HELP NEEDED!!!!!

[lenjoy28]

HELP!!!
If you have a circle with a sector cut out of it (using angle thedda as the cut out of the circle), and you wanted to create a cone out of the remaning sector, how you find the angle thedda that would create the maximum volume in the cone?? *Assume radius of circle equals 1. :?: :?: :?: [/tex][/code]

 

[stapel]

The Greek letter is spelled "theta", and is pronounced "THAY-tuh".

If the sector has a central angle of \(\displaystyle \theta\), then what is the arc-length that has been cut out of the circle's circumference? So what then will be the circumference of the base of the resulting cone?

If the radius of the circle is r = 1, then what is the slant-height of the resulting cone? How might this relate to the radius and height of the resulting cone?

Please reply showing what you can come up with. Thank you.

Eliz.

 

[lenjoy28]

The whole slant/height and radius of the cone is what I need help with!!!!!!!

 

[stapel]

Have you drawn a picture of the situation? The slant height should be immediately obvious. If you can't picture this, then I would suggest that you grab a piece of paper and a scissors, and do a quick cut-out to see the relationship.

Eliz.

 

[jacket81]

problem

ignore

 

[galactus]

This is an interesting little max/min problem.

As we know, the volume of a cone is \(\displaystyle V=\frac{{\pi}}{3}r^{2}h\)

Letting R=the slant height of the cone.

Since the slant height of the cone is the radius of the circle we have:

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

\(\displaystyle r^{2}+h^{2}=1\)

Now, remembering that \(\displaystyle s=r\theta\) and relating the cone to the circle we

have:

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

\(\displaystyle 2{\pi}r=1(2{\pi}-\theta)\)

Express volume as a function of h:

\(\displaystyle r^{2}+h^{2}=1\)

\(\displaystyle r^{2}=1-h^{2}\)

Sub into volume equation:

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

\(\displaystyle \frac{dV}{dh}=\frac{{\pi}}{3}(1-3h^{2})\)

Setting to 0 and solving for h, we get \(\displaystyle h=\frac{1}{\sqrt{3}}\)

Subbing back in above we find that \(\displaystyle r=\frac{\sqrt{6}}{3}\)

Sub into \(\displaystyle (2{\pi})\frac{\sqrt{6}}{3}=2{\pi}-\theta\)

\(\displaystyle \theta=1.15298598653\) radians or \(\displaystyle 66.0612308659\) degrees.

You will find, no matter the radius, the value of theta to max out volume will be the same.