What is the radius of smallest circular disk large enough to cover every acute isosceles triangle of a given perimeter $L$? What is the radius of smallest circular disk large enough to cover every acute isosceles triangle of a given perimeter $L$?
Let $a,a,b$ are the sides of the isosceles triangle whose perimeter is constant $L$.
$\Rightarrow a+a+b=L$
How to formulate an equation involving radius of circular disk which covers every acute isosceles triangle?
 A: Let $\alpha$ be the angle at the base of each of the sides of length $a$, and $\theta$ be the other angle.
Since the perimeter is constant, we also have $$b=L-2a \tag{1}$$
Also, from the sum of angles in a triangle $$\theta=\pi-2\alpha \tag{2}$$ 
For the triangle to be acute we require that $\alpha\in(\frac{\pi}{4},\frac{\pi}{2})$. 
Now observe that for any acute triangle, the minimum radius covering circle is the circumcircle of that triangle. [Exercise: Prove that the circumcircle cannot cover all three sides if it is translated anywhere else in the plane]. 
Applying the Sine Law is therefore appropriate:
$$\frac{a}{\sin\alpha}=\frac{b}{\sin\theta}=2R$$
Using the results of (1) and (2) this reduces to
$$\frac{a}{\sin\alpha}=\frac{L-2a}{\sin(\pi-2\alpha)}=2R$$
From the first equality and the fact that $\sin\alpha\neq0$:
$$\frac{a}{\sin\alpha}=\frac{L-2a}{\sin(2\alpha)}=\frac{L-2a}{2\sin\alpha\,\cos\alpha}\implies 2a\cos\alpha=L-2a \implies a=\frac{L}{2(1+\cos\alpha)}$$
So then
$$R=\frac{L}{4\sin\alpha\,(1+\cos\alpha)} \tag{3}$$
Define the function $f:\mathbb{R}\to\mathbb{R}$ as
$$f(\alpha)=\sin\alpha\,(1+\cos\alpha)$$
We need to find the minimum of $f$ on $(\frac{\pi}{4},\frac{\pi}{2})$ so that $R$ is maximised (the worst case).
Taking the derivative 
$$\begin{align}
f'(\alpha)&=\cos\alpha\,(1+\cos\alpha)-\sin^2\alpha \\
          &=\cos\alpha+\cos^2\alpha-(1-\cos^2\alpha) \\
          &=2\cos^2\alpha+\cos\alpha-1 \\
          &=(2\cos\alpha-1)(\cos\alpha+1) \\
\end{align}$$
So
$$f'(\alpha)=0\text{ and }\alpha\in(\frac{\pi}{4},\frac{\pi}{2}) \implies \alpha=\frac{\pi}{3}\text{ only}$$
Note that $$f''(\alpha)=-\sin\alpha(1+4\cos\alpha)<0\text{ for }\alpha=\frac{\pi}{3}$$
So, $\alpha=\frac{\pi}{3}$ gives a local maximum of $f(\frac{\pi}{3})=\frac{3\sqrt3}{4}$. Interpretation: when the triangle is equilateral, it is most easily covered by a circle.
At the endpoints, $f(\frac{\pi}{4})=\frac{1+\sqrt2}{2}$ and $f(\frac{\pi}{2})=1$.
So, $\alpha=\frac{\pi}{2}$ gives the global minimum of $1$.
Therefore, by (3) the maximum radius is:
$$R_{max}=\frac{L}{4\cdot 1}=\frac{L}{4}$$
which is the radius of the circle that will cover all acute isosceles triangles of perimeter $L$.
