Optimization (Min-Max)

Date: 01/21/2001 at 04:54:22
From: Asim Shah
Subject: Differentiation

Dear Sir, 

This question is given in a book _Core Course for A Levels_, by 
Bostock and Chandler (chapter 5):

If a piece of string of fixed length is made to enclose a rectangle, 
show that the enclosed area is greatest when the rectangle is a 
square.

Thank you and best regards,
Asim

Date: 01/21/2001 at 10:33:01
From: Doctor Paul
Subject: Re: Differentiation

This is a nice application of optimization (sometimes called max and 
min) problems in calculus. Here we want to maximize the area. Let's 
start by writing down everything we know.  

We know that we're going to have a rectangle of some sort with fixed 
perimeter. And we know that the fixed perimeter is going to be equal 
to the sum of the four sides.  That is, 

   P = 2W + 2L

But we also know that the area of any rectangle is given by A = L*W.

That is all we need. We want to maximize the area, so we need to 
differentiate A and set it equal to zero (an application of the first 
derivative test).

The question then becomes, do we differentiate A with respect to L 
or W? It turns out that neither will help us, so what we need to do is 
use the other information we know to eliminate either L or W.  The 
choice is arbitrary. Let's pick L. So we want to solve P = 2W + 2L 
for L. We get L = (P-2W)/2.

Subsitute this value for L into A = L*W  and distributing yields:

   A = (PW - 2W^2)/2

We still have two variables (P and W), but remember that P is just a 
number. We fixed P when we began the problem. So we differentiate A 
with respect to W and just treat P like a number (because that's all 
it is).

   A' = (P-4W)/2

Now set A' = 0 and solve for W.  We get W = P/4.

In words, this says that the area is maximized when the width is 
one-fourth the perimeter.

Substitute this value for W back into P = 2W + 2L and now solve for L.  

You get L = P/4, which completes what you were asked to show.

If L and W are the same (they're both P/4), then the rectangle is 
actually a square.

Please write back if you still have trouble.

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/