This week’s assignment will show how two-variable inequalities can be used in real-world scenarios by using independent and dependent variables. This week’s assignment will use graph representations and show how the two-variable inequalities can be incorporated into several problems to show how many of each item trucks can ship without going over their weight limit. The first problem that I will be doing is #68 on page 539 (Dugopolski, 2012).
Below, the graph shows the maximum number of TVs the 18-wheeler can hold without refrigerators, and the maximum number of refrigerators the 18-wheeler can hold without TVs.
On the X-axis, the graph shows the refrigerators, and on the Y-axis, the graph shows the TVs that the 18-wheeler can carry at a time. To find the slope of the line, I will use the two points that are on the graph, (0,330) and (110,0).
The slope is m= y1 – y2 = 0 – 330 = -330 = -3, so the slope is -3.
x1 – x2 110 – 0 110
To make it easier to find how many refrigerators and how many TVs can fit in the 18-wheeler, it would be best to have a linear equation. To find the linear equation, the point-slope form can be used. y – y1 = m(x – x1)This is the point-slope form.
y – 330 = -3 (x – 0)The slope is substituted for m and (330,0) is substituted for x and y. y = -3x + 330Distributive property is used and 330 is added to both sides. y+3x ≤ 3303x is added to both sides and the less than or equal to symbol has been changed. The graph shown above has a
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solid line instead of a dotted line because the points that are on the line are a part of the solution set. Anytime the inequality symbol has the “equal to” bar, the line will be solid rather than dotted. There are two more parts to problem #68 on page 539.
1. Will the truck hold 71 refrigerators and 118 TVs?
Since the x-axis represents the refrigerators, and the y-axis represents TVs, the point would look like (71,118).
This would be called a test point and is used to make sure that the shading would fall on the correct side of the line, and test if the truck would be able to hold 71 refrigerators and 118 TVs. To test if the truck could hold 71 refrigerators and 118 TVs, the points would be plugged into the inequality. y + 3x ≤ 330The inequality that was created.
(118) + 3(71) ≤ 330The points are plugged into the inequality. 118 + 213 ≤ 330
331 ≤ 330Since this is a false statement; 331 is not less than 330, the truck would not be able to hold 71 refrigerators and 118 TVs. 2. Will the truck hold 51 refrigerators and 176 TVs?
The given question is another question and will be worded the same way as the 1st problem. y + 3x ≤ 330The inequality that was created.
(176) + 3(51) ≤ 330The points are plugged into the inequality. 176 + 153 ≤ 330
329 ≤ 330This answer is true, 329 is less than 330, so it would be possible for the truck to hold 51 refrigerators and 176 TVs.
The next problem uses the same inequality that was created and requires a minimum and/or maximum range to be found. The Burbank Buy More store is going to make an order which will include, at most, 60 refrigerators. What is the maximum number of TVs that could also be delivered on the same 18 wheeler? To find the maximum number of TVs that the truck could hold, 60 would be plugged into the linear inequality to solve for y. y + 3x ≤ 330The inequality that was created.
y + 3(60) ≤ 33060 is plugged in for x
y + 180 ≤ 330180 is subtracted from both sides
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y ≤ 150
The maximum number of TVs that the truck could hold in addition to the 60 refrigerators is 150. The graph would have a horizontal line at y = 150 and a vertical line at x = 60, and the rectangular shape that is made by the horizontal and vertical lines would be shaded. The only possible shipment amounts would be in the shaded area of the rectangle as shown in the graph to the right.
The next day, the Burbank Buy More decides they will have a television sale so they change their order to include at least 200 TVs. What is the maximum number of refrigerators that could also be delivered in the same truck? This problem will be like the last problem, only I will be solving for x instead of for y. To find the maximum number of refrigerators that could also be delivered in the same truck, 200 TVs would be plugged in to solve for x. y + 3x ≤ 330The inequality that was created.
(200) + 3x ≤ 330200 is plugged in for y 3x ≤ 130200 is subtracted from both sides, and both sides are divided by 3. 43(rounded because people don’t want to buy partial refrigerators)
If 200 TVs were shipped, then the maximum number of refrigerators that could also be shipped is 43. The graph would now have a horizontal line at y = 200, and a vertical line at x = 43. After the horizontal and vertical line makes a rectangle, there is a triangle that is enclosed above the line at the top of the rectangle. The possible number with the minimum number of recliners would fall inside the shaded portion of the triangle as depicted in the graph to the right.
Inequalities can be used in many instances in the real world, such as in the above problems to quickly find out how many items could ship with other items. Just by finding the inequality, I could plug in variables and find out how many refrigerators could be shipped if there were a certain number of TVs and vice versa. If there was a bigger truck that had a bigger capacity, the line would be parallel to the line that was shown in the above graphs. It would have nearly the same linear equality, but since the truck has a higher capacity, the maximum number of TVs and the maximum number of
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refrigerators would be at different points than the given problems. I think that using inequalities in real-world scenarios could prove to be very efficient and effective.
Reference
Dugopolski, M. (2012).
Elementary and intermediate algebra (4th ed.).
New York, NY: McGraw-Hill Publishing
