College Algebra

\(\mbox{A rectangular area has perimeter }2(L+W) \mbox{ and area }L W\\ \mbox{so }2(L+W)=2000\\ L+W=1000\\ L=1000-W\)

\(A=LW = (1000-W)W = 1000W-W^2\)

\(A=-(W^2-1000W) = -((W-500)^2-250000)=250000-(W-500)^2\)

\(\mbox{It should be clear that the max area is thus }250000 sq. yds\\ \mbox{and this occurs at }W=500 yds, L=500yds\)

Thanks, Rom for that good answer.....!!!!

I might add that, for any given perimeter, P, the area is always maximized when the side of the rectangle [square, actually ]   = P / 4 ........to see why this is, note:

P = 2 ( L + W)  →  P/2 = L + W  → P/2 - L  = W

And

Area  = L * W  →  L *  [ P/2 - L ]   →  PL/2 - L^2

And taking the derivative of the area with respect to L  [remember, P is a constant] , we have

A '  =  P / 2 - 2L

And setting this to 0, we have

P / 2  = 2L

P /4  = L

And the second derivative = -2....so the area is  maximized   when L =  P / 4

And the width, W  = [ P / 2  -  L]  = [  P/2 - P/4  ]  =  P/4

So.....the area is maximized when  L = W  = P/4

So......a square with a side of P / 4   maximizes the area for any given perimeter, P

.....as a side note,  if he wanted to enclose the MAXIMUM area with the fencing he has, he should make a CIRCLE with the fence . 

pi x d = 2000

d = 2000/pi = 636.62

pi r^2 = area = pi (636.62/2)^2 =~ 318309 sq yds