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