A right circular cylindrical container with a closed top is to be constructed with a fixed surface area.

analysis/cylinder New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

A right circular cylindrical container with a closed top is to be
constructed with a fixed surface area.  Find the ratio of the height to
the radius which will maximize the volume.

-- Bill Smith <baron@voyager.net>

As a follow up to this calculus problem - you should now be able to
deduce the height to radius ratio of an open-topped cylindrical can
(for maximum volume) -- only logic and not calculus required!

-- K. <postmaster@litak.demon.co.uk>

Show solution

The volume is pi*r squared*h.  The area is 2*pi*r*h + 2*pi*r squared = K

dV/dr = 2*pi*r*h + pi*r squared*dh/dr

dA/dr = 2*pi*h + 2*pi*r*dh/dr + 4*pi*r = 0
from the last equation we get dh/dr = (-2*r - h)/r

substitute in to the dV equation and we get

0 = 2*pi*r*h + pi*r(-2*r - h)

finally we get h = 2r

These dimensions are optimum.

-- Arun Prakash <aprakash@cloudnet.com>

The logical solution is as follows:

 Take two topless cans of maximum volume        |       | |       |
                                                |  1st  | | 2nd   |
                                                |_______| |_______|

Take the 2nd one (hopefully empty) - turn it upside down and place it on
top of 1st.
         _______
        |  2nd  |       You now have a closed cylindrical can of maximum
        |       |       volume (as is made up of 2 max. volume halves) -
        |_ _ _ _|       and which we now know has a height:radius ratio
        |  1st  |       of 2:1.
        |       |       Therefore the height:radius ratio of either
        |_______|       topless(??) cylindrical can is 1:1.

-- K. <postmaster@litak.demon.co.uk>