A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. The total volume of the solid is 6 cubic centimeters. Find the radius of the cylinder that produces the minimum surface area.
I have found that my answer is between 0.985<x<1.241 but cann't find the exact number for 6 cubic centimeters

Let the radius of the cylinder be r and the height be h.
Volume of the cylinder is πhr^2
Volume of the two hemispheres is (4/3)πr^3
So the volume of the solid is
6 = πhr^2 + (4/3)πr^3
Solve this for h and get
h = (6/πr^2) - (4/3)r

The surface area of the two hemispheres is 4πr^2
The surface area of the side of the cylinder is 2πhr

The surface area of the solid is then
4πr^2 + 2πhr

Substitute the above expression for h to get surface area as a function of r alone

S (r) = (4/3)πr^2 + 12/r

Take the derivative

S' (r) = (8/3)πr - 12/r^2

Set this equal to zero and solve for r
and get r = (9/2π)^1/3

Some elementary calculus shows that this gives the absolute minimum for S(r) = surface area


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I can answer questions from the standard four semester Calculus sequence. I am not prepared for questions on Tensor Calculus. Everything else is welcome. Derivatives, partial derivatives, ordinary differential equations, single and multiple integrals, change of variable, vector integration (Green`s Theorem, Stokes, and Gauss) and applications.


Ph.D. in Mathematics and many years teaching Calculus at state universities.

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