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

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SUMMARY

The discussion focuses on determining the maximum surface area of an open right circular cylinder inscribed in a sphere of radius 'a'. The surface area function is defined as f(h,r) = 2πrh. Participants clarify the constraints for the cylinder's dimensions, confirming that the relationship a² = r² + h² must hold true. The conversation emphasizes the importance of correctly applying the constraint when differentiating the function to find optimal dimensions.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and optimization techniques.
  • Familiarity with geometric relationships involving spheres and cylinders.
  • Knowledge of Lagrange multipliers for constraint optimization.
  • Basic proficiency in mathematical notation and functions.
NEXT STEPS
  • Study the method of Lagrange multipliers for constrained optimization problems.
  • Explore geometric properties of inscribed shapes, particularly in spheres.
  • Learn about surface area calculations for various geometric figures.
  • Investigate applications of optimization in real-world engineering problems.
USEFUL FOR

Mathematicians, engineering students, and anyone interested in optimization problems involving geometric shapes will benefit from this discussion.

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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