Max Volume Cylinder in Sphere

[p***@gmail.com]

Hi All

I've come about another question which I'm curious about. This is how
it reads:-

A circular cylinder is cut out of a sphere with radius 10cm. Calculate
the maximum possible volume for the cylinder.

So I looked on the net for a solution and found the following page:-

http://library.thinkquest.org/C0110248/calculus/changemax.htm

The only difference with the question on the above page and the one in
the book, is that the one in the above page specifies that the height
of the cylinder is 4x cm, while the one in the book does not. However,
the results on the above page agree with the answer in the text book,
so presumably the methods are valid.

What I couldn't figure out was why the height of the sphere should be
taken as 4x cm?

I would be most greatful for any advice.

Regards,

Paul

[Stuart Williams]

Post by p***@gmail.com
Hi All
I've come about another question which I'm curious about. This is how
it reads:-
A circular cylinder is cut out of a sphere with radius 10cm. Calculate
the maximum possible volume for the cylinder.
So I looked on the net for a solution and found the following page:-
http://library.thinkquest.org/C0110248/calculus/changemax.htm
The only difference with the question on the above page and the one in
the book, is that the one in the above page specifies that the height
of the cylinder is 4x cm, while the one in the book does not. However,
the results on the above page agree with the answer in the text book,
so presumably the methods are valid.
What I couldn't figure out was why the height of the sphere should be
taken as 4x cm?
I would be most greatful for any advice.
Regards,
Paul

I didn't make any assumption about the dimensions of the cylinder, but
worked on the angle between the diagonals of a rectangle representing a
cross-section through the widest part of the cylinder. (So half of each
diagonal is equal to the radius of the sphere.)

I think the instruction to take the height to be 4x is just intended to
be helpful, providing you with a starting point and an easy bit of
manipulation.

Stuart Williams

[p***@gmail.com]

I see what you mean. So:-

V=pi(2sqrt(25-h^2))^2h

because by pythagoras

radius = sqrt(10^2-(h/2)^2)
which = 2sqrt(25-h^2)

Thanks again!

[p***@gmail.com]

Post by p***@gmail.com
I see what you mean. So:-
V=pi(2sqrt(25-h^2))^2h
because by pythagoras
radius = sqrt(10^2-(h/2)^2)
which = 2sqrt(25-h^2)
Thanks again!

Let me correct myself, it should be:-

radius = sqrt(10^2-(h/2)^2)
which = sqrt(100-h^2/4)
which = sqrt(400/4-h^2/4)
which = 1/2sqrt(100-h^2)