The area of the conference table in Mr. Nathan’s office must be no more than 175 ft2. If the length of the table is 18 ft more than the width, x, which interval can be the possible widths?
boy what the hell boy
To find the possible widths of the conference table in Mr. Nathan's office, we need to set up an equation and solve for x.
Let's denote the width of the table as x.
According to the given information, the length of the table is 18 ft more than the width, so we can express the length as x + 18.
The equation for the area of a rectangle is length times width. So, the area of the conference table is given by (x + 18) * x.
We are told that the area must be no more than 175 ft^2, so we can set up the following inequality:
(x + 18) * x ≤ 175
To solve this inequality, we multiply out the left side:
x^2 + 18x ≤ 175
Rearranging the inequality to make it easier to solve:
x^2 + 18x - 175 ≤ 0
Now we need to solve the quadratic inequality. We can either factor it or use the quadratic formula. However, in this case, factoring may not be straightforward. So, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In our case, a = 1, b = 18, and c = -175. Substituting these values into the quadratic formula:
x = (-18 ± √(18^2 - 4(1)(-175))) / (2(1))
Simplifying:
x = (-18 ± √(324 + 700)) / 2
x = (-18 ± √1024) / 2
x = (-18 ± 32) / 2
x = (-18 + 32) / 2 or x = (-18 - 32) / 2
x = 14 / 2 or x = -50 / 2
x = 7 or x = -25
Since width cannot be negative in this context, the possible widths are x = 7 ft.
Therefore, the only possible width for the conference table in Mr. Nathan's office is 7 ft.
x(x+18) <= 175
x^2+18x-175 <= 0
(x+25)(x-7) <= 0
So, -25 <= x <= 7
Or, for real-world tables,
0 <= x <= 7
There is probably a minimum positive width for a useful table.