# How to find maximum volume

#### mimie

##### New member

I need someone to point out my mistake. Thanks.

#### MarkFL

##### Super Moderator
Staff member
I would begin as you did, by using similarity:

$$\displaystyle \frac{h}{R}=\frac{h-r}{r}$$

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:

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

#### MarkFL

##### Super Moderator
Staff member
You appear to be trying to optimize the volume of the cone, instead of the volume of the inscribed sphere.

#### mimie

##### New member
$$\displaystyle \displaystyle r=\frac{Rh}{h+R}$$
How to get the above expression?

#### MarkFL

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

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

Which implies:

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

#### mimie

##### New member
$$\displaystyle \displaystyle \displaystyle r=\frac{Rh}{h+R}$$

I dont know to to change to $$\displaystyle \displaystyle \displaystyle r=\frac{Rh}{h+R}$$. Please guide me. Thanks

#### MarkFL

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