Posted by CJ on .
24 m of fencing are available to enclose a play area. what would the maximum area be if the fencing only went on 3 sides because the wall is used as the 4th side.
Please show you work

Math 
JJ,
l + 2w = perimeter
l + 2w = 24
l = 242w
Area = l times w
(242w)w = Area.
24w 2w^2 = Area.
If you are doing this using calculus, you would take the derivative of the area and set it equal to zero to find w. 
Math 
Reiny,
let the side parallel to the wall by y m
let the other two sides be x m each
so we have y + 2x = 24 or y = 242x
area = xy
= x(242x
= 2x^2 + 24x
let's complete the square to find the vertex of this parabola
= 2(x^2  12x + 36  36)
= 2( (x6)^2  36)
= 2(x6)^2 + 72
the vertex is (6,72)
So the maximumum area is 72 ^2 , when x = 6 and y = 12
or
by calculus
d(area) = 4x + 24 = 0 for a max area
x = 6
then y = 24  12 = 12
max area = 6(12) = 72