# Thread: Cone volume maximization problem

1. ## Cone volume maximization problem

I've done maximization problems before but I'm finding this one though because there're no numbers to work with and when I think I'm getting somewhere I get all confused. I would really apreciate some help.
Thanks...

...A sector is removed from a circular disk of radius L and subsequently the two radial edges of the remainder of the disc are joined together to form a right circular cone. Find the angle of the cut-out sector that maximizes the volume of the cone..."

2. Originally Posted by Arturo_026
I've done maximization problems before but I'm finding this one though because there're no numbers to work with and when I think I'm getting somewhere I get all confused. I would really apreciate some help.
Thanks...

...A sector is removed from a circular disk of radius L and subsequently the two radial edges of the remainder of the disc are joined together to form a right circular cone. Find the angle of the cut-out sector that maximizes the volume of the cone..."
The volume of a right cone is one third the area of the base times the height.

The radius $r$ of the base is derived from knowing the circumference.

The circumference of the base is the length or the curved boubdary of that part of the circle that the cone is formed from:

$C=2\pi L-\theta L$

( $\theta$ in radians)

The height of the cone is $\sqrt{L^2-r^2}$

CB

3. Ok, this is what i have so far...

r = L(1 - ( Θ/2pi)) and let x = 1 - ( Θ/2pi) ... to make things easier.
then...

V = (1/3)pi * [L^2* x^2] * sqrt[L^2 - L^2 * x^2]

then I know i have to differentiate but I'm having trouble doing that?
any ideas?

4. Originally Posted by Arturo_026
Ok, this is what i have so far...

r = L(1 - ( Θ/2pi)) and let x = 1 - ( Θ/2pi) ... to make things easier.
then...

V = (1/3)pi * [L^2* x^2] * sqrt[L^2 - L^2 * x^2]

then I know i have to differentiate but I'm having trouble doing that?
any ideas?
Product rule:

$\frac{d}{dx} x^2 \sqrt{1-x^2}=2x\sqrt{1-x^2}-\frac{x^3}{\sqrt{1-x^2}}$

CB

5. Ok, this is what I do next:
from setting f'(x)=0 i get...

(3x^2) - 2x = 0
x(3x^2 - 2)= 0