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Maximum Surface Area

Date: 07/03/2003 at 09:13:59
From: John Hickman
Subject: Maximum Surface area of shape (ellipse?) within a rectangle

I have a rectangle, X long by Y wide. Within that rectangle, I wish 
to draw a shape that is no more than X across in any direction, but 
which has the *largest* surface area possible within the confines of 
my rectangle.

Specifically I'm after a solution where X = 20 inches, and Y = 12 

I think the shape might be an ellipse, but I don't know how to 
calculate its parameters. If it is an ellipse, then I know how to 
draw it if the 2 fixed points and distance of the loci from them can 
be calculated. If it isn't an ellipse, then I'm stumped.

Date: 07/03/2003 at 16:58:23
From: Doctor Douglas
Subject: Re: Maximum Surface area of shape (ellipse?) within a 

Hi John,

Thanks for writing to the Math Forum.

Can you clarify the problem a bit?  What do you mean by surface area?  
Do you mean perimeter of the confined shape? Does the confined shape 
have to be convex?

- Doctor Douglas, The Math Forum 

Date: 07/04/2003 at 06:09:18
From: John Hickman
Subject: Re: Maximum Surface area of shape (ellipse?) within a 

Thanks for the reply. By maximum surface area, I want the shape to 
occupy as much space within the confines of the rectangle as possible.

As for convex, I didn't specify that originally, but I would prefer 
the shape to be curved for its entire perimeter.
For instance, I could take a circle, X diameter (where X > Y), and 
put that over the rectangle, snip off the parts that would fall 
outside it, and that might give me a solution that would have the 
largest surface area. But it would have some flat sides. 

Apologies for my inability to specify this problem well.


Date: 07/04/2003 at 16:29:46
From: Doctor Douglas
Subject: Re: Maximum Surface area of shape (ellipse?) within a 

Hi John,

You are looking for a shape that lies within the rectangle R and
possibly touching its sides, and has a maximum area A <= R. Of course, 
the solution to this problem is R itself. But this doesn't sound like 
what you want.

Now, let's add the restriction that the shape is curved (never flat)
over its entire perimeter. This will force the shape to touch the
sides of the rectangle at isolated tangent points, rather than
over straight line segments.

Let me also modify the problem a little bit so that the mathematics
becomes a little easier - it will be relatively easy to go back to
the original problem.  Let the rectangle be centered on the origin 
(0,0) and have length 2A and width 2B. Then the top right point is at 
(x,y) = (A,B). Let's shift the point (A,B) by scaling in the x and y 
directions so that this becomes (1,1):

   f(x,y) = (x/A,y/B)

Now the problem is to work in the square S, which is defined by

   S = {-1 < x < 1, -1 < y < 1}.

Note that an ellipse that touches the original rectangle R at the
midpoints of the four sides becomes a circle with its center at the
origin and radius 1, within the square S, in this transformed picture.  
Furthermore, let me also just consider the first quadrant, as we can 
do the same treatment symmetrically in the other three quadrants.  

You've already found that a circle/ellipse is not a bad shape to start 
with: x^2 + y^2 = 1. In the first quadrant, this algebraic relation 
generates a quarter-circle, which of course has area = pi/4.  But we 
can do better:

   x^3 + y^3 = 1           [or y = (1-x^3)^(1/3)]

yields a curve that lies inside the square, but outside the circle and 
thus with more area than the circle. And x^4 + y^4 = 1 will do even 
better at surrounding more area, while touching the square only at the 
points (0,1) and (1,0). In fact, if you choose

   x^n + y^n = 1

with some very large integer n, you will get a shape that fills out
most of the square/rectangle, while still touching the border at the 
rectangle midpoints. Furthermore, this family of curves is never flat 
(except for n=1, where we get a diamond shape). For n>1, the curve 
yields a convex region. By choosing n sufficiently large, you can get 
an area that approaches the entire area of the square S (and by 
rescaling, the rectangle R).

I hope this helps.  Please write back if you have more questions
about this.

- Doctor Douglas, The Math Forum 

Date: 07/04/2003 at 18:13:49
From: John Hickman
Subject: Re: Maximum Surface area of shape (ellipse?) within a 

Thanks for the answer. I've had the aha moment now. I can do the 
scaling and increase n until it gets the shape I want.

Associated Topics:
College Conic Sections/Circles
College Triangles and Other Polygons
High School Conic Sections/Circles
High School Triangles and Other Polygons

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