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
