maximize cone volume

[Sendell]

The slant hight of a cone is 3 m. How large should the indicated angle be to maximize the cone's volume? Accompanied with the problem is a picture: A right triangle with hypotenuse 3, height h and length of the radius of the base r. The angle is formed by the intersection of the slant height and h.

Now, I suppose I have to maximize arccos(h/3). I know pi/2>h/3>0, because of the domain of arccos. The angle has to be greater than 0, and less than pi/2 because of common sense. Now, the greatest value of arccos(h/3) happens when h = 3. However, I don't believe it is possible to have a cone with slant height and height equal. Is this right?

The answer in the back of the book says the answer is arccos(1/sqrt(3)). How is this possible?

 

[galactus]

Let's try it this way:

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

\(\displaystyle \L\\r^{2}+h^{2}=3^{2}=9\)

So, \(\displaystyle \L\\r^{2}=9-h^{2}\)

\(\displaystyle \L\\V=\frac{1}{3}{\pi}(9-h^{2})h=\frac{1}{3}{\pi}(9h-h^{3})\), \(\displaystyle 0\leq{h}\leq{3}\)

\(\displaystyle \L\\\frac{dV}{dh}=\frac{1}{3}{\pi}(9-3h^{2})\)

\(\displaystyle \L\\\frac{1}{3}{\pi}(9-3h^{2})=0\)

Solving, we find that volume is max when \(\displaystyle h=\sqrt{3}\)

Now, use cosines to find that \(\displaystyle \L\\cos({\theta})=\frac{\sqrt{3}}{3}\Rightarrow{{\theta}}=\fbox{cos^{-1}(\frac{1}{\sqrt{3}})}\)

 

[Sendell]

Very smart... Thank you!