Euler's Trail

I found this video on YouTube, and I thought it was very interesting. The challenge is to draw a single curve that crosses through every single line in the picture only one time. I tried it a few times myself, but was unable to solve it. Finally, the video revealed that completing this puzzle is mathematically impossible!

Solving Matrices Using Gauss Jordan Elimination

Today we learned how to solve systems of equations using augmented matrices and two methods: Gaussian Elimination and Gauss Jordan Elimination.

I will explain the process for Gauss Jordan Elimination below, where the goal is to achieve a leading one in each row of the matrix while every other number in the matrix (except for the rightmost constants) is equal to zero.

To start, here are the problems from Step 1 of our assignment today; in these, I just solved for each variable as usual with systems of equations:

Now, I will demonstrate how to solve such problems using matrices and the Gauss Jordan Elimination method:

First, convert the system of equations into an augmented matrix. Next, perform any of the three elementary matrix operations on the augmented matrix: either switch rows, add rows, or multiply rows. Try to get a leading one in each row while making every other number a zero, except for the constant on the right. Also, it is recommended to work column-to-column from left to right.

Below, I have attached an image of two problems (#1 and #2) from Step 3 from our assignment today, which I have worked out. I also have listed the operation I performed on each matrix to produce it in the right margins.

Link to Prezi

Follow the link below to go to the Prezi presentation!
http://prezi.com/zbgesjwavrrw/?utm_campaign=share&utm_medium=copy

Was Pi Really Going to Be Changed to 3.2?

I just saw a video that I thought was both very strange and also very funny. The video talked about a man named Edward Goodwin who thought he had solved one of the greatest math problems that involved a circle, but there was one problem: for his solution to work, pi had to be 3.2. Even though this is ridiculous and pi is actually ~3.14, he wanted to patent this idea for fear of somebody else using it!

Below, I have included a picture of the problem called "squaring the circle," which is referred to in the video.

Partial Decomposition of Fractions

In this section we covered a variety of topics, and, since I had not created a post for Partial Decomposition of Fractions, I will make that the subject of this post. Below, I have listed the steps for solving such a problem:

1. Identify the factors of the denominator:

 5x - 4    
x² - x - 2

 5x - 4      
(x-2)(x+1)

2. Create a new fraction for each of those factors:

   A  
(x-2)
+
   B  
(x+1)

3. Multiply this new set of fractions by the denominator of the first fraction:

A(x+1) + B(x-2)

4. Simplify and combine like terms:

Ax + A + Bx - 2B

5. Factor:

(A+B)x + A - 2B

6. Equate the coefficients to their counterparts in the numerator of the first equation, and solve for their values:

A + B = 5
A - 2B = 4

7. Solve as a system of equations:

A = 14/3
B = 1/3

8. Substitute these values into the equation in Step 2, and that will be the answer:

  14  
3x-6
+
   1   
3x+3

Inflation Calculator

I decided to do my free blog post this week on an inflation calculator I found online. I discovered this calculator when I heard my friend talking about how much he thought a certain item would cost in the early 1900's in terms of today's money. When I looked it up, I found the site very interesting.

For example, if you would like to know how much $5.00 in 1913 would be worth in 2013, you select 1913 in the first option box and 2013 in the second. Once you hit calculate, it will give you the result.

I was surprised to see that what seemed to be a very small amount of money, $5.00, was worth over $117!

Here is the link to the site if you would like to try it out yourself.

Linear Programming

Linear programming is used to determine the maximum or minimum value of a set of inequalities.
The easiest method of demonstrating this is to provide an example, so I will use #39 on page 581 of our textbook.
It reads: A merchant plans to sell two models of compact disc players at costs of $250 and $400. The $250 model yields a profit of $45, and the $400 model yields a profit of $50. The merchant estimates that the total monthly demand will not exceed 250 units. The merchant does not want to invest more than $70,000 in inventory for these products. Find the number of units of each model that should be stocked in order to maximize profit.

We must first set up the inequalities and determine that x is to represent the $250 model and y is to represent the $400 one. The first inequality will relate the dollar amounts of the items, while the second will relate the number stocked. Also, we must include two "hidden" inequalities, which dictate that there cannot be a negative number of either model of CD players.

250x + 400y ≤ $70,000
x + y             ≤ 250
x                   ≥ 0
y                   ≥ 0

We should also create an equation to find the maximum profit. We will set z equal to the profit of the $200 model times x plus the profit of the $400 model times y:

z = 45x + 50y

Next we should put the first two inequalities in slope-intercept form (y = mx + b) so that we can graph them (since the last two inequalities are already in an easily graph-able state, we can leave them as-is). This produces:

y ≤ (5/8)x + 175
y ≤ -x + 250

To find the vertices, or the points of intersection, we can use elimination or substitution and solve the inequalities as we would with systems of equations. The x and y values we get as a result will be formatted as an ordered pair (x, y).
We now get the following ordered pairs as vertices:

(0, 0), (250, 0), (0, 175), (200, 50)

All we have to do now is substitute each of these ordered pairs into the equation we created earlier for z and determine which gives the greatest profit:

45(0) + 50(0) = $0
45(250) + 50(0) = $11,250
45(0) + 50(175) = $8,750
45(50) + 50(200) = $11,500

Given that the ordered pair (200, 50) produced the greatest profit, our answer will be:

200 of the $250 CD players
50 of the $400 CD players

Solving Systems of Inequalities

We recently learned how to solve systems of inequalities; as opposed to systems of equations, whose solution was the intersection of the lines involved, the solution of a system of inequalities is the graph itself.

Here is a very simple example:

x > 3

y > 5

We can see that x is greater than 3 and y is greater than 5.

To graph this, we can locate the point on the x-axis where x = 3 and draw a vertical line through it. Then we can locate the point on the y-axis where y = 3 and draw a horizontal line through it.

The type of line we need to draw is dependent on the inequality.

1. If the inequality has a less than or greater to sign, we need to draw a dashed line because the region does not include the line itself. 
2. If the inequality has a less than or equal to sign or a greater than or equal to sign, we need to draw a solid line because the region does include the line itself.

Since both of the inequalities contain a greater than sign, both lines will be dashed.

To determine which side of the line to shade, we can substitute any ordered pair into the inequality, and, if the ordered pair satisfies it, we shade that side. If it does not, we shade the opposite side.

The easiest ordered pair to substitute is usually (0, 0), so that is what we will use.

0 > 3 (false)

0 > 5 (false)

Now we see that neither of these inequalities have been satisfied, so we must shade the side that does not include (0, 0).

Points of intersection of the lines are called vertices, and these will be useful in linear programming. Nevertheless, the solution to the system of inequalities is the region which all inequalities in the system include.
Any point within this region is a valid solution to the system.
Therefore, for our example problem, any point whose x position is greater than 3 and whose y position is greater than 5 will be considered a solution.

The "Golden Spiral"

Above is a video that I found on YouTube and that really interested me.
The video talks about a numerical sequence that, when represented as a spiral, can be seen all throughout nature in shells, leaves, and even the formation of galaxies.
The sequence is called the Fibonacci sequence, and looks like this:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34 . . .

You can continue this sequence easily, since the next number is always the sum of the two before it; for example, the 10th number in the sequence is 34 because the two numbers before it, 21 and 13, add up to 34. Therefore, if you wanted to determine the number after 34, you would add 34 and 21 to get 55.
I thought this was very interesting, so I wanted to make it my third blog post for this week.

Solving Systems of Equations That Contain 3 Unknown Variables

The following is a link to the solution of Problem #4 in 7.3 Problems on Edmodo.
http://www.educreations.com/lesson/view/problem-4/15521447/?s=bt50Vf&ref=app

In this example, I will show the steps to solving a system of equations with 3 unknown variables:

4x + y - 3z = 11
2x - 3y +2z = 9
x + y + z = -3

First, we can start by eliminating y from the first two equations. To do this, we need to multiply the top equation by 3.

12x + 3y - 9z = 33
2x - 3y + 2z = 9

After adding these two equations together, we are left with the following:

14x - 7z = 42

Now, we can eliminate y from the two bottom equations. To do this, we need to multiply the bottom equation by 3.

2x - 3y + 2z = 9
3x + 3y + 3z = -9

After adding these two equations together, we are left with the following:

5x + 5z = 0

Next we will eliminate z by using the equation we produces earlier. We will need to multiply the first equation by 5 and the second by 7.

5 * (14x - 7z = 42) = 70x - 35z = 210
7 * (5x + 5z = 0) = 35x + 35z = 0

After adding these two equations together, we are left with the following:

105x = 210
x = 2

Now, we can substitute the value of x, which we have determined is 2, into either of the two equations we produced earlier. I will substitute x into the second one.

5x + 5z = 0
5(2) + 5z = 0
10 + 5z = 0
5z = -10
z = -2

Next, we can substitute the values of x and z into any of the original three equations. I will substitute x and z into the third one.

x + y + z = -3
(2) + y + (-2) = -3
y = -3

Finally, we have determined the values of all 3 variables. Before writing the answer down, we should confirm the answer by substituting these values into one of the first three equations. I will substitute them into the first equation.

4x + y - 3z = 11
4(2) + (-3) - 3(-2) = 11
8 - 3 + 6 = 11
11 = 11

Since the values we found for the variables were valid substitutions for this equation, we can conclude that we have found the correct values. The last step is to present our answer in the following format:
(x, y, z)

Therefore, our final answer is:
(2, -3, 2)

Solving Systems of Equations by Using Substitution

Systems of equations are used to solve a set of equations consisting of two unknown variables. In this post, I will explain how to solve such a set of equations using the substitution method.

Example:

x - y = 0

5x - 3y = 10

First, we need to isolate one of the variables in either of the equations. I will substitute x in the first equation by adding y to both sides.

x = y

Next, we can substitute the value of x in terms of y into the second equation. Since x = y, the second equation can be expressed as the following:

5y - 3y = 10

2y = 10

y = 5

Now that we know the value of y, we can substitute this value into either of our two original equations in order to get the value of x. I will use the first equation.

x - y = 0

x - (5) = 0

x = 5

At this point, we have determined that x = 5 and y = 5. We can substitute these values into either of the two original equations in order to determine if we have gotten the correct result.

x - y = 0

(5) - (5) = 0

0 = 0

Through this process, we have concluded that both x and y are equal to 5.

Also, systems of equations can have three different outcomes:

1. One unique solution: the graphs of the equations intersect at one point.

2. No solution: the graphs of the equations never intersect.

3. Infinitely many solutions: the graphs of the equations are the same.