Word Problem/Quadratic? Help!

[hakandragon]

Hi everyone; I hope I'm doing this right LOL. So, it's Principles of Math 11, and I just can't do it (trust me, I've tried). I'm working with Quadratic Functions and that sort of stuff...(also let me add that I can't do math to save my life). I'm the sort of person who needs examples. However, as soon as the question changes, I need a new example. Which sort of sucks...anyways, any help would be soooo great.

The question is: What is the maximum rectangular area that can be enclosed by 120m of fencing if one of the sides of the rectangle is an existing wall (Show algebraically.)

Then there is a drawing of a box, something like this:



I'm not quite sure where to begin; my math "teacher" went over 'y=a(x-h)²+k' with me, and I sort of got that but now I'm stuck again (see, new problem, needs new example LOL). So far all I've got is:

A=LxW
L=120-2x
W=x
A=(120-2x)(x)

And I doubt that's even close to being on the right planet, let alone the right track.
:(
I think that's everything (for now)...
Thanks so much for any tips.

[stapel_eliz]

What is the maximum rectangular area that can be enclosed by 120m of fencing if one of the sides of the rectangle is an existing wall

The "perimeter" is 120 meters, consisting of one length and two widths. Letting the widths be "x", then:

. . . . .120 = L + 2x

. . . . .120 - 2x = L

Then the area is:

. . . . .A = Lw

. . . . .A = (120 - 2x)(x) = 120x - 2x²

The maximum of this negative quadratic will be at the vertex (h, k), where h = -b/2a:

. . . . .h = -(120)/(2(-2)) = 120/4 = 30

Then the width will be x = 30, the length with be L = 120 - 2(30), and the maximal area A will be the value of A when x = 30. (So plug "30" in for "x".)

[hakandragon]

Thank you so much!!!