Circumscribed Cone about a Sphere

[Adviolin]

Find the dimensions of minimum volume of a right circular cone that
circumscribes a sphere of radius 4.

I set up a triangle. And drew an inscribed circle that is tangent to the sides
and base. Drew a diamater parallel to the base. The length of the diameter is
8. Then used the big half of the triangle and the little half of the triangle
for similar figures.
The height of the big one is h and the height of the smaller is h-4 (the radius
of the circle. r on the big is r and the radius of the smaller "cone" is 4

h/r=(h-4)/4

I put it into the equation and when setting the derivative to zero, I found
answers of 12, 4, neither of which match the book's answer?

Any suggestions?

Adam

[Bruce Coughtrey]

Your 12 is correct but the cone radius is 6.

Adviolin <advi...@aol.com> wrote in message
news:20011018140318...@mb-cu.aol.com...

[Virgil]

In article <20011018140318...@mb-cu.aol.com>,
advi...@aol.com (Adviolin) wrote:

I do not folow your construction using a "diameter parallel to the base".

I find that the line through (4,0) and (0,12) is not tangent to the
circle x^2 + (y-4)^2 = 16, which leads me do suspect your model is
incorrect.

Projected to the xy-plane, the sphere outline becomes a circle,
x^2 + (y-4)^2 = 16, and the cone outline becomes an iscoceles triangle
with horizontal base on the x-axis with its other (equal) sides tangent
to the circle.

Construct a tangent line to the circle at x = a, 0<a<4.

The height of the cone is the y on the tangent line for which x = 0.

The radius of the cone's base is the x on the tangent line for which
y = 0.

Find an expression for the volume in terms of a, differentiate to find
the minimal volume, then find height and radius.

When I did this, I found h = 16 and r = 4*sqrt(2).

[Virgil]

In article <ePFz7.33444$ng6.1...@brie.direct.ca>,
"Bruce Coughtrey" <lizb...@idirect.com> wrote:

> Your 12 is correct but the cone radius is 6.
>
>
> Adviolin <advi...@aol.com> wrote in message
> news:20011018140318...@mb-cu.aol.com...
> > Find the dimensions of minimum volume of a right circular cone that
> > circumscribes a sphere of radius 4.
> >
> I

> found
> > answers of 12, 4, neither of which match the book's answer?
> >
> >
> > Any suggestions?
> >
> > Adam

The line through (6,0) and (0,12) is not tangent to the circle
x^2 + (y-4)^2 = 4^2., so r = 6 and h = 12 cannot be correct.

[Adviolin]

>Construct a tangent line to the circle at x = a, 0<a<4.
>
>The height of the cone is the y on the tangent line for which x = 0.
>
>The radius of the cone's base is the x on the tangent line for which
>y = 0.
>
>Find an expression for the volume in terms of a, differentiate to find
>the minimal volume, then find height and radius.

how do I go about doing this? And the book does this before implicit
differentiation. I know implicit differentiation so feel to use it, but the
book may have intended a less analytic approach.

[Virgil]

In article <20011018163101...@mb-cb.aol.com>,
advi...@aol.com (Adviolin) wrote:

Since the point of tangency to the circle x^2 + (y-4)^2 = 4^2 will
always be on the upper semi-circle, y = 4 + sqrt(4^2 - x^2), one can
find dy/dx without implicit differentiation.

[Adviolin]

I took your advice, and ended up with about three pages of work where there is
probably a mistake, and am left with multiplying a trinomial by a trinomial.
You got the same answer as the book, but did you really spend an hour doing
this, or am I missing something?

I had the volume as
V=(-64*pi/3)[((Sqrt(16-a^2)-4)/a)^2(1+(4/Sqrt(16-a^2)))] and then I had to
differentiate and set to zero. Just to get to that took awhile. Please if
you have any help please tell me. And if implicit differention makes it easier
then please include it.

[Virgil]

In article <20011019175606...@mb-cu.aol.com>,
advi...@aol.com (Adviolin) wrote:

An alternate method is much easier:

The line through (r,0) and (0,h), where r is the radius of the base and
h is the altitude of the vertex of the cone, must pass at a distance 4
from the center (0,4) of a circle of radius 4, x^4 + (y-4)^2 + 4^2, in
order to be a tangent line to the circle.

The line can be given in the two intercept form as: x/r + y/h = 1, or,
more commonly, as h*x + r*y - r*h = 0.

The distance from an arbitrary point, (x,y), to such a line is given by

d(x,y) = |h*x + r*y - h*r|/sqrt(r^2 + h^2)

From [d(0,4)]^2 = 4^2, one can deduce that r^2 = 16*h/(h-8).

Then Volume = pi*r^2*h/3 = pi*16*h^2/(3*(h-8)), and the rest is easy.

[Adviolin]

I found a much easier way (well for me at least) The book used a geometric
approach in this section and the the answer is really quite elegant. I can't
believe it took so long to figure out. Drawing the flat figure, and then
dropping from the center of the circle to the base for 4, and from the center
to the left point of tangency forms two congruent triangles. Using half of
the BIG triangle I get

h^2+r^2=k^2 where k is the leg of the big triangle.

and then with the little ones, the corresponding part of the left one = the
part of the bottom one = r. With the line up the center, and the lines
already drawn a new triangle appears up at top, and it is also a rt triangle,
and therefore

4^2+(k-r)^2=(h-4)^2

and with a system of equations fairly quickly I came up with the answers of
(h,r)=(16, 4*Sqrt(2))

Did no one notice that, or is it easier to use algebra. You all seemed to
turn a geometry problem into an analytic one? Are you more accustomed to the
latter?

Thanks to all who helped.

Adam

[Virgil]

In article <20011019221317...@mb-fm.aol.com>,
advi...@aol.com (Adviolin) wrote:

I follow your geometry, as far as you explain it, but not your algebra.

You seem to have done the algebra part using 2 equations to solve for 3
unknowns, which is bloody miraculous.

Are you forgetting to mention that you are using the similarity of your
small left triangle and your large left triangle to get a third equation?

[Adviolin]

You just need an equation in terms of h and r. And then solve for one of then
and put it in

V=pi*r^2*h/3

so k=sqrt(r^2+h^2)
and 16+(sqrt(r^2+h^2)=(h-4)^2

surely from that you can solve for r or h and minimize the volume.

Adam

[Adviolin]

sorry it should be

16+(sqrt(h^2+r^2)-r)^2=(h-4)^2

[christin...@gmail.com]

what if the radius has an unknown value?

[Peter Percival]

christin...@gmail.com wrote:
> what if the radius has an unknown value?

You may need to be more explicit.

--
Sorrow in all lands, and grievous omens.
Great anger in the dragon of the hills,
And silent now the earth's green oracles
That will not speak again of innocence.
David Sutton -- Geomancies

[Ken Pledger]

In article <4352b873-71a7-427b...@googlegroups.com>,

christin...@gmail.com wrote:

> what if the radius has an unknown value?

That doesn't describe your problem properly, but I'll guess something
about it.

Three appropriate measurements are:
h = the height of the cone,
R = the radius of the cone, and
r = the radius of the sphere.

There's an algebraic connection between the three. Have you been
asked to find that? As a result, if any two of h, R, r are known then
the third one can be calculated.

HTH

Ken Pledger.