Problem 1
w = width
w+36 = length, because it's 36 feet longer compared to the width
=======================================================
Problem 2
The area w(w+36) is less than 2040, and it's also larger than 0.
We can write that as [tex]0 \le w(w+360) \le 2040[/tex] which is the same as [tex]0 \le w^2+360w \le 2040[/tex]
If you wanted to drop the first part, then you can say either [tex]w(w+360) \le 2040[/tex] or [tex]w^2+360w \le 2040[/tex]
I used the formula area = length*width
=======================================================
Problem 3
There are many possibilities here. Let's say w = 10 feet. If so then the length would be w+36 = 10+36 = 46 feet. This 10 by 46 rectangle has an area of 10*46 = 460 square feet which is under the 2040 sq ft limit.
Another possibility is that w = 20 ft and w+36 = 56. This 20 by 56 rectangle has an area of 20*56 = 1120 sq ft.
It turns out you can pick any value of w between 1 and 30, assuming you only restrict yourself to integers. You can find the largest possible value of w by solving the equation w(w+36) = 2040. The positive solution to this equation is roughly w = 30.62
Side note: even though something like w = 1 is possible, it's not very realistic. A conference hall that's only 1 ft wide won't be able to fit a person comfortably unless they don't mind being sandwiched between two walls and have to walk sideways.
=======================================================
Problem 4
If you solved w(w+36) = 360 with a graphing calculator or the quadratic formula, then you would find the positive solution for w is roughly 8.15
If your teacher is considering positive real numbers, then this is realistic; however, if they are only considering positive integers, then something like w = 8.15 isn't possible.