Let w represent the width of the fenced lot. That will be the distance between the street and the fence that is parallel to the street. The area will be given by

area = w(344 - 2w) = -2w^2 + 344w

The extreme value of aw^+bw is obtained when w is -b/(2a). For your numbers, that value is

w = -(344)/(2*(-2)) = 344/4 = 86

The area of the enclosed lot will be 86*172 ft^2 =

For future reference in working such problems, note that the width is 1/4 of the perimeter fence length. Maximum area of a

area = w(344 - 2w) = -2w^2 + 344w

The extreme value of aw^+bw is obtained when w is -b/(2a). For your numbers, that value is

w = -(344)/(2*(-2)) = 344/4 = 86

The area of the enclosed lot will be 86*172 ft^2 =

**14,792 ft^2**For future reference in working such problems, note that the width is 1/4 of the perimeter fence length. Maximum area of a

*fully enclosed*lot is obtained when the lot is square (or circular). Maximum area of a lot*enclosed on 3 sides*is obtained when the long side is parallel to the "open" side and is twice the length of the short side.