Minimizing the amount of material to make a cup

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SUMMARY

The discussion focuses on optimizing the dimensions of a conical cup to minimize material usage while maintaining a specified volume "V". The objective function for surface area is defined as S(r,h) = πr√(r²+h²), with the volume constraint given by V(r,h) = (1/3)πr²h. Participants confirm that the optimal dimensions can be derived using calculus, leading to the conclusion that R = H = (V/π)^(1/3) minimizes material usage.

PREREQUISITES
  • Understanding of calculus, particularly optimization techniques.
  • Familiarity with geometric properties of cones and cylinders.
  • Knowledge of the Pythagorean theorem as it applies to conical shapes.
  • Basic understanding of volume and surface area formulas for geometric shapes.
NEXT STEPS
  • Study the derivation of the surface area formula for conical shapes.
  • Learn about optimization techniques in calculus, including Lagrange multipliers.
  • Explore practical applications of minimizing material usage in manufacturing.
  • Investigate the implications of geometric constraints on design in engineering contexts.
USEFUL FOR

Mathematicians, engineers, industrial designers, and anyone involved in optimizing material usage in product design will benefit from this discussion.

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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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