From the Venn Diagram, there are 60 businesses that offer healthinsurance but not dental insurance.

16

Problem 10B Application Problem:

(a) The data is as follows:

Double Tent 5-Person Tent Available Weekly Hours

Assembly 4 7 1100Packaging 1 3 450

Profit per Tent $55 $95

(b) Define variables:

Let x1 = number of double tents.

Let x2 = number of 5-person tents.

(c) The objective function is P = 55x1 + 95x2. The problem states thatyou should maximize P.

(d) Constraints are as follows:

4x1 + 7x2 ≤ 1100

x1 + 3x2 ≤ 450

x1, x2 ≥ 0

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