Question 151799
Let's start with the formulas for the area and the perimeter of a rectangle.
{{{A = L*W}}}
{{{P = 2(L+W)}}}
The perimeter is given as 250 feet, so we can write:
{{{2(L+W) = 250}}} Dividing both sides by 2, we get:
{{{L+W = 125}}} Rewrite this as:
{{{L = 125-W}}} and substitute it into the formula for the area (A = L*W),
{{{A = (125-W)*W}}} Simplifying this, we get:
{{{A - 125W-W^2}}} Now this is a quadratic equation and if you were to graph this, you see a parabola that opens downward, thus there would be a maximum value of A (also known as the vertex).
You can find the value of W at this vertex by:
{{{W = (-b)/2a}}} This comes from the standard form for a quadratic equation:{{{y = ax^2+bx+c}}}
In this problem, a = -1 and b = 125, so, making the appropriate substitutions, we get:
{{{W = (-125)/2(-1)}}}
{{{W = 62.5}}} Now this is the value of W (the width) that would make the area (A) a maximum.
The length (L) is:
{{{L = 125-W}}}
{{{L = 125-62.5}}}
{{{L = 62.5}}}
So the length and the width of the rectangle would be L = 62.5 feet and W = 62.5 feet.  In other words, the rectangle is a square.