Mathematics: Arectangular 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

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Arectangular playground is to be fenced off and divided

doll1234

12.03.2022

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 ft. and width is 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 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 , 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.

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