Minimizing the amount of material to make a cup

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Discussion Overview

The discussion revolves around finding the optimal dimensions for aluminum cups shaped as "V"-shaped straight circular cylinders open at the top, with the goal of minimizing the amount of material used. Participants explore various mathematical approaches and formulations related to the volume and surface area of the cups.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that the volume of a cylinder is given by the formula \(V=\pi r^2h\), while others suggest that the cups may actually be conical, leading to different interpretations of the problem.
  • A participant mentions the need to clarify the objective function for minimizing material, questioning whether it pertains to the surface area of the cups.
  • There is a discussion about the relationship between the dimensions of the cone, with references to the Pythagorean theorem and the slant height of the cone.
  • One participant suggests the objective function as the lateral surface area of a cone, \(S(r,h)=\pi r\sqrt{r^2+h^2}\), while another proposes the surface area of a cylinder open at one end, \(S(r,h)=\pi r^2+2\pi rh\).
  • Several participants express confusion regarding the shape of the cups, with conflicting interpretations of "V"-shaped versus cylindrical forms.
  • There are claims regarding the dimensions derived from the volume, with participants stating \(R = H = (V/\pi)^{1/3}\) and seeking verification of their results.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the cups are cylindrical or conical, leading to multiple competing views on the appropriate objective function and constraints for the problem. The discussion remains unresolved regarding the correct interpretation and formulation of the problem.

Contextual Notes

There are limitations in the discussion regarding the clarity of the problem statement, particularly the definition of "V"-shaped and its implications for the geometry involved. Additionally, the mathematical steps and assumptions underlying the proposed functions are not fully resolved.

leprofece
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An industrial makes a volume aluminum cups given "V" shaped straight circular cylinder open at the top. Find the dimensions that use less material.

Answer H = R = r = Cubic sqrt (3V/pi)

V = pir2h
And from here i don't know if it is right
r^2 = R2+ h2
as usual
I want to make sure
 
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Re: max and min 7

leprofece said:
An industrial makes a volume aluminum cups given "V" shaped straight circular cylinder open at the top. Find the dimensions that use less material.

Answer H = R = r = Cubic sqrt (3V/pi)

V = pir2h
And from here i don't know if it is right
r^2 = R2+ h2
as usual
I want to make sure

Yes, the volume of a cylinder is given by:

$$V=\pi r^2h$$

However, you say "V"-shaped leading me to believe the cups are cones instead. You are trying to minimize the amount of material used, to what aspect of the cups will be our objective function?
 
Re: max and min 7

to a part of it so i infer is the pitagorean one or maybe by thales I mean R/r = h-r/R
 
Re: max and min 7

leprofece said:
to a part of it so i infer is the pitagorean one

Yes, the slant height of the cone is related to the radius and height of the cone via the Pythagorean theorem. Can you state the objective function and its constraint?
 
Re: max and min 7

V = pir2h Objetive
And from here i don't know if it is right
r^2 = R2+ h2
as usual
I want to make sure Constraint so it must be
h2= r2-R2
 
Re: max and min 7

I'm confused about how a cylinder can be "V"-shaped...I think the cups are conical, and so the objective function is:

$$S(r,h)=\pi r\sqrt{r^2+h^2}$$ (the lateral surface area of a cone)

subject to:

$$V(r,h)=\frac{1}{3}\pi r^2h$$ (the volume of a cone, which is a constant here)
 
Re: max and min 7

MarkFL said:
I'm confused about how a cylinder can be "V"-shaped...I think the cups are conical, and so the objective function is:

$$S(r,h)=\pi r\sqrt{r^2+h^2}$$ (the lateral surface area of a cone)

subject to:

$$V(r,h)=\frac{1}{3}\pi r^2h$$ (the volume of a cone, which is a constant here)

An industrial makes a volume aluminum cups given "V" shaped straight circular cylinder open at the top. Find the dimensions that use less material.

oh my frind remenber i live in venezuela where people speaks spanish
so sorry it must be An industrial makes aluminum cups of a volume given "V" whose form is a straight circular cylinder shaped and open at the top. Find the dimensions that use less material.So it must be how you say
 
Re: max and min 7

Okay, when you say "V" shaped, this to me sounds like a conical cup. But hey, we will get to the bottom of it yet! (Star)

So, what we are trying to minimize is the surface of a cylinder open at one end. Hence, our objective function is:

$$S(r,h)=\pi r^2+2\pi rh$$

Subject to the constraint:

$$V=\pi r^2h\implies h=\frac{V}{\pi r^2}$$

So, substitute for $h$ using the constraint, into the objective function, and you will have the surface area $S$ as a function of one variable, $r$, and then you may then minimize the function in the usual way.
 
Re: max and min 7

I got R = H= (v/pi)1/3
 
  • #10
Re: max and min 7

leprofece said:
I got R = H= (v/pi)1/3

May you check please?
If The answer mine is right
 
  • #11
Re: max and min 7

leprofece said:
May you check please?
If The answer mine is right

I would be glad to check your work...otherwise I have to work the problem to see if your result is correct. :D
 

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