# A rectangular playground is to be enclosed by a fence, and divided into three sections with fencing parallel to two sides. (Below...)If 800m of fence are used to enclose a maximum area, what are...

A rectangular playground is to be enclosed by a fence, and divided into three sections with fencing parallel to two sides. (Below...)

If 800m of fence are used to enclose a maximum area, what are the overall dimensions of the playground?

thank you very much!

### 1 Answer | Add Yours

Let x and y be the length of the rectangle.

We assume the parallel to width , the rectangle is trisected. So the required fencing is x+x+x+2y = 800 meter. So y = (800-3x)/2.

Therefore the area A(x) = x*y = x(800-3x)/2.

To setermine x = c for which A(x) = x (800-3x)/2 is maximum.

So we have to determine x = c for which A'(c) = 0 and A''(c) < 0 so that A(c) becomes maximum .

We differentiate A'(x) and solve for A,(x) = 0 to get c.

A'(x) = x(800-3x)/2 = (1/2) (800x-3x^2)' = (1/2)(800 - 6x).

Therefore A'(x) = 0 gives (800 -6x ) = 0. Or 6x = 800, or x = 800/6 = 400/3 meter.

A"(c) = (1/2) (800 -6x)' = (1/2)(-6) < 0. So A(400/3) is the maximum possible area.

Therefore the maximum area A(c) = (400/3) {800 - 400) = 400^2/3 = 53333.33 sq m.