Minimizing the surface of a sphere and cylinder

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SUMMARY

The discussion centers on optimizing the surface area of a composite body formed by a straight circular cylinder and a hemisphere, given a constant volume V. The optimal dimensions are derived as R = H = cubic root(3V/5π), indicating that the height of the cylinder must equal the radius of the hemisphere. The total surface area is expressed as A = 2πrH + 2πr², and the volume is defined as Vt = πr²H + (2/3)πR². The participants clarify the relationship between the dimensions and the volume constraint.

PREREQUISITES
  • Understanding of calculus, specifically optimization techniques.
  • Familiarity with geometric properties of cylinders and hemispheres.
  • Knowledge of implicit differentiation and volume calculations.
  • Basic algebra for manipulating equations involving π and cubic roots.
NEXT STEPS
  • Study optimization problems in calculus, focusing on surface area minimization.
  • Learn about implicit differentiation techniques and their applications.
  • Explore geometric properties of composite solids, particularly cylinders and hemispheres.
  • Investigate real-world applications of volume and surface area optimization in engineering.
USEFUL FOR

Mathematicians, engineering students, and professionals involved in design and optimization of geometric structures will benefit from this discussion.

leprofece
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279) A body is formed by a straight circular cylinder which ends up in a hemisphere. What are the dimensions that should have this body so the total surface area is minimal, if your volume is

answer Cubic sqrt( 3V/5 pi)

i tried to post an image of my notes and i couldnot i will type later

Vt = pir2H +2/3piR2 constant
A= 2pirH+ 2pir2

Derive implicit 2piH + pir(dH) + 4pir
dh = (-2r-h)/(r)

derive volume 2pirH + pir2(dH) + 2pir2

2pirH + pir2(-2r-h) + 2pir2 = 0
And I got piRh = 0

So h = 0 and this is not the answer
 
Last edited:
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Re: Max and min 6

If the image of your notes is on your hard drive and you are using Windows, you may upload it as an attachment as follows:

1.) Click the View attachment 1977button on the toolbar (you will see "Insert Image" when you hover your mouse cursor over it).

2.) Click the "From Computer" tab.

3.) Click the "Browse" button.

4.) Locate the file on your hard drive (or other available media), then double-click it to select it.

5.) Click "Upload File(s)" and the image will be uploaded and inserted inline in your post.
 

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Re: Max and min 6

MarkFL said:
If the image of your notes is on your hard drive and you are using Windows, you may upload it as an attachment as follows:

1.) Click the View attachment 1977button on the toolbar (you will see "Insert Image" when you hover your mouse cursor over it).

2.) Click the "From Computer" tab.

3.) Click the "Browse" button.

4.) Locate the file on your hard drive (or other available media), then double-click it to select it.

5.) Click "Upload File(s)" and the image will be uploaded and inserted inline in your post.

Right I going to try
NO it appears (! in red)
 
Re: Max and min 6

leprofece said:
Right I going to try
NO it appears (! in red)

What is the file's extension?
 

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Re: Max and min 6

leprofece said:
279) A body is formed by a straight circular cylinder which ends up in a hemisphere. What are the dimensions that should have this body so the total surface area is minimal, if your volume is

answer Cubic sqrt( 3V/5 pi)

i tried to post an image of my notes and i couldnot i will type later

Vt = pir2H +2/3piR2 constant
A= 2pirH+ 2pir2

Derive implicit 2piH + pir(dH) + 4pir
dh = (-2r-h)/(r)

derive volume 2pirH + pir2(dH) + 2pir2

2pirH + pir2(-2r-h) + 2pir2 = 0
And I got piRh = 0

So h = 0 and this is not the answer

From what I can make out, you are given a body made up of a cylinder inscribed within a hemisphere, and the volume of the two components added together must remain constant. Which measure are you asked to find in terms of this constant volume? The radius of the hemisphere, the radius of the cylinder, or the height of the cylinder.
 
Re: Max and min 6

OHH Maybe I forgot to write that R = H
so it is radius and height that are equals to a cubic sqrt= and it is not zero as I found out

R = H = Cubic sqrt( 3V/5 pi)
 
Re: Max and min 6

leprofece said:
OHH Maybe I forgot to write that R = H
so it is radius and height that are equals to a cubic sqrt= and it is not zero as I found out

R = H = Cubic sqrt( 3V/5 pi)

Does this mean the height of the cylinder must be equal to the radius of the hemisphere? When you post a question, I would ask you to make sure the problem is clearly stated so that we know what you are asking. I am still unclear what you are being asked to do here.
 
Re: Max and min 6

MarkFL said:
Does this mean the height of the cylinder must be equal to the radius of the hemisphere? When you post a question, I would ask you to make sure the problem is clearly stated so that we know what you are asking. I am still unclear what you are being asked to do here.

volume is "V" So it is a problem of max and minimun and i am asked the minimal area of total surface of the body that is the sum of two as you said
 
  • #10
Re: Max and min 6

volume is "V" So it is a problem of max and minimun and i am asked the minimal area of total surface of the body that is the sum of two as you said

Are you still confused ??
Do you want a copy of the spanish problem in your mail ??'
 
  • #11
Re: Max and min 6

leprofece said:
...
Are you still confused ??
Do you want a copy of the spanish problem in your mail ??'

I don't speak or read Spanish, but yes I am still unsure what you are being asked to do. You have not answered my question:

Does this mean the height of the cylinder must be equal to the radius of the hemisphere?
 
  • #12
Re: Max and min 6

Yes According to answer both of them are equals
 
Last edited:

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