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### Understanding Rectangle Area and Perimeter

```Date: 11/08/2002 at 16:50:32
From: Mark
Subject: Area

Dear Dr.Math,

I am a sophmore in high school and we are reviewing area and
perimeter. My teacher gave us a problem and told us to prove it true
or false. I know that the answer is false, but I need help
understanding and explaining why it's wrong.

Here is the problem:

We have a given square and we see that if we increase the perimeter,
area increases as well in our new rectangle. Is that always true? Give
a thorough explanation and clear examples.

Given Square:
------
3ft |      |
|      |
|      |
------
3ft

P = 12ft and A = 9 sq.ft

New Rectangle:
---------
|         |
3ft |         |
|         |
---------
4ft

P = 14ft and A = 12 sq.ft

I do see that the theory that "as perimeter increases, the area as
well increases" is not always true, even though it works for the
given problem. My example is that if our new rectangle looks like
this, then our perimeter increased but our area decreased, so the
theory is wrong or at least not always right.

----------------------
1ft |                      |
----------------------
6ft

P = 14ft and A = 6ft

I know how to do the math and figure out perimeter and area, but I
don't know how to explain why it doesn't work if the one side of the
rectangle is one, and why it did work for the given problem.

Sincerely,
Mark
```

```
Date: 11/09/2002 at 17:22:49
From: Doctor Rick
Subject: Re: Area

Hi, Mark.

You are correct that a rectangle with perimeter greater than 12 (the
area of the given square) does not necessarily have an area greater
than 9 (the area of the given square). When you are told to disprove
a statement, all you need to do is to provide a counterexample - a
case that satisfies the premises (rectangle with perimeter greater
than 12) and does not fit the conclusion (area greater than 9). You
have done this, so you have a thorough explanation with a clear
example.

You don't need to explain "why" the statement is not true in any
deeper sense than "because here is an example in which it is not
true." But I understand your desire to understand it on a deeper
level.

The fact is that if you require that the perimeter of a rectangle be
THE SAME as that of the square, the area of the rectangle will ALWAYS
be LESS than that of the square (unless the rectangle is the square
itself). In other words, for a given perimeter, the rectangle that
has the largest area is a square.

Let's consider a rectangle with length L and width W. Its perimeter is
2(L+W) and its area is LW. If we require that the perimeter be some
particular value P, then we can find what the width must be when we
know the length of the rectangle:

2(L+W) = P
L + W = P/2
W = P/2 - L

Then the area will be

LW = L(P/2 - L)
A = (P/2)L - L^2

Thus the area of the rectangle is a quadratic function of the length
L. Its graph is a parabola that opens downward, and the greatest area
occurs at the vertex of the parabola. You can prove that the value of
L at the vertex is P/4, so that the width is also P/4 and the
rectangle with greatest area is a square.

me. "Why" questions can lead in lots of different directions, and
only you can tell me when we have hit on an explanation of the type
that will satisfy you. Of course, a true mathematician will never be
completely satisfied, there are always more directions to explore.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Triangles and Other Polygons
Middle School Algebra
Middle School Triangles and Other Polygons

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