A rectangular lot has a perimeter of 320meters determine the maximum area of the lot

Answer 1

width --- x
length -- y
2x+2y = 320
y = 160-x

area = xy = x(160-x)
= -x^2 + 160x

this is a downwards opening parabola with a vertex at (80, 6400)
(I assume you know how to find the vertex)

the max area is 644 m^2 , when x = 80m
that is, when the rectangle is a square, which shows that for any given perimeter, the largest rectangle is obtained when it is a square.