Question 421354
A dog owner has 250ft of fencing to enclose a rectangular run area for his dog. If he wants the maximun possible ares, what should the length and width of the rectangle be? 
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The max area for a rectangle is a square.
The perimeter = 250 ft = 2L + 2W
--> L + W = 125
W = 125 - L
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Area = L*W = L*(125 - L)
{{{f(L) = 125L - L^2}}} is a parabola.  The max is at the vertex
{{{f(L) = - L^2 + 125L}}}
The line of symmetry is L = -b/2a
L = 125/2 = 62.5
The vertex is on the line of symmetry at f(-b/2a) = f(62.5)
That proves that 62.5 ft gives the max area for the given perimeter.
Max area = 62.5^2 = 3906.25 sq ft