Finding the maximum surface area of an inscribed cylinder in a sphere

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Homework Help Overview

The problem involves finding the dimensions of an open right circular cylinder that maximizes surface area while being inscribed in a sphere of radius a. The original poster expresses uncertainty about the constraints related to the sphere and the cylinder's positioning.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the cylinder's dimensions and the sphere's radius, with attempts to formulate the constraint equation. There are questions about the validity of the constraint and its implications for the problem setup.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the constraint. Some guidance has been offered regarding combining the constraint into the function for differentiation, but there is no consensus on the correct formulation yet.

Contextual Notes

Participants are grappling with the correct geometric relationships and constraints that apply to the cylinder inscribed in the sphere, indicating potential misunderstandings in the setup.

Mindscrape
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The problem is to find the radius and height of the open right circular cylinder of largest surface area that can be inscribed in a sphere of radius a. What is the largest surface area?

The open cylinder's surface area will be
f(h,r) = 2 \pi r h

I am not really sure about the sphere, because I'm not really sure about the constraints that would apply. It looks like it would just be
a^2 = r^2 + h^2 + r^2 = 2r^2 +h^2, but this would only be if the cylinder is centered about the origin. But I guess that since it is a sphere, and perfectly symmetrical, then trying to squeeze the cylinder in diagonally would be the same as along the origin. Right?
 
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Mindscrape said:
... But I guess that since it is a sphere, and perfectly symmetrical, then trying to squeeze the cylinder in diagonally would be the same as along the origin. Right?

Yes, it is the same. Combine the constraint into the function and try to differentiate.
 
Huh, I got something funny:

\nabla f = \lambda \nabla g
\nabla f = (2 \pi r)i + (2 \pi h)j
\nabla g = (4r)i + (2h)j
so
2 \pi r = \lambda 4r
2 \pi h = \lambda 2h
which means
\lambda = \frac{ \pi}{2} = \pi?
 
Last edited:
No, I'm pretty sure \pi\ne \frac{\pi}{2}!
 
I think the constraint you set is wrong. It should be a^2 = r^2 + \left(\frac{h}{2}\right)^2, unless I'm missing something.
 
Ah crap, you are right. I just realized that I built the constraint wrong. But this constraint doesn't work either. The only one that would work would be a^2 = r^2 + h^2.
 
Last edited:

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