Expert: Scotto - 3/18/2011

Question
Hi,
I am not sure how to start on this project. My professor mentioned we can use Maple or Mathlab. Can you please help me get started with the project?  Thanks!

The following is the problem:

You are the mathematics consultant for a donut company which makes donuts which
have a thin layer of chocolate icing covering the entire donut. One day you decide to
point out that the company might cut costs on chocolate icing if they keep the volume
(and hence weight) of the donut fixed but adjust the shape of the donut to minimize the
surface area. Alternatively, they could advertise extra icing by maximizing the surface
area. You need to write a report presenting your ideas which can be read by both the
company president and the technical engineers.

A donut has the shape of a torus which is specified by giving a big radius a from the
center of the hole to the center of the ring and a small radius b which is the radius of the
ring, as shown in the figure.

Your job is to determine the values of a and b which extremize the surface area while
keeping the volume fixed at the volume of the typical donut mentioned above. This
original donut has a = 5 cm and b = 3 cm.

Answer
The surface area is (2*pi*R)(2*pi*r) where R is the radius of the circle that is on the outside of the torus and r is the radius a circle that is in the torus.  Thus, if we had a torus that was made from a 2" diamter peace of plastic pipe and had a outside radius of 11", the area would be (2*pi*2)(2*pi*11) = 44pi² in.².

The volume of a torus is 2pi²Rr².  Since, in this example, R = 11" and r = 2", the volume is
88pi² in^3.

For the torus here, it seems like (witn no picture) that R = 5" and r = (5-3)/2 = 1".
The volume is then (2pi)(10pi) = 20pi² and the surface area is 10*pi², so the ratio is 2:1.

To maximize the volue with respect to the surface area, make the inside 0, so r would be given
by r = R/4.  Since the volume is (2*pi*r)(2*pi*R), that is the same as (2*pi*r)(2*pi*4r).
It works out to 16*pi²*r².  For this to be 88pi², we need r² = 5.5, so r = √5.5 = √22 / 2.

The ratio of surface area to volume is (2*pi*R)(2*pi*r)/(2pi²Rr²) = (4Rr)/(2Rr²) = 2/r.
Thus, the smaller we make r, the bigger the volume gets.  This means you would want to make the  donut with a small r, generating a big R.  Physically, however, R could only be so large.