How to Find the Maximum Area of a Rectangle

One of the most famous and most important problems from introductory differential calculus is the "farmer and the fence" problem.  There are many variations of this problem, but one version or another of this problem has appeared in every Calculus 1 course I've tutored and in every Calculus 1 text I've read.

The problem is this:

A farmer wants to build a rectangular pen with one side against a river. He has 200 feet of fence at his disposal.  What are the dimensions of the rectangle that maximize the area?

The first step in this problem is to write two algebraic equations.  The first is for the perimeter of the rectangle, which is

p=2w+l

This is because one side of the pen is against the river.

The second equation is for the area of the rectangle, which is

a=w*l

The next step is to solve the perimeter equation for l.  Since we know the farmer has a total of 200 feet of fence we can use p=200.  Solving for l, the equation should read

200-2w=l

The next step is to rewrite the area equation in terms of one variable.  The revised area equation should read

a=w*(200-2w)

We then distribute, and the equation should read

a=200w-2w^2

The next step is where we use calculus to maximize the area function, and we do this by taking its derivative.  The differentiated equation should read

a'=200-4w

We now set the equation to 0 and solve for w

0=200-4w
-200=-4w
w=50

We now know the value of w that maximizes the area, so we can substitute this value directly into the perimeter equation.  This equation should read

200=(2*50)+l
200=100+l
l=100
 

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