Find the magnitude of the vertex angle of an isosceles triangle of the given area $A$ Find the magnitude of the vertex angle $\alpha$ of an isosceles triangle with the given area $A$ such that the radius $r$ of the circle inscribed into the triangle is maximal.

My attempt:


Suppose that the base of our trangle is $2a$ then the coresponding height is $A/a$ and the length of the side is $b=\sqrt{a^2+A^2/a^2}$, and finally the semi perimeter of the triangle is 
$$p= a+\sqrt{a^2+\frac{A^2}{a^2}}$$
Now because $pr=A$ the maximal value of $r$ corresponds to the smallest value of $p$. So, we need to find the minimum of $p$ as function of $a$.
To simpify things we introduce a new variable $x=a/\sqrt{A}$ so that
$$f(x)=\frac{p}{\sqrt{A}}= x+\sqrt{x^2+\frac{1}{x^2}}$$
Clearly,
$$\eqalign{f'(x)&=1+\frac{x^2-1/x^2}{\sqrt{x^4+1}}=\frac{\sqrt{x^4+1}+x^2-1/x^2}{\sqrt{x^4+1}}\cr
&=\frac{3x^4-1}{x^4\sqrt{x^4+1}(\sqrt{x^4+1}-x^2+1/x^2)}
}$$
Thus $f(x)$ is decreasing on $(0,1/\sqrt[4]{3})$ and increasing on $(1/\sqrt[4]{3},\infty)$. Thus $f$ attains its minimum when $x=1/\sqrt[4]{3}$. The minimum corresponds to 
$$ 2a=2\sqrt{\frac{A}{\sqrt{3}}},\qquad b=\sqrt{a^2+A^2/a^2}=2a.$$
So, our triangle when $r$ is maximum is equilateral, and the desired apex is equal to $\frac{\pi}{3}$.
