# You have a 500 metre roll of fencing and a large field. You want to construct a rectangular playground area. What is the largest area?

*print*Print*list*Cite

Let the sides of the playground be x an y

Then the area :

A = x *y

But we know that the perimeter should be 500 m

==> 2x + 2y = 500

==> x + y = 250

==> y= 250 - x ..........(1)

Now substiute in A:

A = (250 - x) *x

==> A = 250x - x^2

Now we need to find the maximum area. Then we need to calculate the maximum value for A.

First we need to find the derivatives zero.

==> A' = 250- 2x = 0

==> x = 250/2 = 125 m

==> y= 250 -125 = 125 m

Then the maximum area is when the sides are 125 and 125

**THEN MAXIMUM AREA IS:**

**A = 125*125 = 15,625 m^2 **

For a rectangle: when the given perimeter is a constant, the area of the rectangle is the largest when all the sides are equal in length or the figure takes the form of a square. A square is a special form of a rectangle anyway.

Here the length of the fencing that is available is 500m.

500 divided by 4 gives 125 m. So each side of the square can be 125 m. This gives the area of the playground as 125*125 = 15625 m^2.

**So the largest area that you can enclose with the fencing of 500m is 15625 m^2.**

500 meter of fencing implies the perimeter of the rectangular play ground is 500 meter.

To determine the largest area of the playground.

We know that the rectangle with perimeter 500 m can have the 4 sides as x , 500/2-x, x and 500/2-x respectively.

Then area A = (500/2-x)x

A = (500/2)x -x^2

A = (500/4)^2 - (500/4)^2 + (500/2)x - x^2 , as (500/4)^2 added and subtracted.

A = (500/2)^2 - (500/2 - x)^2. Here (500/2-x)^2 is always > =0.

So can be maximum when (500/4 -x)^2 = 0

Therefore A = (500/4)^2 is the maximum area when x = 500/4.

So A = 15625 sq meter is the area of the play ground.