a right circular cone is inscribed in a sphere of radius 15 cm. find the dimensions of the cone that has the maximum volume.

i just cant seem to figure this one out! please help!

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- Jun 10th 2010, 08:00 PMemmiexoxMaximum volume of cone in sphere.
a right circular cone is inscribed in a sphere of radius 15 cm. find the dimensions of the cone that has the maximum volume.

i just cant seem to figure this one out! please help! - Jun 10th 2010, 10:51 PMmr fantastic
- Jun 16th 2010, 11:11 AMmfetch22
Remember that:

Volume of a sphere:

$\displaystyle V = \frac{4}{3} \pi r^3$

Volume of a cone:

$\displaystyle V = \frac{1}{3} \pi r^2h$

I know this has to deal with finding the relationship between how far up you are on a sphere vertically, and the radius of a circle of a horizontal slice at that vertical point on the sphere. If I can find that, I know how to optimize. And I think I just figured it out. Consider the base of the cone, it makes a circle within a sphere. Now consider looking at the sphere a from the side so the cone is straight up and down. Now draw x and y axis in your mind. We now have a circle and a triangle. To find the hieght realation to the base radius of the cone we can model the circle as the equation:

$\displaystyle r^2 = x^2 + y^2$

We want to know the hieght can be negitive or positive, but that the hieght from (0, 0, 0) is certainly the y value. The radius at any give y value is the distance from (0, 0) to the circle, horizontally, which is x. In our original definition we know that the radius of the sphere is 15, and thus the vertical circle also. I will rearange the equation to represent a circle with center at (0, 15) so that we can talk about hiegth in strictly positive terms. So we now need to solve the follwoing for x:

$\displaystyle 225 = (y-15)^2 + x^2$

$\displaystyle 225-(y-15)^2 = x$

$\displaystyle x = \sqrt{15^2-(y-15)^2}$

X represents the radius of the cone, so lets call it r. Y represents the hieght of the cone so lets call it h. Pugging these constriants into the formulat for the volume of a cone gives:

$\displaystyle V = \frac{1}{3}\pi(\sqrt{15^2-(h-15)^2)(h)}$

Now I beielive this is the correct forumula that needs to be maximized, there was allot of agebraic geomerty involved though, so I could've made a mistake. Check my equations and steps to make sure. If you agree that this is the right equation to use then all you need to do is diffirentiate and find the zeros, all the normal steps of maximization. There might be a simpler way to describe this in terms of polar coordinates, but I'll leave that to a more ambitious poster. Good luck - Jun 16th 2010, 11:21 AMWarrenx
- Jun 16th 2010, 12:10 PMmfetch22
Correction, the formula is:

$\displaystyle V = \frac{1}{3}\pi\sqrt{15^2-(h-15)}(h)$ - Jun 16th 2010, 12:18 PMmfetch22
I diffrentiated this and got the following:

$\displaystyle \frac{dV}{dh} = \frac{\pi(30-2h)}{6\sqrt{30h-h^2}}$

This function doesnt have any zeros, so either I differentiated incorrectly, or the method I laid out above is incorrect. My apologies, looks like I screwed up. Atleast I tried - Jun 16th 2010, 12:42 PMWarrenx