What Dimensions Minimize Material in a Paper Cup Designed for 8oz?

[tjc9]

Homework Statement

company manufactures paper cups that are designed to hold 8 fluid ounces
each. The cups are in the shape of a frustum of a right circular cone (so the top and
bottom of the cup are circles, not necessarily of the same size, and the side profile is
that of a trapezoid). What are the dimensions for a paper cup that minimizes the
amount of material used?

Homework Equations

The volume of the cup would be pi/3(R^2+Rr+r^2)h
The surface area would be pi(r)^2+pi(R+r)sqrt((r-R)^2+h^2​

The Attempt at a Solution

Know that I'd need to hold volume constant at V=8 fl. oz. or 14.4375 in.^3. Know I need to minimize the surface area of the cup. I solved for h in the volume equation then substituted that into the area equation. I don't know where to go from there because if you take the partial derivatives it becomes way too complicated.

 

[bolometer]

I believe you should use the lagrange multiplier method, minimizing the function of the area using the constant volume function as a constraint. Are you familiar with such thing?

 

[tjc9]

I know sort of how to but not with that function. Also, what constraint would the volume function create

 

[Ray Vickson]

I know sort of how to but not with that function.

The problem can range from almost-impossible to nasty but do-able, depending on how you represent it. Your representation is just about impossible with your choice of variables, although it can be solved easily enough using a numerical optimization package.

I found it much better to represent the frustrum as an actual difference between two cones of radii $R$ and $r$ and of heights $H$ and $h.$ We have $R = H t$ and $r = h t$, where $t$ is the tangent of the half-angle at the apex of the cone. It is best to leave it as just $t$, rather than as $\tan( \theta/2).$

When I did all of that the Lagrange multiplier method became just barely practical.