The Skimpy Donut I've come across this problem on several calculus tutorials but can't find any solutions for it.  Can someone please explain how to figure these questions out?
Link to "The Skimpy Donut" problem

For question #1 I found the link below that helped me figure it out:
Volume of a Torus: the Washer Method
ANSWER for question #1:


But I can't find any help on how to solve problems 2, 3, and 4 based on the instructions.  


Item 2
The derivation of the surface area formula actually exists already on this site, here. You can also see it done in this YouTube video.
Item 3
We want to find the minimum surface area for the fixed volume of $90 \pi^2$. We know that the volume $V$ and the surface area $S$ is given by the formulas
$$V=2 \pi^2 a b^2 = 90 \pi^2 \tag {1}$$
$$S = 4 \pi^2 ab \tag {2}$$
From (1) we can find $$a = \frac {45}{b^2} \tag {3}$$
Inserting this in (2) gives $$S=\frac {180 \pi^2}{b} \tag{4}$$
We want $S$ as small as possible so $b$ must be as large as possible. But for a torus $b \le a$, otherwise it overlaps itself. So $b = a$, which using (3) gives $a = b = \sqrt[3]{45}$, and the minimum surface area is then $$S_{min} = 4 \pi^2 45^{\frac 2 3 } \approx 50.6 \pi^2$$
Item 4
Equation (4) shows there is no maximum surface area as $S \to \infty$ as $b \to 0$. 
