Question

Alex has 520 yards of fencing to enclose a rectangular area.

Answer 1

Perimeter =2w+2L= 520.
We can solve this by understanding that the area is maximized by a square
Therefore L=w.

p=2w+2w=520=4w
w=130

Area
 
A=wL=130(130)= 16900 square yards

Answer 2

The first thing we are going to do for this case is define variables.

We have then:

w: width

l: length

The perimeter is given by:

$2w + 2l = 520$

The area is given by:

$A = w * l$

The area as a function of a variable is:

$A (w) = w * (260-w)$

Rewriting we have:

$A (w) = -w ^ 2 + 260w$

To obtain the maximum area, we derive:

$A '(w) = -2w + 260$

We equal zero and clear the value of w:

$-2w + 260 = 0\\2w = 260$

$w = \frac{260}{2}\\w = 130$

Then, the length is given by:

$l = 260 - w\\l = 260 - 130\\l = 130$

Finally, the maximum area obtained is:

$A = w * l\\A = 130 * 130\\A = 16900$

Answer:

A retangle that maximizes the enclosed area has a length of 130 yards and a width of 130 yards.

The maxium area is 16900 square yards