# Max/Min problem - Find proportions of a right circular cylinder

Find the proportions of a right circular cylinder of greatest volume which can be inscribed inside a sphere of radius $R$. There's a poor image I made of what I think it looks like..

Using Pythagoras, I got this:

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

$$r^2= R^2 - (\dfrac{h}{2})^2$$

I then subbed that into the volume of a cylinder formula:

$$π(R^2 - (\dfrac{h}{2})^2)h$$ $$πR^2h - \dfrac{πh^3}{4}$$

I differentiated that with respect to $h$ and put it equal to zero to solve for $h$:

$$\dfrac{dV}{dh}=πR^2 - \dfrac{3πh^2}{4}$$ $$πR^2 - \dfrac{3πh^2}{4}=0$$ $$h=\dfrac{2R}{\sqrt{3}}$$

I then subbed that value back into Pythagoras to solve $r$:

$$r^2= R^2 - (\dfrac{\dfrac{2R}{\sqrt{3}}}{2})^2$$ $$r^2= R^2 - \dfrac{\dfrac{4R^2}{3}}{4}$$ $$r^2= R^2 - \dfrac{4R^2}{12}$$ $$r^2= R^2 - \dfrac{R^2}{3}$$ $$r^2= \dfrac{2R^2}{3}$$ $$r= \dfrac{\sqrt{2}R}{\sqrt{3}}$$

$$h:r$$ $$\dfrac{2R}{\sqrt{3}}:\dfrac{\sqrt{2}R}{\sqrt{3}}$$

Then I divided both sides by $\dfrac{\sqrt{2}R}{\sqrt{3}}$ and got this:

$$\sqrt{2}:1$$

Is this correct? it's the same answer as the back of the book, but I just wanted to make sure...I found this quite difficult...

• looks mostly fine to me! – danimal Jun 22 '15 at 16:15

Note that you could make your $h/2$ your $r$ - the radius of the cylinder and have $R$ be the radius of the circle and perhaps make what you called $r$ instead $h/2$ or $H$ to have your variables align more with what they are in the diagram. But the choice of variables is always yours!