Running Head: Problem Number Six
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Problem Number Six
Student Name
Class
February 28, 2013
Problem Number Six
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The number of TVs is represented by ‘V’. The number of refrigerators is represented by ‘R’.
The slope of the straight line represented by the graph is:
(330-0)/(0-110) = -3 and intercept is 330.
The slope-intercept form of the equation of straight line is:
V = 330 - 3R, which is equal to V+3R= 330.
The colored area to the left of the straight line can be represented by the linear inequality:
3R + V ≤ 330.
The colored region under the straight line is represented by the less than portion of the
inequality where the solid line is represented by the equal to portion of the inequality. If the
colored area and a dashed line were the only constituents of the graph, then the equality part
would be eliminated and only the < sign could have been used.
To check if the size of the truck is enough for 71 refrigerators and 118 TVs, we could substitute
the numbers in the inequality
From the inequality, this is what we see 3*71 + 118 = 331 > 330.
Problem Number Six
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The inequality is not satisfied and therefore the truck cannot hold the two items. By using
substitution method
R=51 and V=176, we get 51*3+176 = 329 < 330.
The inequality stays the same, thus the truck’s dimensions is enough for the refrigerators.
If there is a stipulation that the order would contain at most 60 refrigerators, it would be
indicated by the inequality R ≤ 60. This will be represented by a solid line when graphed at
R=60. This line will be parallel to the y-axis and the colored area will be to the left of the vertical
carried alongside the 150 Tv’s which is the extreme number of, which can be carried alongside
60 refrigerators. Yet, the new scenario can also mean zero refrigerators so the maximum is 330
TVs. Taking into consideration another example, let’s assume that the order will include at least
200 TVs. It would be represented by the inequality V ≥ 200.
When graphing a solid line parallel to the x-axis will be procured at V=200 and the colored region
will be above the straight horizontal line. The maximum number of refrigerators that can be
transported will now be (330-200)/3 = 43 this is after rounding the number off.