Two-Variable Inequalities
Shaun P Bowlin
MATH 222 Week 2 Assignment
Instructor: Mark Marino
February 4, 2014

Two-Variable Inequalities In keeping with the syllabus and overall progression of our study of functions and relationships of variables, this week we took a look at Two-Variable inequalities. The assignment was a real world application of using these types of equations to solve shipping problems. We can use these equations to figure out the solutions to common problems and provide graphs to show a range of correct and incorrect answers that is easier to read. The problem that was given, #68 on pg. 539 shows a graph that illustrates how many TVs and refrigerators a semi-truck can hold at any given time in any combination of the two. By looking at the graph, it shows that you can have a maximum load of 330 TVs with 0 fridges or 110 fridges with 0 TVs. It drawn as a solid line to show that any point on that line could also be a solution, as opposed to a dashed line which would denote not including those values into the solution. Also, because there are so many solutions, the area is shaded to include all viable numbers.

You can substitute y for the number of TVs and x for the number of refrigerators for the purpose of simplifying the equation. There are already 2 points on the graph, (0,330) and (110,0) so that is where we will start to compute the slope of the line. m=y1-y2 = 330-0 = 330= 3 x1-x2 0-110 -110 -1
The slope is determined to be -3/1

Now we can go ahead and use the point slope form to write the actual equation

y-y1=m(x-x1) This is the point slope form, now its just plug and solve

y-330= -1/3(x-0) Substitute the slope for m and (330,0) for x,y

330+y-330=-3/1x+330 Add 330 to both sides with distributive property

y=-3x+330 Multiply both sides by 1 to eliminate the fraction

y+3x≤330 Add 3x to both sides and you get the finished equation.

There are 2 more questions that were asked in this section:

1. Can the truck hold 71 fridges and 118 TVs?

This could be called a test point (71,118), they are used to make sure that the shading of the solutions on the graph will actually fall on the correct side of the line. The pair is used to substitute into the linear inequality to determine if it is an actual solution.

y+3x≤330

118+3(71)≤330

331 So, no, the combination will not work.

2. Can the truck hold 51 refrigerators and 176 TVs?
This test point will be worked in the same way as before.

y+3x≤330

176+3(51)≤330

329 which is less than or equal to 330 so yes this combination will work.

The following is a scenario using the inequality from before that was used to solve #68 on page 539 (Dugopolski,2012): The Burbank Buy More store is going to make an order which will include, at most 60 fridges. What is the maximum number of TVs that could be delivered on the same semi?

y+3x≤330

y+3(60)≤330 Substitute 60 for x

y≤330-180 Subtract from both sides

y≤150 So, there may be up to 150 TVs in the truck

The next day, The Burbank Buy More decides they will have a TV sale so they change their order to include at least 200 TVs. What is the maximum number of fridges that could also be delivered? y+3x≤330 200+3x≤330 Substitute 200 for y

3x≤330-200 Subtract from both sides

x≤330/3 or 100 So, there could be 100 fridges on the truck as well as the 200 TVs.

This assignment further solidified the relationships between variables and functions using graphs to depict the solutions as well as relating the subject to real life situations. The graphs were used to demonstrate how a group of solutions can be easily identified by sight.

Reference

Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY: McGraw-Hill Publishing.