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 $r$:
$$r^2= R^2 - (\dfrac{\dfrac{2R}{\sqrt{3}}}{2})^2$$
$$r^2= R^2 - \dfrac{\dfrac{4R^2}{3}}{4}$$
$$r^2= R^2 - \dfrac{4R^2}{12}$$
$$r^2= R^2 - \dfrac{R^2}{3}$$
$$r^2= \dfrac{2R^2}{3}$$
$$r= \dfrac{\sqrt{2}R}{\sqrt{3}}$$
Then I wasn't sure about this bit...
$$h:r$$
$$\dfrac{2R}{\sqrt{3}}:\dfrac{\sqrt{2}R}{\sqrt{3}}$$
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...

This looks good to me! 
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!
The important thing is that the work is correct. Well done!
