Question 21789
Let x = width of the rectangle
40-x = length of the rectangle


Area = W*L
A = x(40-x)
A = 40x - x^2
A = -x^2 + 40x


This graph is a parabola (because of the x^2) that opens downward (because of teh negative coefficient of x^2), and you need to find the HIGHEST POINT, which is the vertex of the graph.  


For a parabola {{{y = ax^2 + bx + c}}}, the vertex is always at {{{x= -b/(2a)}}}.


In this case, {{{x= -b/(2a) = -40/(-2) = 20}}}
width = x = 20 feet
length = 40-x = 20 feet


It makes sense because a rectangle of maximum area is a square.


It can also be solved by graphing methods.  Find the highest point on the graph {{{y = -x^2 + 40x}}}

{{{graph (300,300, -50, 50, -100, 500, -x^2 + 40x) }}}

As you  can see, the vertex (the highest point) of this parabola is at x= 20. 


R^2 at SCC