Expert: Paul Klarreich - 3/13/2011

Question
A cone is made by cutting a sector of angle (theta) from a circle with a radius of 1 and gluing the edges together.

What are expressions for the volume of the cone?
Which angle of (theta) gives maximum volume?
What shape is the maximal cone?

Lastly,

The sector not used is part 1 is also used to make a cone
Which angle of (theta) maximises the sum of the volumes of the cones

Thanks for the help, im a bit stuck, I tried solving it using related rates of change but I don't think that was the right way to approach it.

Answer
Questioner: Alan
Category: Advanced Math
Private: No
Subject: Volume of a Cone
Question: A cone is made by cutting a sector of angle (theta) from a circle with a radius of 1 and gluing the edges together.

What are expressions for the volume of the cone?
Which angle of (theta) gives maximum volume?
What shape is the maximal cone?

Lastly,

The sector not used is part 1 is also used to make a cone
Which angle of (theta) maximises the sum of the volumes of the cones

Thanks for the help, im a bit stuck, I tried solving it using related rates of change but I don't think that was the right way to approach it.
................................
No, it wasn't.  This is a max-min problem,

(See this link for those:)

http://en.allexperts.com/q/Calculus-2063/2009/11/Maximum-minimum-problem-41.htm

not a related rates problem; see this link:

http://en.allexperts.com/q/Calculus-2063/2009/11/Related-Rates-87.htm

..................................

You will want to determine the dimensions of the cone (height and radius of base) from theta [which will be typed as 't']

Now your radius of 1 will be the SLANT HEIGHT.

The base of the cone is a circle;

its circumference is the remaining arc of the disc:

C = arc length = theta * radius = (2pi - t) * 1 = 2pi - t

[Note: t would be your radius for the cone made from the 'cutout'.]

its radius is found from:  

2 pi r = 2pi - t

So  r = (2pi - t)/2pi = 1 - t/2pi

Now you want h.  Use the Pythagorean Theorem:

SH^2 = r^2 + h^2, with SH = 1,  r = 1 - t/2pi

h^2 = 1 - (1 - t/2pi)^2

Finally, the volume is  1/3 pi r^2 h

V = 1/3 pi (1 - t/2pi)^2 sqrt(1 - (1 - t/2pi)^2)

This should get you started.

Let me know how it goes.