- Thread starter mimie
- Start date

- Joined
- Nov 24, 2012

- Messages
- 2,720

\(\displaystyle \frac{h}{R}=\frac{h-r}{r}\)

Solve for \(r\):

\(\displaystyle r=\frac{Rh}{h+R}\)

Now, the volume of the cone is:

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

Hence:

\(\displaystyle r=\frac{R\cdot\dfrac{3V}{\pi R^2}}{\dfrac{3V}{\pi R^2}+R}=\frac{3RV}{\pi R^3+3V}\)

Maximizing the volume of the sphere is equivalent to maximizing the radius of the sphere, and so we find:

\(\displaystyle r'=\frac{3V(3V-2\pi R^3)}{(\pi R^3+3V)^2}=0\)

From this, we obtain the critical value:

\(\displaystyle R=\sqrt[3]{\frac{3V}{2\pi}}\)

Now, the first derivative test easily allows us to state:

\(\displaystyle r_{\max}=r\left(\sqrt[3]{\frac{3V}{2\pi}}\right)=\sqrt[3]{\frac{4V}{9\pi}}\)

And so the maximum volume of the sphere is:

\(\displaystyle V_{S_{\max}}=\frac{4}{3}\pi r_{\max}^3=\frac{16V}{27}\)

With \(\displaystyle V=10\text{ m}^3\), we have:

\(\displaystyle V_{S_{\max}}=\frac{160}{27}\text{ m}^3\)

How to get the above expression?\(\displaystyle \displaystyle r=\frac{Rh}{h+R}\)

- Joined
- Nov 24, 2012

- Messages
- 2,720

Looking back now, it appears I bungled the use of similarity. I should have begun with:How to get the above expression?

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

Which implies:

\(\displaystyle r=\frac{h}{\sqrt{\left(\dfrac{h}{R}\right)^2+1}+1}\)

- Joined
- Nov 24, 2012

- Messages
- 2,720

That was a mistake I made. Your image shows an expression for \(r\) equivalent to the latest one I posted.I dont know to to change to \(\displaystyle \displaystyle \displaystyle r=\frac{Rh}{h+R}\). Please guide me. Thanks