Question 629991
The perimeter, P = 2(l+w)
Expressing the length in terms of the width, we have
l = P/2 - w
The area, A = w*l = w(P/2 - w) = (P/2)w - w^2
The area is maximized when dA/dw = 0:
dA/dw = 0 = P/2 - 2w -> w = P/4, which means the area is maximized when l=w, ie. a square
Inserting the value for P gives w = 440/4 = 110 yd
A = 110*110 = 12,100 sq. yd.