Question 330225
1.{{{P=W+L+W=2W+L=500}}}
2.{{{A=LW}}}
From eq. 1, 
{{{L=500-2W}}}
Substitute into eq. 2,
{{{A=(500-2W)W=500W-2W^2}}}
To find the maximum value of A, convert the equation to vertex form,
{{{A(W)=a(W-h)^2+k}}} where (h,k) is the vertex.
THe value of k is the maximum value for the function A(x).
{{{500W-2W^2=-2(W^2-250W)}}}
{{{A(W)=-2(W^2-250W+15625)+2(15625)}}}
{{{A(W)=-2(W-125)^2+31250}}}
So the maximum area of 31250 ft^2 occurs when W=125. 
From eq. 1,
{{{250+L=500}}}
{{{L=250}}}
The rectangle has the dimension of 125 ft wide by 250 ft long.