maximizing volume of cone

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This is part of a long problem involving maximizing the size of a cone:

Basically, a circle of radius R has a sector of angle theta cut out of it. I found what the equation for the area of the sector would be: (pi/3)r^3(1-(theta/2pi))(1-(1-(theta/2pi)^2)^.5). The maximum angle is 1.840 radians.

The question now is how to maxamize the sum of the volume of the two cones created. One cone is of the sector removed, and the other is of the other sector. I thought the answer would be the original formula times 2, but this gives me the same answer.

Thoughts? Thanks!
 
amejr9999 said:
This is part of a long problem involving maximizing the size of a cone:

Basically, a circle of radius R has a sector of angle theta cut out of it. I found what the equation for the area of the sector would be: (pi/3)r^3(1-(theta/2pi))(1-(1-(theta/2pi)^2)^.5). The maximum angle is 1.840 radians.

I don't believe this is correct. The max. angle is 1.15 radians.


Here's my approach:

\(\displaystyle V=\frac{{\pi}}{3}r^{2}h\)

Since the slant of the cone is the radius of the circle, R:

\(\displaystyle r^{2}+h^{2}=R^{2}\)

Using \(\displaystyle s=r{\theta}\)

\(\displaystyle 2{\pi}r=R(2{\pi}-{\theta})\)

\(\displaystyle V(h)=\frac{{\pi}}{3}(R^{2}h-h^{3})\)

\(\displaystyle V'(h)=\frac{{\pi}}{3}(R^{2}-3h^{2})\)

\(\displaystyle V'(h)=0\) when h=\(\displaystyle \frac{R}{\sqrt{3}}\)

Subbing back in gives \(\displaystyle r=\frac{\sqrt{6}R}{3}\)

Sub this into \(\displaystyle 2{\pi}r=R(2{\pi}-{\theta})\) and solving

for theta gives \(\displaystyle \frac{{-2}(\sqrt{6}-3){\pi}}{3}=1.15\) radians



The question now is how to maxamize the sum of the volume of the two cones created. One cone is of the sector removed, and the other is of the other sector. I thought the answer would be the original formula times 2, but this gives me the same answer.

Thoughts? Thanks!
 
Thoughts without math. The angle has to be 0 or pi. As the angle increases one cone gets bigger, the other gets smaller. They are equal at pi, then the bigger gets smaller & vice versa as you go to 2 pi, so pi must be either a max or a min. At 0 or 2pi both have zero volume so pi is a max. I wanted to get this in quickly. I'll try the math if I get time.
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Gene
 
So much for thoughts without math. The darn thing has a saddle. Galactus correctly found the max for the big one but it has to be done again using \(\displaystyle {\theta}\) instead of \(\displaystyle 2{\pi}-{\theta}\) for the small one and the sum of the derivitives set = 0. Unfortunatly I didn't look at his and used r instead of h in my work so we can't mix them.
I got \(\displaystyle {\theta}\) = 2.036 for the final answer.
 
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