Find the maximum dimensions that will strengthen rod I did a quiz and we got the following question:

A cylindrical rod has a diameter of 2cm and on the cylinder is an isosceles triangle. The strength of the rod is proportional to the product of the base and the height squared. What are the maximum dimensions that will strengthen the cylinder.

We also got the following image:

I would really like to know the solution to this problem because no one in my class got the problem or had a thorough understanding of how to go about solving the problem. 
Thanks a lot!
 A: Consider a circle of radius $R$ with an isosceles triangle inscribed, with base $b$ and height $h$.  Draw lines from the center of the circle to the endpoints of the base to make another isosceles triangle.  Draw the altitude of this triangle to the base to produce a right triangle having legs of $h-r$ and $b/2$, and hypotenuse $R$.  Then
$$(h-R)^2 + \frac{b^2}{4} = R^2$$
from which you may deduce that
$$b=2 \sqrt{2 R h -h^2}$$
Your merit function is then
$$f(h) = b h^2 = 2 h^2 \sqrt{2 R h -h^2}$$
Of course, take $f'(h)=0$ and solve for $h$.  You will get a linear equation with one solution.  Proving that this solution is a max means evaluating $f''(h)$, which is a bit messy.
I will leave the mess to you.  I get for the max strength:
$$f_{max} = \frac{50 \sqrt{5}}{27} R^3$$
A: If we denote height by $x$, then from Pythagorean theorem we obtain a formula for base
$$y(x)=2\sqrt{r^2-(x-r)^2}=2\sqrt{2xr-x^2}$$
where $r$ is the radius of cylinder. Thus the required strength value will be
$$f(x)=x^2*y(x)=2x^2\sqrt{2xr-x^2}$$
To maximise this value you'll need to differentiate it and look at function values in critical points (where derivative becomes 0 or is not defied). Also, take into account that $0<x<2r$.
A: The radius of the cross section of the cylinder is equal to 1 cm.
If we draw three lines connecting the vertices of the triangle with the centre, we get one additional isosceles triangle with two sides 1 cm, and one side being the base of the bigger isosceles triangle.
Let the base be side AB, the top vertex be C and the centre of the circle be O.
$\angle{AOB} = 2\angle{ACB}$
Let $\angle{AOB} = \theta$
Then, $AB = 2\times r\times\sin\dfrac{\theta}{2} = 2\sin\dfrac{\theta}{2}$
The height (say h) of the triangle becomes:
h = 'Base to centre' + r
$H = r\cos\dfrac{\theta}{2} + r$
You want to maximise $bh^2$, so, you need to maximise:
$$\left(2\sin\dfrac{\theta}{2}\right)\left(\cos\dfrac{\theta}{2} + 1\right)^2$$
