you are correct, here's one (long) explanation

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Posted by T.Gracken on October 21, 2002 at 18:01:07:

In Reply to: Re: Cylinder to Sphere Ratio posted by T.Gracken on October 21, 2002 at 16:47:01:

: : IF the cylinder of largest possible volume is inscribed in a given sphere, ratio of the volume of the sphere to that of the cylinder is?

: : I know the answere is 3^(1/2):1

There is probably a nice short sweet proof, but I can't seem to see the forest through the trees... so here's one (long) attempt at an explanation.

Let r = radius of a sphere, and let V_s be the volume of the sphere.

Then V_s = (4/3)pi*r³

+++++++++++++++++++++++++++

Now, let k = radius of a right circular cylinder inscribed within a sphere of radius r where h is the height of the cylinder.

Then the volume of the cylinder, (I’ll use V_c), is V_c = pi*k²*h.

Note that the distance from the center of the cylinder (same as center of sphere) to a point on the rim of the cylinder (also on sphere) is r. Note also that the (shortest) distance from the center of cylinder to the side of the cylinder is k. Note even further, that the distance from the point on the side (just mentioned) to a point on the rim if cylinder is h/2. A diagram can help you see that the points mentioned above form a right triangle. Label the central angle of this triangle as T.

Then sin(T) = h/(2r) and cos(T)=k/r

this gives h = 2r*sin(T) and k = r*cos(T)

So, we can rewrite the volume of the (inscribed) cylinder as

V_c = pi*r²cos²(T)*2r*sin(T)

giving V_c = 2pi*r³cos²(T)sin(T)

...Maximize V_c.

differentiate with respect to T (remember that r is considered constant here).

V_c’ = 2pi*r³cos(T)*[cos²(T) - 2sin²(T)] ...this is after some simplifying/rearranging.

Keep in mind that T must be between 0 and pi/2 exclusive (refer to your diagram)...

critical values will be when V_c’ = 0.

this should lead you to cos²(T) - 2sin²(T) {since the other factor will have no critical values in the desired range}

further, this gives tan(T) = sqrt(2)/2 {remember there are range restrictions which allow us to disregard other possibilities}

This gives sin(T) = sqrt(3)/3 and cos(T) = sqrt(6)/3.
Substituting the values into our volume equation for a cylinder gives

V_c = 2pi*r³cos²(T)sin(T)

= (4/9)sqrt(3)pi*r³ as the maximum volume of an inscribed cylinder

phew... {and I even left out some of the stuff}

so, the ratio of volume of sphere to maximum volume of inscribed cylinder is

V_s to V_c

or [(4/3)pi*r³] / [(4/9)sqrt(3)pi*r³]

which yields sqrt(3) to 1 after simplifying.

{...so your given answer was correct}

sure hope you followed all that ‘cause i’m tired of typing

Follow Ups:

I knew there'd be a typo, use this post. T.Gracken 18:04:50 10/21/02 (2)
- A short way Brad Paul 20:00:04 10/21/02 (1)
  - Re: A short way T.Gracken 10:11:45 10/22/02 (0)

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: : : IF the cylinder of largest possible volume is inscribed in a given sphere, ratio of the volume of the sphere to that of the cylinder is? : : : I know the answere is 3^(1/2):1 : There is probably a nice short sweet proof, but I can't seem to see the forest through the trees... so here's one (long) attempt at an explanation. : Let r = radius of a sphere, and let Vs be the volume of the sphere. : Then Vs = (4/3)pi*r3 : +++++++++++++++++++++++++++ : Now, let k = radius of a right circular cylinder inscribed within a sphere of radius r where h is the height of the cylinder. : Then the volume of the cylinder, (I’ll use Vc), is Vc = pi*k2*h. : Note that the distance from the center of the cylinder (same as center of sphere) to a point on the rim of the cylinder (also on sphere) is r. Note also that the (shortest) distance from the center of cylinder to the side of the cylinder is k. Note even further, that the distance from the point on the side (just mentioned) to a point on the rim if cylinder is h/2. A diagram can help you see that the points mentioned above form a right triangle. Label the central angle of this triangle as T. : Then sin(T) = h/(2r) and cos(T)=k/r : this gives h = 2r*sin(T) and k = r*cos(T) : So, we can rewrite the volume of the (inscribed) cylinder as : Vc = pi*r2cos2(T)*2r*sin(T) : giving Vc = 2pi*r3cos2(T)sin(T) : ...Maximize Vc. : differentiate with respect to T (remember that r is considered constant here). : Vc’ = 2pi*r3cos(T)*[cos2(T) - 2sin2(T)] ...this is after some simplifying/rearranging. : Keep in mind that T must be between 0 and pi/2 exclusive (refer to your diagram)... : critical values will be when Vc’ = 0. : this should lead you to cos2(T) - 2sin2(T) {since the other factor will have no critical values in the desired range} : further, this gives tan(T) = sqrt(2)/2 {remember there are range restrictions which allow us to disregard other possibilities} : This gives sin(T) = sqrt(3)/3 and cos(T) = sqrt(6)/3. : Substituting the values into our volume equation for a cylinder gives : Vc = 2pi*r3cos2(T)sin(T) : = (4/9)sqrt(3)pi*r3 as the maximum volume of an inscribed cylinder : phew... {and I even left out some of the stuff} : so, the ratio of volume of sphere to maximum volume of inscribed cylinder is : Vs to Vc : or [(4/3)pi*r3] / [(4/9)sqrt(3)pi*r3] : which yields sqrt(3) to 1 after simplifying. : {...so your given answer was correct} : sure hope you followed all that ‘cause i’m tired of typing

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