A recatangular yard is to be made with 120 feet fencing. The yard is divided into 4 equal parts. And an existing property fence will be used for one long side.

If x represents the width of the fence exeorsss its area A(x) in terms of x

Determine the dimensions of the rectangle that will make area maximum

5 answers

width --- x
length --- y

If I understand your wording correctly, we will have
5x + 2y = 120 ----> y = 60 - 5x/2

A(x) = xy = x(60 - 5x/2) = 60x - (5/2)x^2
d A(x)/dx = 60 - 5x = 0 for a max of A(x)
5x = 60
x = 12
so the whole rectangle is 12 by 30 and has a max area of 360 ft^2
Ok so for the first part where it asks to express the area of width of the fence in terms of x is that part like this
A(x)=x(60-5x/2)
?
one long side is already there

5x + y = 120 ... y = 120 - 5x

area = x (120 - 5x^2) = 120x - 5x^2

the max is on the axis of symmetry ... x = -120 / (2 * -5) = 12

y = 60 ... area = 720
I'm confused now lol whose correct? 2 very different conclusions
The first answer has two long sides made of new fencing.
The second answer assumes that only one long side is new fencing.
By the way I saw a third way with three long sides and three short sides.