Inequalities of the university to the college

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INEQUALITIES OF THE UNIVERSITY TO THE COLLEGE.

BOGOTA-COLOMBIA

INEQUALITIES OF THE UNIVERSITY TO THE COLLEGE. In this workshop you will work some examples of inequalities that work at the University level, but which can perform them with the students of secondary education. When we are in the top-level teaching we have problems with students who come to us at this level, since in the vast majority of them do not bring concepts and minimal mathematical knowledge that we wish that they bring. But this is because we see a large gap between higher education and the secondary education, when it should not have this vacuum but be a bridge to continue learning and teaching.

A student in secondary education learn math concepts, but 90% do not understand them and is where starting to arise the difficulties of understanding the new topics. One of the issues that we face are inequalities, in the secondary level use it find intervals, and solving inequalities in the plane on some occasions. But we need them as the basis of the linear programming in economic science students at the University level.

We as teachers don't want our students to learn subjects memorizing a series of rules that achieve internalize the concepts, and these we do doing exercises that have to do with the reality.

For this the needs to practice more on exercises that achieve an internalization of mathematical concepts. And if accompanied of technology in the learning processes, these processes are learned and understand in a better way.

In part one we are going to put a line in the Cartesian plane and ask ourselves how many parties divided the straight to the plane? In our calculator we take the option of algebra and we write our equation

Now we copy our equation on the right side and put it in a new window, it should be with the graphics option.

We already have the straight and can be seen in how many parts divided us plane, we can see that it is divided into three parts: to the right of the line to the left of her and herself. We can write this as follows: A.) y< 3 x + 5 B) y > 3 x + 5 C) y=3 x + 5 The third option is that we have represented on our screen.

We pushed the keys CTRL + G and we are going to the f1 function and delete the equal sign, when we do a menu of equalities and inequalities, appears to us in this case take the option of > y 路 The two remaining regions Graficamos.

Y>3x+5

y<3x+5

We note that we can have several regions with points which are to the right of the line, to the left and those who are on the straight, but we also have the region points to right of the straight with the points on the line, left with the points of the straight and the three regions, right - left and the same straight or right - left and without the straight. We realize these situations in our calculators.

Now let's see how we can internalize each of those situations in our students. For this we are going to solve a problem that has to do with any real situation. For example:

(a) draw the enclosure formed by the points that meet the following y  3 conditions:  y  x  1 y  3 x  0 

(b) indicates if the points (0, 0), (2, 1) and (1, 2) are part of the previous system solutions.

Solution: a) Draw thelas lines presenta mos rectas

y  3  y  x  1  y  x  1 y  3 x  0  y  3 x 

b) Now draw the regions

y  3  y  x  1 y  3 x  0 

Now that we have the straight lines we find regions The requested region is:

Is the solution of three inequalities.and if we look at the above chart, we have that (0, 0) and (2, 1) are not solutions of the system, but (1, 2) Yes it is.

And the reality that serves us? This is what we teach in our secondary education programs. Now we will see the same thing but with an application in reality. This is what we call a graphic solution to a linear programming problem, which is seen in 3 or 7 semester of University.

We have 210 000 dollars to invest in stock market. They recommend two types of actions. A type that pay 10% and those of type B that pays 8%. We decided to invest a maximum of 130 000 dollars in kind to and at least 6 000 $ in the type B. In addition, we want investment in type A is less than or equal to double investment in B. What do you have to be the distribution of the investment to obtain maximum annual interest?

Solution: We call x the money we invest in shares of e to type and to which invest in the type B. We summarize the data in a table: (insert a calculation sheet ~→4→6 ))

The restrictions are

� + � ≤ 210000 � ≤ 130000 � ≼ 6000 � ≤ 2� �≼0 �≼0

z  0,1x  0,08 y 

1 10 x  8 y   2 5 x  4 y   1 5 x  4 y . 100 100 50

The function which gives us the total yield is: z  0,1x  0,08 y 

1 10 x  8y   2 5 x  4y   1 5 x  4y . 100 100 50

We must maximize this function, subject to the above restrictions. We draw the corresponding restrictions enclosure

The region is:

Now in a window of algebra are the cut-off points

The maximum is reached at the point (130000, 80000). Where z takes the greater value. Therefore, we need to invest $130 000 in shares of type A and $80 000 in the type B. In this case, the annual benefit will be of

In this way we see that it is not difficult to solve with education students average this kind of problems, which are a reality and in this way the student application internalizing the concepts of inequalities, lines, and maximum and minimum values as well as the solution of linear equations. I hope that this small example will serve as much to see how we can become of contents of university education. We can now solve more applications.

Exercises 1. True manufacturer produces two articles, and (b), which requires the use of two sections of production: Assembly section and paint section. Article requires an hour of work in the Assembly section and two in the painting; and article B, three hours in the Assembly section and an hour in the painting. The single mounting section can be in operation nine hours a day, while the paint only eight hours each day. The benefit obtained producing article B is 40 euros and the to is 20 euros. Calculates the daily production of items A and B that maximizes the benefit.

2. A kiosk sells to 20 euro cents pens and notebooks to 30 euro cents. We carry 120 euro cents and intend to buy the same notebooks than pens, at least. What is the maximum number of pieces that we can buy? 3. A goldsmith manufactures two types of jewelry. The type A need 1 g of gold and 1.5 g of silver, selling at 40 euros each. For the manufacture of the type B used gold 1.5 g and 1 g of silver, and sells them for $ 50. Goldsmith has only 750 g of each of the metals workshop. It calculates how many jewels has discontinued each class to obtain maximum benefit.

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