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
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
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)
?
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
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.
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.