Respuesta :
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 [tex]l[/tex] ft. and width is [tex]w[/tex] 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 [tex]w[/tex] ft.
So, the total length of the fence needed [tex]=(2l+3w) ft.[/tex]
It is given in the question that 480 feet of fencing is used. That means...
[tex]2l+3w= 480\\ \\ 2l= 480-3w\\ \\ l=240-\frac{3}{2}w ...............................(1)[/tex]
Now, the area of the playground....
[tex]A= l*w\\ \\ A=(240-\frac{3}{2}w)*w\\ \\ A= 240w-\frac{3}{2}w^2[/tex]
Taking derivative on both side in respect of [tex]w[/tex] , we will get...
[tex]\frac{dA}{dw}=240-\frac{3}{2}(2w) \\ \\ \frac{dA}{dw}=240-3w[/tex]
A will be maximum when [tex]\frac{dA}{dw}=0[/tex] . That means...
[tex]240-3w=0\\ \\ 3w=240\\ \\ w= \frac{240}{3}=80[/tex]
Now plugging this [tex]w=80[/tex] into equation (1)...
[tex]l= 240-\frac{3}{2}(80) \\ \\ l=240-120=120[/tex]
So, for the greatest total area, length will be 120 ft. and width will be 80 ft.
