Biggest Cuboid in a Sphere

Date: 03/20/2003 at 07:35:21
From: Mitsy
Subject: What is the biggest cuboid which fits inside a sphere r=10cm

I have a sphere with radius of 10cm. I have to find the biggest cuboid 
that fits in that sphere.

I don't know how I can relate the width, the length and the height of 
the cuboid. I can find the length and the height, but I don't know how 
to find the width.

For my length and height of the cuboid in the sphere, I drew a cross-
sectional diagram of a cuboid in a sphere. It looks like a square in a 
circle. The diametre of the circle (the sphere) is equivalent to the 
diagonal of the square (the cuboid) in the diagram, so I could use the 
Pythagorean theorem to find the length and the height. 

The sum of the squares of length and height must not go over 400, as 
the diametre of the sphere is 20, and the square of 20 is 400.

But then, there is a problem. Even if I tried different lengths and 
heights, I would still have no idea of what sort of width the cuboid 
would have. Is there a formula that could help me to find the width of 
the cuboid, without knowing the volume, but knowing the length and the 
height? Is there a formula that relates the length, height and width 
of a cuboid, without the need to know the volume or the cuboid?

Mitsy

Date: 03/21/2003 at 09:38:51
From: Doctor Ian
Subject: Re: What is the biggest cuboid which fits inside a sphere r=
10cm

Hi Mitsy,

This is really just the same problem in three dimensions that you
already solved in two dimensions. The largest rectangle that you can
fit in a circle is a square. Similarly, the largest cuboid that you
can fit in a sphere is a cube.  

So if you have a cube with side length L, you can use the Pythagorean
theorem (or just the formula for the distance between two points, if
you've learned that) to show that the distance from the center of the
cube to any corner will be 
        ____________________________
  D = \| (L/2)^2 + (L/2)^2 + (L/2)^2

And since the cube will touch the sphere at its corners, this is just
the radius of the sphere. In this case, you know D, so you need to
solve for L.  

Does that make sense? 

So, how do you know that there isn't a bigger cuboid, one that isn't a
cube, that will fit? This is easier to see in two dimensions, but the
idea works in three dimensions, too. 

Suppose we have a square, with side length L. The area of the square 
is  

  A = L * L

and the distance D from the center to any corner is given by 
        
  D^2 = (L/2)^2 + (L/2)^2

      = 2 * (L/2)^2

But D is the radius of the smallest circle that will fit around the
square. 

Now, suppose we keep the area the same, but make one of the sides
longer.  The other side has to get shorter, right?  If we increase the
first side by a factor of k, we have to decrease the other by the same
factor. Now we have

  A = (kL) * (L/k) 

    = L * L * (k/k)

So the area is fixed. But what is the distance from the center to any
corner? It's now given by 

  D^2 = ((kL)/2)^2 + ((L/k)/2)^2
  
      = k^2(L/2)^2 + (1/k)^2(L/2)^2

      = (k^2 + (1/k)^2) * (L/2)^2

How does this compare to the old distance? Since we made one side
longer, we know that k > 1. I'll leave it as an exercise for you to
show that 

  for k > 1,  (k^2 + (1/k)^2) > 2
  
which means that as we change the square into a rectangle with the 
same area, the distance from the center to any corner gets larger, so 
the smallest surrounding circle must get larger. So for a given area, 
the rectangle with the smallest surrounding circle is a square.  

You can use the same argument (although it's a little messier) to show
that the cuboid with the smallest surrounding sphere is a cube.  Which
means that the largest cuboid for any given sphere must be a cube. 

Does this help? 

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

Date: 03/22/2003 at 14:54:49
From: Mitsy
Subject: Thank you (What is the biggest cuboid which fits inside a 
sphere r=10cm)

Doctor Ian,

Thank you SO much for helping me; it was VERY kind of you!

I've worked them out, and they all work perfectly. Your explanations 
were really clear and were very easy to understand.

I'd like to thank you again, for bearing with someone like me.
It was _really_ kind of you.

Mitsy