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

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

2) Determine the dimensions of the rectangle that will make area maximum

In my previous solution, I missed the part about needing only one long side.

So it is actually easier with no fractions.

long side ---- y ft
short side --- x ft each
y + 5x = 120
y = 120-5x

A(x) = xy = x(120-5x) = 120x - 5x^2
A'(x) = 120 - 10x
= 0 for a max of A(x)

10x = 120
x = 12 , same as before, but
y = 120-60 = 60

state the required conclusions

To find the answers to these questions, we need to follow a few steps:

1) Express the area A(x) in terms of x:
Since we are given the total length of the fencing, and it is divided into 4 equal parts, each part will have a length of 120/4 = 30 feet. The width of the fence is represented by x. We have one long side with the existing property fence. Therefore, the length of the rectangle will be 30 feet, and the width will be x. The formula for the area of a rectangle is A = length × width. Substituting the values, we get A(x) = 30 × x = 30x.

2) Determine the dimensions of the rectangle that will maximize the area:
To find the dimensions that will maximize the area, we need to find the value of x that gives the maximum value for A(x). We can do this by finding the critical points of the function A(x) = 30x. Since this is a linear function, the maximum value is obtained when x is at its highest value, which is when the long side aligns with the existing property fence. Therefore, the dimensions of the rectangle that will maximize the area will be a length of 30 feet and a width of x.

To summarize:
1) The area A(x) of the rectangle in terms of x is given by A(x) = 30x.
2) The dimensions of the rectangle that will maximize the area are a length of 30 feet and a width of x.

as always in this type of question, maximum area is achieved when the fence is divided equally among lengths and widths.

So, with only one length, that means it is 120/2 = 60.
The other 60 is divided into two widths, of 30 each.

The dimensions are this 60x30 for a maximum area of 1800.