Hi, there--

YOUR PROBLEM:
The perimeter of a rectangle is 50 yards. What are the dimensions that will produce a maximum area of such a rectangle?

A SOLUTION:
Let x be the width of the rectangle.
Let y be the length of the rectangle.

The formula for the perimeter is P = 2x + 2y. The perimeter is 50 yards, so
2x +2y = 50

Rewrite the perimeter equation in terms of x. (Write it in "y=" form.)
2y = 50 - 2x
y = 25 - x

In other words, the length of this rectangle will always be equivalent to 25 minus its width.

Now look at the area of this rectangle. The formula for the area of a rectangle is A = x * y. 
Substitute 25 - W for L in the area formula.

A = x * (25-x)

Simplify.

A = 25x - x^2
A = -x^2 + 25x

This is a quadratic equation. Recall that the graph of a quadratic equation is a parabola. Since 
the coefficient of the W-squared term is negative (-1) the parabola opens down, and its vertex
is the maximum point (vertex) on the graph.

We can use the vertex formula find this maximum point.

The x-term of the vertex is -b/2a. In our area equation, a is the coefficient of the x-squared 
term, or -1, and b is the coefficient of the x-term, or 25.

x = -b/2a = -(25)/[(2)*(-1)] = 12.5

In the context of this problem x = 12.5 means that the width of the rectangle is 12.5 yards. 
The length y is 25-x, or 12.5. We see that the rectangle with the maximum area is a square 
with  length and width of 12.5. The area is (12.5*(12.5) = 156.25 square yards.

Hope this helps! Feel free to send a follow up email if you have questions about this 
explanation.

Mrs.Figgy
math.in.the.vortex@gmail.com