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.  

 A: Let's assume the formulae $V= 2\pi^2ab^2,$ $A = 4\pi^2 ab$ are known. Then, as noticed by others, $A = (2V)/b .$ Thus if we assume $V$ is fixed at some $V_0,$ we are trying to minimize $(2V_0)/b.$ That is the same as maximizing $b$ given the constraints of the situation.
One of the constraints is $0\le b\le a.$ The other one is
$$\tag 1 b= \left ( \frac{V_0}{2\pi^2a}\right)^{1/2}.$$
Looking at the $a$-$b$ plane, we want to consider the intersection in the first quadrant of the curve $b = V_0/(2\pi^2a)^{1/2}$ with the region $0\le b\le a.$ The intersection point with the largest $b$ occurs where the curve intesects the line $b=a.$ Thus we only need solve $V_0 = 2\pi^2a^3,$ which gives $a=b= (V_0/(2\pi^2))^{1/3}$ as yielding the minimum surface area.
If we take $b$ tiny and $a$ large so as to have $V_0= 2\pi^2ab^2,$ it is clear that $A = (2V_0)/b$ can be made arbitrarily large. Thus there is no maximum surface area for a fixed volume.
As for the surface area formula $A = 4\pi^2 ab,$ we can obtain this by noticing the upper half of the torus is the solid swept out by revolving the upper half of the circle $(x-a)^2 + y^2= b^2$ $360$ degrees about the $y$-axis. Solving for $y$ in the upper half circle gives
$$y = \sqrt { b^2 - (x-a)^2}.$$
The desired surface area is thus
$$2\cdot 2\pi \int_{a-b}^{a+b}x\cdot \sqrt {1+ (dy/dx)^2}\, dx.$$
This works out nicely. Ask if you have questions on it.
A: I will let you solve for the volume and surface area any way you choose. Here I'll just give the results using Pappus's centroid theorems. The solutions are
$$
V=2\pi R A=2\pi^2 Rr^2\\
S=2\pi R C=4\pi^2 Rr
$$
where $R$ is the centroid of the revolving circle, and $A=\pi r^2$ and $C=2\pi r$ are its area and circumference, respectively. Here, $R$ and $r$ correspond to $a$ and $b$ in the problem.
Then we find that we can express
$$S=\frac{2V}{r}$$
If $V$ is fixed, then the maximum surface area coincides with smallest radius. In this problem, the smallest radius is equal to $R$, i.e., the doughnut with no hole has the largest surface area. As the hole increases in size (for a fixed volume) the surface area necessarily decreases.
A: 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 $b \to 0$. 
