Ice Cream Cone Challenges (New)

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Suppose you wanted to make an ice cream cone that would hold as much ice cream as possible (do not assume ice cream comes in spheres).

Challenge I
Cut a wedge from a circle and remove it. From the remaining piece of the circle into a cone. Find the angle of the wedge that produces the cone with the greatest volume.

Challenge II
You can make a second cone from the removed wedge. Find a formula for the volume of this second cone in terms of theta, the angle of the wedge. Find the angle of the wedge that produces the maximum total value of the two cones.
 
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Nice challenges, so how would you begin?
 
Let the radius of the circle be R, and the angle, after removing the wedge, be t radians.
Let the radius of the cone be r.
Let the height of the cone be h.
Let the volume of the cone be V.

2(pi)r = 2(pi)R * (t / [2(pi)])
r^2 + h^2 = R^2
V = (1/3)(pi)(r^2)hSolve each equation for a single variable, and substitute accordingly, into the equation for Volume, and then derive the Volume equation. Then, take the derivative of that equation and set it to zero.

For Challenge Two:
Let the radius of the circle be R, and the angle of the wedge be s radians.
Let the radius of the cone be r.
Let the height of the cone be h.
Let the volume of the cone be V.

Solve for variables s and R, plug into the Volume equation, and solve, finding the wedge angle that produces the maximum total volume of the two cones.
 
Well I haven't verified the work or anything, but was this an actual challenge to us? I thought you were trying to get us to do your homework. Sorry!
 
I'm not. This was a challenge posed to me, by my calc teacher. I need some help, and decided to come here. Please understand that this isn't homework, it's not even extra credit, it's just a challenging problem.

I could really use some help, if you, or anyone, could offer it.
 
AlexGmu said:
Let the radius of the circle be R, and the angle, after removing the wedge, be t radians.
Let the radius of the cone be r.
Let the height of the cone be h.
Let the volume of the cone be V.

2(pi)r = 2(pi)R * (t / [2(pi)])

I don't know if you already had to answer these or whether they were given to you to work on over holiday break. I would ask whether you understand what each of these equations represents and also to ask you to start with a picture of how the cones are constructed from these pieces of the original circular disk.

I ask this because, for the first problem, I already see a difficulty with this first equation. You will be interested in the circumferences of the circular bases of the two cones in order to find the radii of their bases. However, I believe what you have here,

2 \pi r = 2 \pi R · (t /2\pi) = tR ,

gives the base circumference for the cone in the second problem. What will be the circumference of the base then for the first cone?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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