Question

A rectangular playground is to be fenced off and divided into two parts by a fence parallel to one side of the playground. 480 feet of fencing is used. find the dimensions of the playground that will enclose the greatest total area

Answer 1

For the greatest total area, length will be 120 ft. and width will be 80 ft.

Explanation

Lets assume, the length of the playground is $l$ ft. and width is $w$ ft.

Suppose, the playground is divided into two parts by a fence parallel to width. That means the length of the divider fence will be $w$ ft.

So, the total length of the fence needed $=(2l+3w) ft.$

It is given in the question that 480 feet of fencing is used. That means...

$2l+3w= 480\\ \\ 2l= 480-3w\\ \\ l=240-\frac{3}{2}w ...............................(1)$

Now, the area of the playground....

$A= l*w\\ \\ A=(240-\frac{3}{2}w)*w\\ \\ A= 240w-\frac{3}{2}w^2$

Taking derivative on both side in respect of $w$ , we will get...

$\frac{dA}{dw}=240-\frac{3}{2}(2w) \\ \\ \frac{dA}{dw}=240-3w$

A will be maximum when $\frac{dA}{dw}=0$ . That means...

$240-3w=0\\ \\ 3w=240\\ \\ w= \frac{240}{3}=80$

Now plugging this $w=80$ into equation (1)...

$l= 240-\frac{3}{2}(80) \\ \\ l=240-120=120$

So, for the greatest total area, length will be 120 ft. and width will be 80 ft.