Optimization—Finding the Area of the Largest Isoceles Triangle 
I managed to solve $(a)$. Since the area of a triangle is determined by $\frac{1}{2}$ base $\times$ height, and we already know the height, we just have to solve for the base. Using Pythagorean theorem, we can deduce base as $\sqrt{36-h^2}$. Therefore, the area is $(6+h)(\sqrt{36-h^2})$. To find $h$, we will take the derivative of the area equation and set that equal to zero. It turns out $h = 3$. Plugging $h$ into the area equation gives us an area of $27\sqrt{3}$.
Can you guys help me do $(b)$? I believe for $(c)$, the largest type of triangle of maximum area is an equilateral.
Thanks.

For solving $(b)$, we can find the total height $(h_T)$ of the isosceles triangle by adding the value for $h$ found in $(a)$ to the other height $6$: 
$$h_T=6+3=9.$$
So, we can say that the total height of the triangle is $9$ units. Then we can find the base: 
$$b=\sqrt{36-3^2}=\sqrt{36-9}=\sqrt{27}=3\sqrt{3}.$$ 
We can then divide the base by $2$ to get the length of the base for half the triangle (by the $AAS$ Similar Triangles Theorem) to get the base to be $\frac{3\sqrt{3}}{2}$. We can now focus on the right half of the triangle divided by the dotted line going through the vertex point of the circle. We can tell that the dotted line through the vertex makes a right angle. Thus, we can use Trigonometry to solve for $\alpha$. So, 
$$\alpha=\tan\frac{\text{opposite}}{\text{adjacent}}=\tan\frac{\frac{3\sqrt{3}}{2}}{9}=\tan\frac{27\sqrt{3}}{2}.$$
Since we have found what $\alpha$ is, we can find the function of it with respect to $\alpha$:
$$\boxed{\alpha=\tan\left(\frac{(6+h)(\sqrt{36-h^2})}{2}\right)},$$
or as an incorporated function:
$$\frac12bh=\frac12(6\sin\alpha)(12\cos\alpha)=\boxed{36\sin\alpha\cos\alpha}.$$
And we are done with $(b)$!
For $(c)$, By the Isoperimetric Theorem, it states
Theorem: Among all triangles inscribed in a given circle, the equilateral one has the largest area. 
Therefore, the equilateral triangle has the maximum area.
