Expert: Paul Klarreich - 7/12/2007

Question
This is a optimization problem. A conical paper cup usually made so that the depth is square root of two times radius of rim. Show that this design requires the least amount of paper per unit.

Answer
Questioner:   jose
Category:  Calculus

Question:  This is a optimization problem. A conical paper cup is usually made so that the depth is square root of two times radius of rim. Show that this design requires the least amount of paper per unit.
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Hi, Jose,

To do this, you have to optimize something subject to some constraint.  In this case, you are optimizing the surface area of the cone with a given volume.  To 'model' it, meaning to get some equations or functions that describe it, you need some variables and constants.

------ The area of the cone --------

Take your cone and cut along a straight line from the rim to the vertex.  Now flatten out the paper.  You get something called a SECTOR, which is like a slice of a pie. (Not a pi, a pie, as in apple pie.)

The area of this sector is given by the formula:

Sector Area = (1/2) theta * R^2, where

R = the radius of the sector, AND IS THE SLANT HEIGHT OF THE ORIGINAL CONE.

theta is the central angle of the sector, and I will write @ for it.

(Note that if @ = 2 pi, you have the whole circle and the area is pi R^2.)

And note that the ARC LENGTH of the sector is given by

Arc Length = @ * R

(Note that if @ = 2 pi, you have the whole circumference equal to 2 pi R.)

********** You want to minimize the area. **************

So look at this little diagram.  

NOTE: USE COURIER FONT TO VIEW THIS.
Here is your cone, with the 'rim' at the bottom. (WARNING -- Invert this cone before adding the ice cream.)

      /|\
     / | \
    /  |  \
   /  h|   \ <--  slant height, H = R from above.
  /    |    \
 /     |     \
/      |      \
-----------------
| <-r-> | <-r-> |

So we shall assume that the volume is fixed, say  V = pi/3.  (pi/3 what?  Cubic feet? Quarts?  Gallons?  Liters?  Who cares?  pi/3 is a convenient number to use here.)

For a cone, we have a volume formula:  V = (1/3) pi r^2 h, where  r and h are as in the diagram.  Using our fixed volume:

pi/3 = (1/3) pi r^2 h, so   r^2 h = 1.  << OUR CONSTRAINT.

And since we have a right triangle:

h^2 + r^2 = R^2    << Second constraint.

And a final observation before we get to work:

The arc length of the sector is the same as the circumference of the rim:

@R = 2 pi r

------------- Minimizing the area --------

Now we are about ready to minimize our area:

A = @ R^2

In this, @ and R are both variables.  We need those constraints to reduce this A to a function of one variable:

A = @ R R
A = (@ R) R
A = (2 pi r) R

Now a sneaky max-min trick:  If you have a problem that might involve radicals, which everyone hates, you can eliminate them this way:

Instead of minimizing A, let's minimize A^2:

A^2 = (2 pi r)^2 R^2

A^2 = (4 pi^2 r^2)(h^2 + r^2)

But r^2 = 1/h, from OUR CONSTRAINT.

A^2 = (4 pi^2)(1/h)(h^2 + 1/h)

A^2 = (4 pi^2)(h + 1/h^2)

OKAY, we are in business.  Differentiate A^2 with respect to h:

D(A^2) = 4pi^2 (1 - 2/h^3)

Almost there.  Set that equal to zero and solve: (We can ignore the 4pi^2 factor.)

1 - 2/h^3 = 0
h^3 = 2
h = 2^1/3

Now how about r?  Remember r^2 = 1/h?
        1
r^2 = -------- = 2^-1/3
      2^1/3

So  r = 2^-1/6

Now how about your problem:

"the depth is square root of two times radius of rim"

The depth is h and the radius of the rim is r.

h = square root of two times r

2^1/3 = 2^1/2 times 2^-1/6

2^1/3 = 2^1/2 2^-1/6

2^1/3 = 2^[1/2 - 1/6]

2^1/3 = 2^[1/3]

YESSSSSSSSSSSSSSS!