# Minimizing the amount of material to make a cup

#### leprofece

##### Member
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 dont know if it is right
r^2 = R2+ h2
as usual
I want to make sure

Last edited:

#### MarkFL

Staff member
Re: max and min 7

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 dont 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:

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

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?

#### leprofece

##### Member
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

#### MarkFL

Staff member
Re: max and min 7

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?

#### leprofece

##### Member
Re: max and min 7

V = pir2h Objetive
And from here i dont know if it is right
r^2 = R2+ h2
as usual
I want to make sure Constraint so it must be
h2= r2-R2

#### MarkFL

Staff member
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:

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

subject to:

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

#### leprofece

##### Member
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:

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

subject to:

$$\displaystyle V(r,h)=\frac{1}{3}\pi r^2h$$ (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

#### MarkFL

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

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

Subject to the constraint:

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

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.

#### leprofece

##### Member
Re: max and min 7

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

#### leprofece

##### Member
Re: max and min 7

I got R = H= (v/pi)1/3
I would be glad to check your work...otherwise I have to work the problem to see if your result is correct. 