How can I lose Square Footage and still have the same perimeter?

2(15 + 15) = 2(10 + 20)

A square contains more area than a rectangle having the same perimeter.

it's basic math.  the round 2 dimensional shape always has the greatest area for any given perimeter.  With rectangles, this translates into the nearer a shape is to a square [nearer to a circle], the greater the area for any given amount of fencing [perimeter]

This is a very common math problem. The point of it is to get you to see that the closer the shape is to a square, the larger the area. 

l * m = a

l * 1=l=a

1 * m=m=a

if l=m

l * m=m * m=m^2=a

(l+m)*2=60  in this case

you can see that value of a max when l=m

Different ways to describe it, but let me try this....

Picture a 15 x 15 square as 15 rows with 15 little squares each.

Picture a 10 x 20 rectangle as 10 rows with 20 little squares each.

Now, if you wanted to turn your 15 x 15 square into a 10 x 20 rectangle....

1.  The first thing you'd have to do is get rid of the bottom 5 rows (going from 15 rows to 10 rows).  You pick up all of the little squares in those rows.  That means you have picked up 5 x 15 = 75 squares.

2.  The next thing you would have to do is make the rows longer (from 15 to 20 squares).  So you add 5 squares to the 10 rows (remember, the bottom 5 rows are gone.). You would have to put down 5 x 10 = 50 squares.

However, if you picked up 75 squares and only put down 50, you have 25 left over.  So the areas aren't equal.

As square gives the maximum area for a rectangle of a given perimeter.

Mathematically let s = side of the square and c = the change in length of the side

the area is give by

a = l * w

a = (s + c)(s - c)

a = s² - c² 

therefor any non-zero value for c will reduce the area

If you changed the dimensions to 29 by 1, which is still has a 60 ft perimeter,

the area would be only 29ft²