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Module 2

This is a single, concatenated file, suitable for printing or saving as a PDF for offline viewing. Please

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Module 2 Study Guide and Deliverables

Topics and

Readings

Lecture 3 (Sept 20): Systems of Linear Equations and Matrices

Module 2 Lecture 3 lecture notes

Textbook, Chapter 6

Lecture 4 (Sept 27): Linear Programming and Its Applications

Module 2 Lecture 4 lecture notes

Textbook, Chapter 7 (skip 7.4, 7.6, and 7.7)

Discussions: Module 2 Discussion postings due Oct 3 at 11:59 PM ET

Assignments: Complete the Individual Exercise problems at the end of each Lecture, as well as

selected problems from the textbook.

Midterm Midterm 1 (Oct 4) Covering topics from Lectures 1 through 4

Lecture 3 – Systems of Linear Equations and Matrices

Learning Objectives

After successfully completing the module, students will be able to:

1. Solve systems of linear equations

2. Apply augmented matrices and the Gauss Jordan Method

3. Calculate Matrix operations and Inverses

4. Calculate production matrices from demand and input-output matrices

Systems of Linear Equations

A system of linear equations is simply a group of linear equations on some number of variables, typically denoted . By linear we mean

that the only operations permitted in the system are addition, subtraction, and multiplication by constants. For example,

qualifies as a system of linear equations. On the other hand, none of the following equations would be found in a system of linear equations:

x, y, z

4x + y = z

y = 3x

x + y + z = 12

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Given a system of two linear equations in two variables, the most common ways of solving it are via substitution and elimination.

Substitution

Suppose you’re given

To solve this via the substitution method, isolate one variable, then substitute the new equation from the isolated variable into the other

equation, as follows:

Plugging this into the first equation, we get

We can now solve for . Simplifying:

We can then plug 4.5 in for in either original equation to get

Elimination

The elimination method involves multiplying both sides of an equation by a constant, then adding the resulting equations to cancel a variable.

For example, given the following system of equations:

We could first multiply the bottom equation through by -3 (this would help us to eliminate x from the system), resulting in:

Summing the two equations yields:

Which gives us . Substituting 5 for in either of the original equations returns .

4xy = z

y = + 12x2

y = ex

3x − 4y = 15

2x + 5 = 6y

2x + 5 = 6y ⟹ x = 6y − 5

3 × − 4y = 15 6y − 5

9y − 7.5 − 4y = 15 ⟹ 5y = 22.5 ⟹ y = 4.5

2x + 5 = 6 × 4.5 ⟹ x = (27 − 5)/2 ⟹ x = 11

3x + 4y = 26

x − 7y = −33

3x + 4y = 26

−3x + 21y = 99

0x + 25y = 125

y = 5 y x = 2

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Uniqueness, Dependency, and Consistency

If, like the previous two examples, substitution and elimination, a system has a unique solution it is said to be independent.

Systems of equations don’t always have unique answers. Take the following system of equations:

Dividing the bottom equation through by 3 and bringing over to the other side yields:

You can see that any value of can result in an answer. For example, ( , ) and ( , ) both work. Such a system is

called dependent.

A system is said to be inconsistent if there is no solution that satisfies all of its equations.

Take the following system of linear equations as an example:

Dividing the bottom equation by 4, we get:

This means that has to equal both 40 and 25: an impossibility.

Figure 1 shows, from left to right, graphs of dependent, inconsistent and independent systems of 2 equations.

Figure 1: Dependent, Inconsistent and Independent Systems of 2 Equations

Larger Systems of Linear Equations

x − 3y = 10

3x − 30 = 9y

x − 3y = 10

x x = 16 y = 2 x = 10 y = 0

5x + 6y = 40

20x + 24y = 100

5x + 6y = 25.

5x + 6y

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Elementary Operations

A system of linear equations can contain more than two equations. To solve such an equation, we rely on the elementary operations. These

operations, used correctly, make a system easier to solve without changing the solution. If two linear equations have the same solution, they’re

called equivalent.

The elementary operations are:

1. Change the order of equations

Eq. A

Eq. B

Eq. C

Eq. A’ = B

Eq. B’ = A

Eq. C’ = C

2. Multiply an equation by a nonzero constant

Eq. A

Eq. B

Eq. C

Eq. A’ = A

Eq. B’ = B

Eq. C’ = 2C

3. Replace an equation with the sum of itself and a constant nonzero multiple of another equation from the system

Eq. A

Eq. B

Eq. C

Eq. A’ = A

Eq. B’ = B + 2C

Eq. C’ = C

The following process allows us to use the elementary operations to solve a larger system:

First, set the leading coefficient of the first equation to one using either operation 1 or 2.

Then, eliminate the leading variable of the first equation from subsequent equations using operation 3.

The second equation will have a different leading variable now from equation 1. Repeat the prior steps on the second equation. Do this again

for the third equation, and so on. The last equation will only contain one variable. You can then go backwards through the equations and solve

for each of them.

Example Solve the following system of equations:

First we switch the order so the leading coefficient of the first equation is 1:

Next, we multiply the first equation by 3 and subtract it from the second:

Then we multiply the first equation by 3 and subtract it from the third equation:

Then we multiply the first equation by 4 and subtract it from the last equation:

3x + 2y + z = 10

5x − 2y + 3z = 20

x + y + z = 50

≡

5x − 2y + 3z = 20

3x + 2y + z = 10

x + y + z = 50

3x + 2y + z = 10

5x − 2y + 3z = 20

x + y + z = 50

≡

3x + 2y + z = 10

5x − 2y + 3z = 20

2x + 2y + 2z = 100

3x + 2y + z = 10

5x − 2y + 3z = 20

x + y + z = 50

≡

3x + 2y + z = 10

7x + 5z = 120

x + y + z = 50

3w − 5x − 4y − z = −70

3w + 6x − 2y + 6z = 101

4w − 2x + 3y + 5z = 66

w + x + y + z = 30

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Now we can multiply equation 3 by 3 and add it to equation 2 to set its leading multiplier to 1:

Now we eliminate from equation 3 by subtracting 3 times equation 2:

We go on to eliminate from equation 4 by adding 6 times equation 2

Next we can divide equation 3 by 61:

And add equation 3 × 133 to equation 4:

Solving for in equation 4 and plugging it back into the prior equations we get

We now repeat the process for

Continuing the process, we find and .

Augmented Matrices

You may have surmised in the last module that we can create a more efficient notation to solve linear equations, since the variable names are

superfluous. Given a system of linear equations, we could create a matrix to house the relevant information. For example, instead of writing:

We could create an array of numbers containing the same information, as follows:

We call this an augmented matrix because we include the right side of the equations in the matrix. As before we can apply the elementary

operations to this matrix, only we call them row operations.

Row Operations

Rows can be interchanged:

Multiplied by a nonzero constant:

And exchanged for the sum of itself and a nonzero multiple of another row.

x = 9 w = 5

x + y + z = 7

x − 2y + 3z = 3

2x + z = 9

⎡

⎣

⎢ ⎢ ⎢

−2

⎤

⎦

⎥ ⎥ ⎥

≡

⎡

⎣

⎢ ⎢ ⎢

−2

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

−2

⎤

⎦

⎥ ⎥ ⎥

≡

⎡

⎣

⎢ ⎢ ⎢

−2

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

−2

⎤

⎦

⎥ ⎥ ⎥

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Solving Augmented Matrices

To find the answer to a system using an augmented matrix, you can use the analogous method as for a system of equation: try to use the row

operations to make every element along the diagonal a 1, and everything below the diagonal a 0. Let’s call the 3 rows and .

First, let our new

Then let

Lastly, let

This is equivalent to our first augmented matrix, and represents the following system of equations:

And we see that the solution is .

The Gauss Jordan Method

If instead of back solving the zeroed matrix, we wish to solve the rest of the equation in augmented matrix form, we’d use the Gauss Jordan

method: We use the row operations to resolve the augmented matrix in the following form:

≡

⎡

⎣

⎢ ⎢ ⎢

−2

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

−2

⎤

⎦

⎥ ⎥ ⎥

, ,R1 R2 R3

= −R ′

2 R2 R1

≡

⎡

⎣

⎢ ⎢ ⎢

−2

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

−3

−4

⎤

⎦

⎥ ⎥ ⎥

= × −1/3R ″

2 R

′

≡

⎡

⎣

⎢ ⎢ ⎢

−3

−4

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

− 2 3

7 4 3

⎤

⎦

⎥ ⎥ ⎥

= − 2 :R ‴

3 R

″

3 R1

≡

⎡

⎣

⎢ ⎢ ⎢

− 2

7 4

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

−2

− 2

−1

7 4

−5

⎤

⎦

⎥ ⎥ ⎥

= −3/7( + 2 )R ⁗

3 R ‴

≡

⎡

⎣

⎢ ⎢ ⎢

−2

− 2 3

−1

7 4 3

−5

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

− 2 3

7 4 3

⎤

⎦

⎥ ⎥ ⎥

≡

⎡

⎣

⎢ ⎢ ⎢

− 2 3

7 4 3

⎤

⎦

⎥ ⎥ ⎥

x + y + z = 7

y − z =2 3

4 3

z = 1

x = 4, y = 2, z = 1

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

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To do so, we go through and subtract rows from those above them to zero out the values above the diagonal. In our prior example, starting with

the latest equivalent matrix, add to row to get:

Then subtract rows and from to return:

This is called reduced row echelon form.

Applications of Systems of Linear Equations

In this section, we will cover several applications of linear equations.

Example 1 6.3 Ex 11 Mathematics with Applications, Lial et al. Pretzels cost $3 per pound, dried fruit $4 per

pound, and nuts $8 per pound. How many pounds of each should be used to produce 140 pounds of

trail mix costing $6 per pound in which by weight there is twice as much pretzel as dried fruit?

We can write this as 3 equations with the number of pounds of each ingredient the unknowns: (we

choose as nuts, as pretzels, and as fruit to save some steps)

We put this in augmented matrix form:

Subtract from to get

Add to to get

Solving through, we find , meaning our conditions are satisfied when

we use 80 pounds of nuts, 20 of fruit, and 40 of pretzels.

Going through the same steps with equations:

2/3 × R3 R2

≡

⎡

⎣

⎢ ⎢ ⎢

− 2 3

7 4 3

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

R2 R3 R1

≡

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

x y z

x + y + z = $140

y − 2z = 0

$8x + $3y + $4z = 140 × $6 = $840

8R1 R3

⎡

⎣

⎢ ⎢ ⎢

−5

−2

−4

140

−280

⎤

⎦

⎥ ⎥ ⎥

5R ′

2 R ′

⎡

⎣

⎢ ⎢ ⎢

−2

−14

140

−280

⎤

⎦

⎥ ⎥ ⎥

z = 20, y = 40, and x = 80

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Subtract 8 times equation 1 from equation 3 to get

Add 5 times equation 2 to equation 3 to get

Solving:

Example 2 6.3 Ex 7 Mathematics with Applications, Lial et al. Tickets to a concert cost $5 for adults, $3 for

teenagers, and $2 for preteens. If 570 people attend a concert, total ticket receipts come to $1950,

and ¾ as many teenagers attend as preteens, How many of each age group attended?

Let represent the number adults attending the concert; represent the number of teenagers

attending the conference, and represent the number of preteens attending the conference.

If 570 people attend the conference, then .

If total ticket receipts come to $1950 then .

If ¾ as teenagers attend as preteens, then .

Using an augmented matrix,

Subtract 5 times the first row to the last row to get

Add 2 times the second row to the last row to get

We can conclude that there were 200 preteens, 150 teenagers, and 220 adults in attendance.

Example 3

x + y + z = $140

y − 2z = 0

$8x + $3y + $4z = 140 × $6 = $840

x + y + z = $140

y − 2z = 0 − 5y − 4z = −280

x + y + z = $140y − 2z = 0 − 14z = −280

z = − = 20 −280

−14

y = 2z = 40

x = 140 − 20 − 40 = 80

x y

x + y + z = 570

5x + 3y + 2z = 1950

y = z 3 4

⎡

⎣

⎢ ⎢ ⎢

−0.75

570

1950

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

−2

−0.75

−3

570

−900

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

−0.75

−4.5

570

−900

⎤

⎦

⎥ ⎥ ⎥

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6.3 Ex 24a Mathematics with Applications, Lial et al. An electronics company produces transistors,

resistors, and computer chips. Each transistor requires 3 units of copper, 1 unit of zinc, and 2 units of

glass. Each resistor requires 3, 2, and 1 units of the three materials, and each computer chip requires

2, 1, and 2 units respectively. How many of each product can be made with 810 units of copper, 410

of zinc, and 490 of glass?

Let be our total transistors, resistors, and computer chips produced, respectively.

Since we have 810 units of copper, and transistors, resistors, and computer chips require 3, 3, and 2

units of copper respectively, we can say .

We can apply the same logic to 410 units of zinc: transistors, resistors, and computer chips require 1,

2, and 1 units of zinc so . We put this equation first in our matrix because we want

a leading 1.

Finally, transistors, resistors, and computer chips require 2, 1, and 2 units of glass, so

. We organize these in the following augmented matrix:

So they can make 100 transistors, 110 resistors, and 90 computer chips.

Matrices

We’ve discussed augmented matrices, but these are not the most frequently used version of a matrix. Most of the time, when we use the word

matrix, we’re referring to an m×n array of numbers, where m is the number of rows and n the number of columns. Rows are always listed first.

A typical 2×3 matrix might look like this:

If , then the matrix is called a row matrix. If , then the matrix is called a column matrix. If , the matrix is called a square

matrix.

For convenience, given a matrix , we denote the entry in the row and column as . In the above example, .

A row matrix A colum matrix A square matrix

Matrix Addition and Subtraction

x, y, and z

3x + 3y + 2z = 810

x + 2y + z = 410

2x + y + z = 490

≡ ≡

⎡

⎣

⎢ ⎢ ⎢

410

810

490

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

−3

−1

410

−420

490

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

−3

−1

410

−420

−330

⎤

⎦

⎥ ⎥ ⎥

≡ ≡ ≡

⎡

⎣

⎢ ⎢ ⎢

−3

−1

410

−330

−420

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

410

110

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

100

110

⎤

⎦

⎥ ⎥ ⎥

[ ]41.5 1−4 38 m = 1 n = 1 m = n

A ith jth aij = 8a23

[ ]1 2 3 ⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

−2

⎤

⎦

⎥ ⎥

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If two matrices, and , are both , then (and only then), they can be added to one another. This is a simple matter of going through and

adding or subtracting each element to the corresponding element in the other matrix, as follows:

For example,

Similarly, an matrix can be subtracted from another in the expected manner:

Scalar Multiplication

A matrix can be multiplied by a scalar. A scalar is just a constant. In this case, you just multiply every element of the matrix by the scalar, as

follows:

For example,

Matrix Multiplication

Given an matrix, and an matrix, , we define the product of the row of (a row matrix) and the column of (an

column matrix) as the sum of the products of each corresponding entry. For example:

We can now define the product of and , (written , as the matrix whose entries = the row of × the column of .

Right away, we can see that it’s not always true that . In fact, if , isn’t even defined.

Example

A B m × n

[ ] + [ ] = [ ] a11

a21

a12

a22

a13

a23

b11

b21

b12

b22

b13

b23

+a11 b11

+a21 b21

+a12 b12

+a22 b22

+a13 b13

+a23 b23

+ =

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

m × n

[ ] − [ ] = [ ] 7

c × [ ] = [ ] a11

a21

a12

a22

a13

a23

c × a11

c × a21

c × a12

c × a22

c × a13

c × a23

10 × =

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

m × n A n × k B ith A 1 × n jth B

n × 1

[ ] × = 2 × 1 + 4 × 3 + 6 × 5 = 442 4 6 ⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

A B AB m × k C cij i th A jth B

AB = BA m ≠ k BA

[ ] = [ ] 1

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

(1 × 1) + (2 × 4) + (1 × 5)

(3 × 1) + (2 × 4) + (4 × 5)

(1 × 5) + (2 × 3) + (1 × 2)

(3 × 5) + (2 × 3) + (4 × 2)

(1 × 2) + (2 × 1) + (1 × 1)

(3 × 2) + (2 × 1) + (4 × 1)

= [ ] 14

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Suppose you’re offered a stock portfolio with 1000 shares of Stock A, 500 of Stock B, and 800 of Stock C, or a different portfolio with 850

shares of A, 800 shares of B, and 700 of C. You’re curious about two scenarios outlined in the following table:

Stock A Stock B Stock C

Scenario 1 $50 $60 $40

Scenario 1 $60 $55 $35

Let’s look at this equation in terms of what each row and column means

The Identity Matrix

Consider an matrix , with for and for . For , for example this matrix would look like this, with 1’s

along its diagonals and 0’s elsewhere:

We call such a matrix an identity matrix. This is because, for any square matrix ,

For example,

[ ] × = [ ] 50

⎡

⎣

⎢ ⎢

1000

500

800

850

500

700

⎤

⎦

⎥ ⎥

112, 000

123, 500

100, 500

110, 000

n × n I = 1ijk j = k = 0ijk j ≠ k n = 3

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

n × n A

AI = IA = A

[ ] [ ] = [ ] = [ ] 1

(1 × 1) + (2 × 0)

(3 × 1) + (4 × 0)

(1 × 0) + (2 × 1)

(3 × 0) + (4 × 1)

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Inverse Matrices

Given a square matrix , we say is the inverse of if . It can be proven that if then and vice versa.

What’s more, if and , ie. and are both inverses of , then . Which is to say if a matrix has an inverse, that inverse

is unique.

Not every square matrix has an inverse. As you would expect cannot be inverted but other matrices share this property, such as

and .

To find the inverse of a matrix , create an augmented matrix of the form , then following the Gauss Jordan process, convert it to the

form . The resulting will be the inverse of .

Thus far, we’ve only seen augmented matrices with column matrices on the right side. Having more than one column on the right side is

permissible because it’s equivalent to performing the Gauss Jordan process multiple times at once.

For example, you would perform the same row operations on these two augmented matrices to put the left half in the desired format

So it saves time to do them at the same time, as below:

Example

Find the inverse of the matrix .

Put equations into augmented matrix

Swap rows 1 and 2

Multiply row 1 by –1/2

Subtract 2 × row 1 from rows 2 and 3

A A−1 A A = IA−1 A = IA−1 A = IA−1

AB = I AC = I B C A B = C

[ ] 0

[ ] 1

0 [ ] 1

A [A│I] [I|B] B A

[ ] ,[ ]13 24 10 13 24 01

[ ]13 24 10 01

⎡

⎣

⎢ ⎢

−2

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

−2

⎤

⎦

⎥ ⎥ ⎥

≡

⎡

⎣

⎢ ⎢ ⎢

−2

⎤

⎦

⎥ ⎥ ⎥

≡

⎡

⎣

⎢ ⎢ ⎢

−2

− 1 2

⎤

⎦

⎥ ⎥ ⎥

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Add 2 × row 2 to row 3

Multiply row 3 by –1/3

Add 2 × row 3 to row 2

Pulling the resulting matrix out of our augmented matrix, we find

the inverse of is .

If , , and are matrices with , then . This allows us to solve a system of linear equations if we can

find the inverse of the associated matrix.

Example A bar sells beer, wine, and cocktails. Beer sells for $5.00 a glass, Wine sells for $6.50 per glass, and

cocktails sell for $9.00 a glass. One evening, the bar sells 5 more glasses of beer than wine, and a

total of 61 drinks. If their total revenue is $399, how much of each did they sell?

Let be number of cocktails sold, y be glasses of beer, and z be glasses of wine. Then

We put these equations into a matrix, then find the inverse and multiply by a column matrix containing

our solutions in order to solve the system.

≡

⎡

⎣

⎢ ⎢ ⎢

−2

− 1 2

⎤

⎦

⎥ ⎥ ⎥

≡

⎡

⎣

⎢ ⎢ ⎢

−2

−3

− 1

⎤

⎦

⎥ ⎥ ⎥

≡

⎡

⎣

⎢ ⎢ ⎢ ⎢

−2

− 2 3

− 1 2

−1

− 1 3

⎤

⎦

⎥ ⎥ ⎥ ⎥

≡

⎡

⎣

⎢ ⎢ ⎢ ⎢

− 1 3

− 2 3

− 1 2

−1

− 2 3

− 1 3

⎤

⎦

⎥ ⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢

−2

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

− 1 3

− 2 3

− 1 2

−1

− 2 3

− 1 3

⎤

⎦

⎥ ⎥ ⎥

A B X AX = B X = IX = AX = BA−1 A−1

x + y + z = 61

y = z + 5 ⟹ y − z = 5

$9x + $5y + $6.5z = $399

× = ⟹ = ×

⎡

⎣

⎢ ⎢

−1

6.5

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

399

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

−1

6.5

⎤

⎦

⎥ ⎥

−1 ⎡

⎣

⎢ ⎢

399

⎤

⎦

⎥ ⎥

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So we conclude they sold 16 cocktails, 25 beers, and 20 glasses of wine.

Excel Operations

Excel has functions MINVERSE and MMULT that allow us to quickly determine the inverse and product of matrices, respectively. MINVERSE,

takes an range of cells as its input and output, while MMULT takes an and range as inputs and returns an range as

an output. Note: to get these functions to produce a result, you must select an appropriately sized range and, hold ctrl and shift while

you press enter. Excel functions of this sort are called CSE Functions. Cells that are part of the output range of a CSE function cannot be

edited individually.

≡ ≡

⎡

⎣

⎢ ⎢ ⎢

−1

6.5

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

−4

−1

−2.5

−9

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

−1

−6.5

−9

⎤

⎦

⎥ ⎥ ⎥

≡ ≡

⎡

⎣

⎢ ⎢ ⎢

−1

1.385

−0.615

−0.154

⎤

⎦

⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

1.385

0.385

−0.615

−0.154

⎤

⎦

⎥ ⎥ ⎥

≡

⎡

⎣

⎢ ⎢ ⎢

−1.769

1.385

0.231

0.385

−0.615

0.308

−0.154

⎤

⎦

⎥ ⎥ ⎥

× =

⎡

⎣

⎢ ⎢

−1.769

1.385

0.231

0.385

−0.615

0.308

−0.154

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

399

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

n × n m × n n × k m × k

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Applications of Matrices

We often model an economy with an input output matrix, A, which explains what each sector of the economy demands from each other

sector. Each row and column of the matrix represents one sector and the result is an matrix with each entry representing the number of

units its row requires to produce one unit of its column. If we know the input output matrix and we know what each sector produces (listed in an

n entry column matrix called the production matrix) then we can say that is the number of units the economy demands just to fulfill its

own production. We can further create a matrix which tells us how much demand the economy can fulfill outside its own

production process. is called the demand matrix.

Often the input output matrix and the production matrix are known, but the production matrix required to satisfy the demand is unknown. For

this we can use some matrix algebra:

Thus if is invertible,

Example 6.6 Example 6 Mathematics with Applications, Lial et al. An economy produces 2 basic products:

wheat and oil. It requires .25 metric tons of wheat and .33 metric tons of oil to produce 1 metric ton of

wheat. It requires .08 metric tons of wheat and .11 metric tons of oil to produce 1 metric ton of oil.

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n × n

X AX

D = X − AX

D = IX − AX

D = (I − A)X

I − A (I − A D = X)−1

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Find how much total production is eneded to satisfy a demand of 500 metric tons of wheat and 1000

metric tons of oil.

It takes 0.25 units of wheat to produce a metric ton of wheat, so we put 0.25 in the first row and

column of the matrix. We designate the first row and column of the matrix the wheat row and column

and the second row and column the oil row and column. It takes 0.08 metric tons of wheat to produce

a metric ton of oil, so since input and output translate to row and column in these matrices,

respectively, 0.08 goes in the first row, second column. Similarly, we put 0.33 in the second row first

column and 0.11 in the last row last column. We can then find the inverse of less our input-output

matrix, and take the product of it and our demand matrix to determine our production matrix.

and

Example 6.6 Ex 33c Mathematics with Applications, Lial et al. An analysis of the 1958 Israeli economy is

simplified here by grouping the economy into three sectors: Agriculture, Manufacturing, and Energy,

with the following input output matrix:

The demand matrix is:

In thousands of units, how much must each sector produce to meet this demand?

So they must produce 195,492,000 units of agriculture, 25,933,000 units of manufacturing, and

13,580,000 units of energy.

Example

A = [ ] ⟹ I − A = [ ] 0.25

0.33

0.08

0.11

0.75

−0.33

−0.08

0.89

(I − A = [ ])−1 1.390.514 0.125

1.17

X = (I − A D = [ ] [ ] = [ ])−1 1.39

0.514

0.125

1.17

500

1000

819

1427

⎡

⎣

⎢ ⎢

0.293

0.014

0.044

0.207

0.010

0.017

0.216

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

138, 213

17, 597

1, 786

⎤

⎦

⎥ ⎥

I − A = , (I − A =

⎡

⎣

⎢ ⎢

0.707

−0.014

−0.044

0.793

−0.010

−0.017

0.784

⎤

⎦

⎥ ⎥ )

−1

⎡

⎣

⎢ ⎢

1.414

0.0267

0.0797

1.261

0.0161

0.0274

1.276

⎤

⎦

⎥ ⎥

(I − A D =)−1

⎡

⎣

⎢ ⎢

195, 492

25, 933

13, 580

⎤

⎦

⎥ ⎥

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6.6 Ex 36b Mathematics with Applications, Lial et al. The 1963 economy of the state of Nebraska has

been condensed into six sectors: livestock, crops, food products, mining/manufacturing, households,

and other. The input/output matrix is as follows:

If the demand matrix is, in millions of dollars, is

use Excel to find the dollar amount each sector should produce.

Solution:

Lecture 3 Individual Exercises

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⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

0.178

0.143

0.089

0.001

0.141

0.188

0.018

0.010

0.252

0.156

0.411

0.088

0.035

0.012

0.088

0.103

0.063

0.089

0.255

0.005

0.001

0.060

0.007

0.402

0.008

0.003

0.014

0.124

0.474

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

1980

650

1750

1000

2500

3750

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

javascript:HelpWindowKey('WSSEndUser_troubleshooting')

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The following are some review questions for you to practice. Please read each question, think

carefully, figure out your own answer first, and then click "Show Answer" to compare yours to the

suggested answer or the possible solution.

Systems of Two Linear Equations in Two Variables

Individual Exercise 3.01 Textbook, pg. 276, Chapter 6.1 Exercises, Exercise 30: An apparel shop sells skirts for $45 and

blouses for $35. Its entire stock is worth $51,750, but sales are slow and only half the skirts and two-

thirds of the blouses are sold, for a total of $30,600. How many skirts and blouses are left in the

store?

Larger Systems of Linear Equations

Individual Exercise 3.02 Textbook, pg. 288, Chapter 6.2 Exercises, Exercise 68: The owner of a small business borrows

money on three separate credit cards: dollars on Mastercard, dollars on Visa, and dollars on

American Express. These amounts satisfy the following set of equations:

How much did the owner borrow on each card?

The associated matrix is

We first multiply by 100 to eliminate the decimals:

Interchange and :

The owner borrowed $1275 on Mastercard, $3825 on Visa, and $4900 on American Express.

Applications of Systems of Linear Equations

Individual Exercise 3.03 Textbook, pg. 294-295, Chapter 6.3 Exercises, Exercise 12: An auto manufacturer sends cars from

two plants, I and II, to dealerships A and B, located in a Midwestern city. Plant I has a total of 28 cars

to send and plant II has 8. Dealer A needs 20 cars, and dealer B needs 16. Transportation costs

based on the distance of each dealership from each plant are $220 from I to A, $300 from I to B, $400

x y z

1.18x + 1.15y + 1.09z

3.54x − 0.55y + 0.27z

0.06x + 0.05y + 0.03z

= 11, 244.25

= 3, 732.75

= 414.75

R1 R3

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from II to A, and $180 from II to B. The manufacturer wants to limit transportation costs to $10,640.

How many cars should be sent from each plant to each one of the two dealerships?

Individual Exercise 3.04 Textbook, pg. 295, Chapter 6.3 Exercises, Exercise 16: According to a data from a Texas agricultural

report, the amounts of nitrogen (lb/acre), phosphate (lb/acre), and labor (hr/acre) needed to grow

honeydews, yellow onions, and lettuce are given by the following table:

Honeydews Yellow Onions Lettuce

Nitrogen 120 150 180

Phosphate 180 80 80

Labor 4.97 4.45 4.65

Read each question, think carefully, figure out the solution for each question, and then click each

question to reveal the suggested solution.

► (i) If a farmer has 220 acres, 29,100 pounds of nitrogen, 32,600 pounds of phosphate, and 480 hours of labor, can he use all of his resources completely? If so, how many acres should

he allot for each crop?

Solution: Let

=acres for honeydews

=acres for onions

=acres for lettuce

Farmer has 220 acres in total

Farmer has 29,100 lbs of nitrogen

Farmer has 32,600 lbs of phosphate

Farmer has 480 hours of labor

If you solve these systems of equations using any of the methods we covered in the previous

section, you will see that it is not possible to use all the resources completely.

► (ii) Suppose everything is the same as in part (i), except that 1061 hours of labor are available. Is it possible to use all of his resources completely? If so, how many acres should

he allot for each crop?

Solution: The new set of equations is:

⇒ x + y + z = 220

⇒ 120x + 150y + 180z = 29, 100

⇒ 180x + 80y + 80z = 32, 600

⇒ 4.97x + 4.45y + 4.65z = 480

x + y + z

120x + 150y + 180z

180x + 80y + 80z

4.97x + 4.45y + 4.65z

= 220

= 29, 100

= 32, 600

= 1061

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When you solve this set of equations, you will see that yes, it is possible to use all of the

resources completely. The allotment for each crop should be acres for honeydew,

acres for onions, and acres for lettuce.

Individual Exercise 3.05 Textbook, pg. 297, Chapter 6.3 Exercises, Exercise 27: At a pottery factory, fuel consumption for

heating the kilns varies with the size of the order being fired. In the past, the company recorded the

figures in the table:

=Number of Platters =Fuel Cost per Platter

6 $2.80

8 $2.48

10 $2.24

Read each question, think carefully, figure out the solution for each question, and then click each

question to reveal the suggested solution.

► (i) Find an equation of the form whose graph contains the three points corresponding to the data in the table.

Solution: From the table, we observe that when , . Therefore,

When , :

In summary, we have the following set of equations:

(Eq.1)

(Eq.2)

(Eq.3)

This set of equations can be easily solved by the elimination method. First, subtract Eq.1 from

Eq. 2 and obtain . Then, subtract Eq. 2 from Eq. 3 and obtain

. Now, the system is

(Eq.4)

(Eq.5)

Substract Eq. 5 from Eq. 4 to obtain . Thus, . Substituting a into Eq. 4

gives . Substituting a and b into Eq. 1 gives . Thus, the equation is

x = 150

y = 50 z = 20

x y

y = a + bx + cx2

x = 6 y = 2.8

2.8 = a(6 + b(6) + c)2

x = 8 y = 2.48

2.48 = a(8 + b(8) + c)2

x = 10 y = 2.24

2.24 = a(10 + b(10) + c)2

36a + 6b + c = 2.8

64a + 8b + c = 2.48

100a + 10b + c = 2.24

28a + 2b = −0.32

36a + 2b = −0.24

28a + 2b = −0.32

36a + 2b = −0.24

−8a = −0.08 a = 0.01

b = −0.30 c = 4.24

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► (ii) How many platters should be fired at one time in order to minimize the fuel cost per platter? What is the minimum fuel cost per platter?

Solution: Write the function in the form

The minimum value of is 1.99, occurring when . Thus, 15 platters should be fired at one

time to minimize the fuel cost. The minimum fuel cost is 1.99.

Basic Matrix Operations

Individual Exercise 3.06 Textbook, pg. 304, Chapter 6.4 Exercises, Exercise 35: There are three convenience stores in

Gambier. This week, Store I sold 88 loaves of bread, 48 quarts of milk, 16 jars of peanut butter, and

112 pounds of cold cuts. Store II sold 105 loaves of bread, 72 quarts of milk, 21 jars of peanut butter

and 147 pounds of cold cuts. Store III sold 60 loaves of bread, 40 quarts of milk, no peanut butter,

and 50 pounds of cold cuts.

Read each question, think carefully, figure out the solution for each question, and then click each

question to reveal the suggested solution.

► a) Use a 3×4 matrix to express the sales information for the three stores.

Solution:

► b) During the following week, sales on these products at Store I increased by 25%, sales at Store II increased by one-third, and sales at Store III increased by 10%. Write the sales matrix

for that week.

Solution: Store I:

Bread

Milk

y = 0.01 − 0.3x + 4.24x2

y = a(x − h + k)2

y = 0.01( − 30x) + 4.24x2

y = 0.01( − 30x + 225) + 4.24 − 2.25x2

y = 0.01(x − 15 + 1.99)2

y x = 15

B re

M il

P. B

C o

ld C

I ⎡

⎣

⎢ ⎢ ⎢ ⎢

88 48 16 112

⎤

⎦

⎥ ⎥ ⎥ ⎥

II 105 72 21 147

III 60 40 0 50

⇒ 1.25(88) = 110

⇒ 1.25(48) = 60

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Peanut Butter

Cold cuts .

Store II:

Bread

Milk

Peanut Butter

Cold cuts .

Store III:

Bread

Milk

Peanut Butter

Cold cuts .

The new matrix is

► c) Write a matrix that represents total sales over the two-week period.

Solution: We add the two matrices from part (i) and (ii):

Matrix Products and Inverses

Individual Exercise 3.07 Textbook, pg. 316, Chapter 6.5 Exercises, Exercise 52: The first table shows the number of

employees (in millions) in various sectors of the U.S. economy over a 3-year period. The second

table gives the average weekly wage for per employee (in dollars) in each sector over the same

period. (Data from: Bureau of Labor Statistics.)

2007 2008 2009

Private 114.0 113.2 106.9

Local Government 14.0 14.2 14.2

State Government 4.6 4.6 4.6

⇒ 1.25(16) = 20

⇒ 1.25(112) = 140

⇒ (4/3)(105) = 140

⇒ (4/3)(72) = 96

⇒ (4/3)(21) = 28

⇒ (4/3)(147) = 196

⇒ (1.1)(60) = 66

⇒ (1.1)(40) = 44

⇒ (1.1)(0) = 0

⇒ (1.1)(50) = 55

⎡

⎣

⎢ ⎢

110

140

196

⎤

⎦

⎥ ⎥

+ =

⎡

⎣

⎢ ⎢

105

112

147

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

110

140

196

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

198

245

126

108

168

252

343

105

⎤

⎦

⎥ ⎥

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Federal Government 2.7 2.8 2.8

2007 2008 2009

Private 853 873 868

Local Government 784 813 830

State Government 883 923 937

Federal Government 1248 1275 1303

Read each question, think carefully, figure out the solution for each question, and then click each

question to reveal the suggested solution.

► a) Write the information in the first table as a 4×3 matrix A.

Solution: We just need to write the table in the matrix form.

► b) Write the information in the second table as a 3×4 matrix B.

Solution:

► c) Explain what each of the following entries in AB represents: row 1, column 1; row 2, column 2; row 3, column 3; row 4, column 4. Do any of these entries represent anything

meaningful?

Solution: row 1, column 1 represents the total weekly payroll (in millions of dollars) over the three years of

the private sector.

row 2, column 2 represents the total weekly payroll (in millions of dollars) over the three years of

the local government.

Private

Local Gov.

State Gov.

Federal Gov.

2007 2008 2009

A =

⎡

⎣

⎢ ⎢ ⎢ ⎢

114.0

14.0

4.6

2.7

113.2

14.2

4.6

2.8

106.9

14.2

4.6

2.8

⎤

⎦

⎥ ⎥ ⎥ ⎥

P ri

v at

L o

ca l

G o v.

S ta

te G

o v.

F ed

er al

G o v.

2007

B =

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢

853 784 883 1248 ⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥

2008 873 813 923 1275

2009 868 830 937 1303

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row 3, column 3 represents the total weekly payroll (in millions of dollars) over the three years of

the state government.

row 4, column 4 represents the total weekly payroll (in millions of dollars) over the three years of

the federal government.

► d) What was the total weekly payroll (in dollars) for the federal government over the 3-year period?

Solution: row 4, column 4 represents the total weekly payroll (in millions of dollars) over the three years of

the federal government. Thus, we need to multiply row 4 of matrix by column 4 of matrix :

million.

Applications of Matrices

Individual Exercise 3.08 Textbook, pg. 326, Chapter 6.6 Exercises, Exercise 22: A 100-bed nursing home provides two levels

of long-term care: regular and maximum. Patients at each level have a choice of a private room or a

less expensive semiprivate room. The tables show the number of patients in each category at various

times last year. The total daily costs for all patients were $24,040 in January, $23,926 in April,

$23,760 in July, and $24,042 in October. Find the daily cost of each of the following: a private room

(regular care), a private room (maximum care), a semiprivate room (regular care), and a semiprivate

room (maximum care).

REGULAR-CARE PATIENTS

Month Semiprivate Semiprivate

January 22 8

April 26 8

July 24 14

October 20 10

MAXIMUM-CARE PATIENTS

Month Semiprivate Semiprivate

January 60 10

April 54 12

July 56 6

A B

(2.7)(1248) + (2.8)(1275) + (2.8)(1303) = $10, 588

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October 62 8

Let

We are given that

We can write this set of equations in the matrix form as follows:

Using row operations that we covered in the previous sections of this chapter, we can convert the left

part of this matrix into an identity matrix to obtain the values of , , , and .

Therefore,

Individual Exercise 3.09 Textbook, pg. 327, Chapter 6.6 Exercises, Exercise 32: A much simplified version of Leontief’s 42-

sector analysis of the 1947 American economy has the following input –output matrix:

Read each question, think carefully, figure out the solution for each question, and then click each

question to reveal the suggested solution.

► (i) What information about the needs of agricultural production is given in column 1 of the matrix?

Solution: From the input-output matrix, we see that .245 unit of agriculture, .099 unit of manufacturing, and

.433 units of households are required to produce one unit of agriculture.

► (ii) Suppose the demand matrix (in billions of dollars) is

Find the amount of each commodity that should be produced.

⎡

⎣

⎢ ⎢ ⎢ ⎢

24, 040

23, 926

23, 760

24, 042

⎤

⎦

⎥ ⎥ ⎥ ⎥

x y z w

⎡

⎣

⎢ ⎢ ⎢ ⎢

205

235

248

277

⎤

⎦

⎥ ⎥ ⎥ ⎥

A g

ri cu

lt u

M an

u fa

ct u

ri n

H o

u se

h o

ld s

Agriculture ⎡

⎣

⎢ ⎢ ⎢ ⎢

0.245 0.102 0.051

⎤

⎦

⎥ ⎥ ⎥ ⎥

Manuf acturing 0.099 0.291 0.279

Households 0.433 0.372 0.011

D =

⎡

⎣

⎢ ⎢

2.88

31.45

30.91

⎤

⎦

⎥ ⎥

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Solution: The input-output matrix is given by

The matrix is given by

And we calculate with the method we covered in the previous section and we find:

Finally, calculate

In conclusion, $18.2 billion of agriculture, $73.2 billion of manufacturing, and $66.7 billion of

households should be produced.

Lecture 4 – Systems of Linear Inequalities

Learning Objectives

After successfully completing the module, students will be able to:

1. Deduce the feasible region and corner points for systems of linear inequalities in 2 variables

2. Calculate the maximum and minimum of an objective function in larger systems of linear equations

3. Formulate and solve linear programming problems

4. Use Solver to calculate the maximum and minimum of an objective function in systems of linear inequalities

5. Perform sensitivity analysis for a linear programming problem

Linear Inequalities in 2 Variables

A linear inequality in two variables is an inequality of the form , where stands in for <, ≤, >, or ≥, and A,B and C∈ℝ.

Examples might include or .

A =

⎡

⎣

⎢ ⎢

0.245

0.099

0.433

0.102

0.291

0.372

0.051

0.279

0.011

⎤

⎦

⎥ ⎥

I − A

I − A =

⎡

⎣

⎢ ⎢

0.755

−0.099

−0.433

−0.102

0.791

−0.372

−0.051

−0.279

0.989

⎤

⎦

⎥ ⎥

(I − A)−1

(I − A =)−1 ⎡

⎣

⎢ ⎢

1.45

0.53

0.84

0.29

1.76

0.79

0.16

0.52

1.28

⎤

⎦

⎥ ⎥

X = (I − A D)−1

⎡

⎣

⎢ ⎢

1.45

0.53

0.84

0.29

1.76

0.79

0.16

0.52

1.28

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

2.88

31.45

30.91

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢

18.2

73.2

66.7

⎤

⎦

⎥ ⎥

Ax + By □ C □ 2x + 3y < 5 7x − 4.5y ≥ 2.6

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A solution to an inequality is a set of coordinates which satisfies the inequality. For example, is a solution to , since

Since any one linear equality has infinite such solutions, it is often useful to graph the solutions. We start by graphing the boundary line, or the

linear equation version of the inequality (found by replacing the <, ≤, >, or ≥ with an =). We use a dotted line to indicate that the boundary line is

not contained within the set of solutions, and a full line to indicate that it is. Figure 1 shows and .

Figure 1: Linear Inequality in Two Variables

To graph a linear inequality, first solve for in terms of , then plot the boundary line, then shade the area on the correct side of the line. Once

is solved in terms of , the set of solutions to the inequality will be below the line for < and ≤ and above the line for > and ≥. A good way to

determine if you shaded the correct area is to plug in a set of coordinates and see if they fit the inequality; (0, 0) is often quickest.

Example Graph

Systems of Inequalities

Often times, you may be confronted by more than one inequality. A set of two or more inequalities is called a system of inequalities. The graph

of a system of inequalities is those points which satisfy every inequality in the system. This might be empty, such as:

The region of points satisfying every inequality in a system is called the region of feasible solutions or the feasible region for short.

Example Graph the set of points satisfying the following system of inequalities:

Maximizing or Minimizing an Objective Function: The Graphical Method

(x, y) (1, 2) 2x + 3y > 5

2 × 1 + 3 × 2 > 5

2x + y ≤ 4 x − 3y < 1

y x

2x + 3y ≥ 9

x + y > 2, x + y < 1.

x + 2y < 5

x ≤ 6

2x − y > 3

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A frequent problem is to find the maximum or minimum value of an objective function subject to constraints, typically represented by

inequalities, and which choice of decision variables attains that maximum or minimum. These constraints can be thought of as the domain of

the objective function. Typical examples include maximizing profit or minimizing costs. One way of thinking about this sort of problem is the

graphical method, demonstrated below:

Suppose you want to maximize subject to the constraints:

Figure 4: Maximizing an Objective Function Subject to Constraints

Our objective function in this problem is : the function we seek to maximize. Our constraints are

. Our decision variables are and .

First, graph the system. Next, draw in lines representing for various values of . Figure 4 shows lines representing

for . In this case, as the lines get further from the origin, they represent higher values of .

From this it is clear that is maximized when it touches the corner formed by .

Corner Points, Boundedness

Those points where the boundary conditions intersect and are in the feasible region are called corner points. If there exist and such that

and for all points within the feasible region, then the feasible region is said to be bounded. Otherwise it’s called

unbounded. In plainer English, the feasible region is bounded if it doesn’t go off to infinity or negative infinity in any direction. Figure 5 shows a

bounded and an unbounded region.

Figure 5: Bounded and Unbounded Systems

f (x, y) = 3x + 2y

2y + x ≤ 20, y + 2x ≤ 16, x ≥ 0, y ≥ 0

f (x, y) = 3x + 2y

2y + x ≤ 20, y + 2x ≤ 16, x ≥ 0, and y ≥ 0 x y

C = 3x + 2y C

3x + 2y = C C = 6, 12, 18, 24, 28, and 30 C

f (x, y) 2y + x = 20 and y + 2x = 16

x0 y0

|x| < x0 |y| < y0 (x, y)

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Figure 4 motivates the corner point theorem. The corner point theorem states that if is a linear function of and , and the feasible

region of and is bounded, and the feasible region contains its boundaries (i.e. our inequalities use or instead of and ), then

attains both a maximum and minimum within the feasible region, and they will both occur on corner points. Further, if the feasible region is

unbounded, though may not attain a maximum or a minimum, any maximum or minimum does attain must occur at a corner point.

Example 7.3 Ex 19 Mathematics with Applications, Lial et al. A pension fund manager decides to invest at most

$50 million in U.S. Treasury Bonds playing 4% annual interest and in mutual funds paying 6% annual

interest. He plans to invest at least $20 million in bonds and at least $6 million in mutual funds. Bonds

have an initial fee of $300 per million dollars, while the fee for mutual funds is $100 per million. The

manager may spend no more than $8400 on fees. How much should be invested in each to maximize

annual interest? What is the maximum annual interest?

Our decision variables are the amount he’ll invest in bonds and the amount he’ll invest in mutual

funds, which we’ll label and . Thus our constraints are

And our objective function is .

The set of corner points of this domain is , , and . (Note that the first inequality

doesn’t actually reduce the feasible region.) Trying each set of points yields

So interest is maximized at $2.24 million when he spends $20 on Treasury Bonds and $24 million on

mutual funds.

Test Yourself A health food company advertises that their supplement contains at least 20 grams of protein and 300

milligrams of potassium per serving, and contains less than 200 calories They obtain it by mixing

powdered whey protein and dehydrated bananas. The whey contains .8 grams of protein and 5 mg of

f (x, y) x y

x y ≤ ≥ < > f

f f

x y

x + y ≤ 50

x ≥ 20, y ≥ 6

300x + 100y ≤ 8400 ⟹ 3x + y ≤ 84

f (x, y) = 0.04x + 0.06y

(20, 6) (20, 24) (26, 6)

f (20, 6) = 1.16

f (20, 24) = 2.24

f (26, 6) = 1.4

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potassium per gram, but costs $.03 per gram to make and contains 5 calories per gram; the bananas

contain .1 grams of protein and 40 mg of potassium per gram, but costs $.025 per gram to make and

contains 6 calories per gram. How much of each should they use to minimize the cost of production?

► Identify the decision variables

grams of whey protein, grams of dehydrated banana

► What are the constraints?

let be powdered whey protein and be dehydrated bananas.

► What is the objective function?

► Graph the feasible region and find the corner points

Larger Systems of Inequalities

We can’t graph a system of inequalities with more than two variables, but it turns out that the corner point theorem still applies. If your objective

function and constraints are all linear, you need only check the corners of your feasible region to find values that maximize or minimize your

objective function. To solve problems like this quickly and accurately, we can use Excel’s Solver add-in.

Activating Solver

To activate solver, we must first navigate to Excel’s options, under the file menu. Navigate to the Add-ins section, choose “Excel Add-ins” from

the dropdown menu and click “Go”. Check the box labeled “Solver Add-in” and choose “OK”.

x y

0.8x + 0.1y ≥ 20, 5x + 40y ≥ 300, 5x + 6y ≤ 200

f (x, y) = 0.03x + 0.025y

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A button labeled “Solver” should now appear on the far right of the data ribbon.

The SUMPRODUCT Function

SUMPRODUCT is a very useful Excel function which takes two or more arrays of equal length, and takes the product of the first number of

each array, then multiplies the second number of each array, and so on to the last number of each array, and returns the sum of the list of

numbers as a result. For example,

SUMPRODUCT({1,3,5,7},{2,4,6,8})=1*2+3*4+5*6+7*8=100. (See Figure 6.)

We’ll make extensive use of the SUMPRODUCT function in Excel to set up our linear and nonlinear programming models.

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Figure 6: Using SUMPRODUCT

Setting up a System of Inequalities: Solver

Solver is also a very convenient way to quickly solve programming problems. Let’s use solver to maximize the objective function

Subject to constraints

1. Fill in the coefficients of your constraints and objective function in the rows corresponding to the inequality or function and the columns

corresponding to their variable, as in Figure 7. Here rows 5-7 are our constraint functions, row 9 is our objective function, row 3 is

reserved for our variables, and columns B, C, and D correspond to , , and respectively

Figure 7: Adding Coefficients

2. Add the right side of the inequalities, as in Figure 7

Figure 8: Adding Constants

f (x, y, z) = 8x + 5y + 10z

x + y + z ≤ 10

2x + 3y + 4z ≤ 30

4x + y + z ≤ 25

x, y, z ≥ 0

x y z

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3. In the cells between the coefficients and the constants, take the sumproduct of your variable cells with each constraint cell and copy

them into a new column representing the value of the left side of each inequality and the objective function. At this step, it can be useful

to fill in values for , , and to test your formulas.

Figure 9:Adding SUMPRODUCTs

4. Now we open solver. Set the cell with the sumproduct of our variables and our objective function as the objective. Below this, choose To:

Max. Set the three variable cells as the variable cells under “subject to the constraints” select your array of other sumproducts on the

right, the “<=” in the dropdown menu, and the array of right sides of inequalities on the right. In our example, , , and must be greater

than or equal to 0, so we leave that box checked. Finally, pick Simplex LP as a solving method. This is the fastest, most efficient method

for solving linear programming problems, because it employs an algorithm that only checks corner points and improves its guess each

time it checks one. Click Solve. If all goes well, you should see a window telling you Solver found a solution.

x y z

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Note that solver can also find the minimum value in a feasible region, or find the solution to an equation in a feasible region using the “value of”

field.

Sensitivity Analysis

Suppose a sculptor sculpts statues and furniture from marble. Each statue takes 750 hours and 200 lb of marble and sells for a profit of

$15,000 and each piece of furniture takes 100 hours and 500 lb of marble and sells for a profit of $5000. Suppose he works for at most 2,000

hours per year and has access to 1800 pounds of marble every year. How should he maximize his profit?

Our decision variables are number of statues and number of pieces of furniture made (x and y, respectively in the following equations).

Thus our objective function is

And our constraints are

We can model this problem in Solver:

f (x, y) = 15, 000x + 5, 000y

750x + 100y

200x + 500y

x, y

≤ 2000,

≤ 1800,

≥ 0.

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And graph it as follows:

Figure 10

We see that his profit is maximized when he makes 2.25 statues and 3.15 pieces of furniture per year. We call this point B in Figure 10. We

may ask the question, however, how much would the profit he makes from statues or marble have to change before it becomes worth it to jump

from point B to point A or point C? You can make the graphical argument, for example, that if the slope of the objective function line were

steeper, it would be steep enough that the maximal went through point C instead of point B. In fact, the slope of the time constraint line is -7.5.

The slope of any line that represents the set of points where the objective function equals a constant is -15,000/5,000, or -3. If the 15,000 went

up above 37,500 or the 5000 went down below 2000, then the objective line representing the highest constant in our feasible region would go

through point C, as in Figure 11, and sculpting only statues would be the most profitable choice.

Figure 11

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Similarly, we may ask how much a constraint would have to change before switching points became more profitable. In this instance, two of the

points will slide around as the constraint line moves in or out. In our example, statues are a more profitable use of marble than furniture. If he

had so little marble or so much time that time no longer effectively constrained him, he would focus on building only statues, as demonstrated

in Figure 12. He gets enough time to build 18 statues in the original model, so he’d somehow need 18*750=13,500 hours to stop building

furniture. Alternatively, if he were limited in supply to just 267 lb of marble, he wouldn’t waste any of it on less profitable furniture.

Figure 12

Both of these scenarios fall under the purview of sensitivity analysis. If, after running Solver, "sensitivity" is highlighted under reports, Excel will

generate a report with this information automatically.

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The Allowable Increase and Allowable Decrease fields tell you how much you can change a coefficient or constraint before it affects which

corner point is optimal, or moves one corner point past another and changes the number of corner points. The Shadow Prices are another

useful field which provide insight about the amount relaxing a constraint would be worth. For example, the shadow Price of 19.18 on cell f5 tells

us that the sculptor should be willing to pay up to $19.18 per hour to hire equally skilled labor. The shadow price of 6.16 tells us that he should

be willing to pay $6.16 per additional pound of marble he can find.

Although Solver’s sensitivity analysis is limited to objective coefficients and constraints, the field also includes changing coefficients in

constraint inequalities, and adding new constraints and variables.

Example A farmer plants corn, wheat and sorghum. He’s allotted 100 tons of water, has only $25,000 to spend

on seeds, has 150 acres to plant on, and managed to produce 40 tons of fertilizer.

In addition to whatever rain may fall, he believes each acre of corn will require 1 ton of water, cost

$385 to seed, and require .25 tons fertilizer. Each acre of sorghum will require .65 tons of water, $170

to seed, and requires .35 tons of fertilizer. Each acre of wheat will require .85 tons of water, cost $205

to seed, and require .3 tons of fertilizer.

He can sell wheat for $1000/acre, corn for $1400/acre, and sorghum for $900/acre.

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How much of each should he plant in order to maximize profit (after the cost of seeds)? Run

sensitivity analysis to determin:

how much he should be willing to pay for extra water, fertilizer, acreage, or seed money.

How much more or less would his crops have to sell for in order to change his decision?

We first subtract the cost of seeds from his selling prices to determine his profit.

Define our decision variables as follows: Let x be acres of corn, y acres of wheat, and z acres of

sorghum. Then we have the following constraints:

Running solver we get

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Sensitivity analysis reveals

Which tells us that, for example, the price of corn could increase to $1659 or decrease to $1217 before it’d change his optimal strategy. We can

also see that, for example, investing in one ton of fertilizer would yield $882 in profit, and that the amount of water available would have to

decrease below 86 tons before he’d be inclined to choose a different corner point.

Test Yourself In the example above, what is the range of sorghum prices for which the farmer should maintain his

current allotment of acreage?

Solution: The price of sorghum could fall to $785 or rise to $984 without affecting his planting strategy.

Lecture 4 Individual Exercises

The following are some review questions for you to practice. Please read each question, think

carefully, figure out your own answer first, and then click "Show Answer" to compare yours to the

suggested answer or the possible solution.

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Graphing Linear Inequalities in Two Variables

Individual Exercise 4.01 Textbook, pg. 344-345, Chapter 7.1 Exercises, Exercise 52: A manufacturer of electric shavers

makes two models: the regular and the flex. Because of demand, the number of regular shavers

made is never more than half the number of flex shavers. The factory’s production cannot exceed

1200 shavers per week.

► (i) Write a system of inequalities that describes the possibilities for making regular and flex shavers per week.

Solution Let

= the number of regular shavers per week

= the number of flex shavers per week

The system of inequalities is:

The number of regular shavers made is never more than half the number of flex shavers

The factory’s production cannot exceed 1200 shavers per week

We also need the inequality constraints as the number of shavers produced cannot be negative

► (ii) Graph the feasible region of this system of inequalities.

Solution

The constraints and restrict the feasible region to the first quadrant.

represents the region on or above the solid line .

represents the region on or below the solid line . The feasible

region is the overlap of all these regions.

Linear Programming: The Graphical Method

x y

⟹ x ≤ y or y ≥ 2x 1 2

⟹ x + y ≤ 1200

⟹ x ≥ 0 and y ≥ 0

x ≥ 0 y ≥ 0 y ≥ 2x

y = 2x

x + y ≤ 1200 x + y = 1200

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Individual Exercise 4.02 Textbook, pg. 352, Chapter 7.2 Exercises, Exercise 18: Find values of x≥0 and y≥0 that minimize

z=3x+2y, subject to each of the following set of constraints:

► (i)

Solution

Corner Point Value of

(0, 7/2) 7 (Minimum)

(0, 6) 12

(175/72, 91/36) 889/72

The minimum value is obtained at and .

► (ii)

Solution

Corner Point Value of

(0, 5) 10 (Minimum)

(75/26,20/13) 305/26

(15/2, 0) 45/2

The minimum value of 10 is obtained when and .

10x + 7y ≤ 42

4x + 10y ≥ 35

x = 0 y = 7 2

6x + 5y ≥ 25

2x + 6y ≥ 15

x = 0 y = 5

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► (iii)

Solution

Corner Point Value of

(0, 22/5) 44/5 (Minimum)

(0, 17/2) 17

(5/2, 6) 39/2

(37/7, 16/7) 143/7

The minimum value is obtained when and .

Applications of Linear Programming

Individual Exercise 4.03 Textbook, pg. 356, Chapter 7.3 Exercises, Exercise 4: A hospital dietician has two meal choices: one

for patients on solid food that costs $2.75 and one for patients on liquids that costs $3.75. There is a

maximum of 600 patients in the hospital. Write the constraints in this problem as linear inequalities

and identify all variables used. Note that you may not need the whole information to write the

constraints.

Let

= number of patients on solid food

= the number of flex shavers per week

The first set of constraint is related to the given information “There is a maximum of 600 patients in

the hospital”:

We also need the non-negativity constraints:

2x + 5y ≥ 22

4x + 3y ≤ 28

2x + 2y ≤ 17

x = 0 y = 22 5

x + y ≤ 600

x ≥ 0, y ≥ 0

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Individual Exercise 4.04 Textbook, pg. 356, Chapter 7.3 Exercises, Exercise 12: The hospital in Individual Exercise 4.03

always has at least 100 patients on solid foods and at least 100 on liquids. Assuming the facts in

Individual Exercise 4.03 what number of each type of patient would minimize food costs?

From Individual Exercise 4.03 we have the constraints:

We have another set of constraints based on the new information given:

“The hospital always has at least 100 patients on solid foods”:

“The hospital always has at least 100 patients on liquid foods”:

Now, we need to find an objective function. Based on the information given in Individual Exercise

4.03, the cost function to be minimized is

We first identify the feasible region and the corner points as follows:

The cost function z is minimized at the corner point (100, 100). Therefore, 100 patients on solid foods

and 100 patients on liquid foods will minimize the food costs.

Individual Exercise 4.05 Textbook, pg. 357, Chapter 7.3 Exercises, Exercise 18: Hotnews Magazine publishes a U.S. and a

Canadian edition each week. There are 30,000 subscribers in the United States and 20,000 in

Canada. Other copies are sold at newsstands. Postage and shipping costs average $80 per thousand

copies for the United States and $60 per thousand copies for Canada. Surveys show that no more

than 120,000 copies of each issue can be sold (including subscriptions) and that the number of

copies of the Canadian edition should not exceed twice the number of copies of the U.S. edition. The

publisher can spend at most $8400 a month on postage and shipping. If the profit is $200 for each

thousand copies of the U.S. edition and $150 for each thousand copies of the Canadian edition, how

many copies of each version should be printed to earn as large a profit as possible? What is that

profit?

Let us first identify the variables:

= the number (in thousands) of copies of the U.S edition

= the number (in thousands) of copies of Canadian edition

Then, let us identify the objective function. We would like to maximize the profit gained from both the

U.S. edition and the Canadian edition. If the profit is $200 for each thousand copies of the U.S.

edition and $150 for each thousand copies of the Canadian edition, then the total profit is

x + y ≤ 600

x ≥ 0

y ≥ 0

2.75x + 3.75y

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Finally, let us identify the constraints based on the information given in the question. Postage and

shipping costs average $80 per thousand copies for the United States and $60 per thousand copies

for Canada and the publisher can spend at most $8400 a month on postage and shipping:

Surveys show that no more than 120,000 copies of each issue can be sold (including subscriptions):

… that the number of copies of the Canadian edition should not exceed twice the number of copies of

the U.S. edition.

There are 30,000 subscribers in the United States:

… and 20,000 in Canada:

In summary, we have the following linear programming problem:

Maximize

In order to solve this problem, we first graph the feasible region and identify the corner points.

There are two solutions to this problem: Either and copies of each edition

or copies of the US edition and copies of the Canadian edition provide a

profit of $21,000.

The Simplex Method: Maximization

Please note that the textbook is covering the simplex method in this section. We will not solve the

problems using the simplex method. Instead, we will be using Excel’s Solver to solve the problems in

this section. Please formulate and solve the following problem using Solver.

Individual Exercise 4.06 Textbook, pg. 371, Chapter 7.4 Exercises, Exercise 32:

Maximize

Subject to

200x + 150y

80x + 60y ≤ 8400

x + y ≤ 120

y ≤ 2x

x ≥ 30

y ≥ 20

z = 200x + 150y

x = 60, 000 y = 60, 000

x = 90, 000 y = 20, 000

z = + 5 − 10x1 x2 x3

8 + 4 + 12 ≤ 18x1 x2 x3

+ 6 + 2 ≤ 45x1 x2 x3

5 + 7 + 3 ≤ 60x1 x2 x3

≥ 0, ≥ 0, ≥ 0x1 x2 x3

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We first formulate the problem in Excel as follows (please see the attached Excel sheet for the

formulas inside the cells).

Then, we obtain the following Solver parameters:

When we hit “Solve” we obtain

This means that the optimal values of the variables are as follows:

with an objective function value of 22.5.

Maximization Applications

Individual Exercise 4.07 Textbook, pg. 377, Chapter 7.5 Exercises, Exercise 14: A baker has 60 units of flour, 132 units of

sugar, and 102 units of raisin. A loaf of raisin bread requires 1 unit of flour, 1 unit of sugar, and 2 units

of raisins, while a raisin cake needs 2,4, and 1 units, respectively. If raisin bread sells for $3 a loaf and

a raisin cake for $4, how many of each should be baked so that the gross income is maximized?

What is the maximum gross income?

Let

= number of raisin bread loaves

= number of raisin cakes

Our problem is

Maximize

We set up our Solver spreadsheet as follows (please see the Excel file for details):

Then, we set the Solver parameters as follows:

When we hit Solve, we obtain the following:

This means that producing a maximum gross income of $168.

The Simplex Method: Duality and Minimization

Please note that the textbook is covering the simplex method: duality and minimization in this section.

We will not solve the problems using the simplex method. Instead, we will be using Excel’s Solver to

solve the problems in this section. Please formulate and solve the following problem using Solver.

Individual Exercise 4.08

= 0x1 = 4.5x2 = 0x3

z = 3 + 4x1 x2

= 48, = 6x1 x2

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Textbook, pg. 386-387, Chapter 7.6 Exercises, Exercise 30: Joan McKee has a part-time job

conducting public-opinion interviews. She has found that a political interview takes 45 minutes and a

market interview takes 55 minutes. To allow more time for her full-time job, she needs to minimize the

time she spends doing interviews. Unfortunately, to keep her part-time job, she must complete at least

8 interviews each week. Also, she must earn at least $60 per week at this job, at which she earns $8

for each political interview and $10 for each market interview. Finally, to stay in good standing with her

supervisor, she must earn at least 40 bonus points per week; she receives 6 bonus points for each

political interview and 5 bonus points for each market interview. How many of each interview should

she do each week to minimize the time spent?

Let

= the number of political interviews to conduct

= the number of market interviews to conduct

Minimize

Subject to

We set up our Solver spreadsheet as follows (please see the Excel file for details):

Then, we set the Solver parameters as follows:

Finally, we solve this system and obtain the following solution:

This means that Joan should conduct 8 political interviews and no market interviews. The minimum

time she can spent is 360 minutes.

w = 45 + 55x1 x2