Here's a problem I just came up with:
I considered a triangle $\triangle ABC$ such $|AB|=11$ and $|AC|=5$.
I asked myself the following question:
What must angle A be when the incircle of ABC has a maximum radius?
I performed several tests with GeoGebra which led me to determine: $A=80^\circ$. So, I'm wondering if this is correct and if the exact answer can be found through calculation?
We recall that $$|\triangle ABC| = rs,$$ where $r$ is the inradius and $s = (a+b+c)/2$ the semiperimeter. Also recall that $$|\triangle ABC| = \frac{1}{2}bc \sin \alpha,$$ where $\alpha = \angle BAC$. Thus $$r = \frac{bc \sin \alpha}{2s} = \frac{bc \sin \alpha}{b + c + \sqrt{b^2 + c^2 - 2bc \cos \alpha}}.$$ Since $c = 11$ and $b = 5$ are fixed, we have as a function of $\alpha$ $$r(\alpha) = \frac{55 \sin \alpha}{16 + \sqrt{146 - 110 \cos \alpha}}.$$ All that remains is to compute a derivative and locate the critical point; however, this is computationally tedious and leads to the cubic $$\cos^3 \alpha + 3 \cos^2 \alpha - \frac{347}{55} \cos \alpha + 1 = 0,$$ with numeric solution (in radians) $$\alpha \approx 1.3962369082289012941855528838 \ldots,$$ which in degrees is approximately $$\alpha^\circ \approx 79.99848204191088333239913491 \ldots.$$ The value of the inradius for this angle is attained at the unique positive real root of $$r^6 + 677r^4 + 4211r^2 - 27225 = 0,$$ with approximate value $$r \approx 1.9865967673502241946 \ldots.$$
I arrive at the same conclusion as @heropup, with a different path (I use trigonometry for the very last computation).
Notations : $S$ for the area of triangle $ABC$ ; $s=\frac12(a+b+c)$ for its semi-perimeter, and $b:=AC=5$ and $c:=AB=11$.
I use relationship $$S=rs \ \iff \ r^2=\frac{S^2}{s^2}\tag{1}$$ (see here)
Using Heron's formula :
$$S^2=s(s-a)(s-b)(s-c),\tag{2}$$
we have to maximize $$r^2=\underbrace{\frac{1}{s}(s-a)(s-b)(s-c)}_{f(a)}=\frac{2(a^2 -36)(16 - a)}{16 + a} \ \text{for} \ a \in [c-b,c+b]=[6,16]$$
Fig. 1 : Graphical representation of function $f$ showing that $r^2$ is maximal for a unique value $a=a_0$.
Differentiation gives :
$$f'(a)=\frac{-a^3 - 16a^2 + 256a + 576}{2(16 + a)^2}$$
Equation $f'(a)=0$ has three real roots. Only one of them is positive
$$a_0 \approx 11.264804945818$$
which is therefore the extremal value of $f$ (see fig. 1).
We know now $s=\frac12(a_0+b+c) \approx 13.632402472909$
By definition of $f$, we have $$S=s \sqrt{f(a_0)}\approx 27.082086683899.$$
The final touch : using formula $S=\frac12 bc \sin(\hat{A})$, we get
$$\hat{A}=\arcsin\left(\frac{2S}{bc}\right)\frac{180}{\pi} \approx 79.998482041911°$$
(decimal degrees) : not exactly $80°$ but so near !
Remark : For any triangle with given sides $a$ and $b$, this optimal angle giving a maximal inradius cannot be less than : $$2 \arcsin(\Phi-1) \ \text{radians}$$
(where $\Phi$ is golden ratio)
$\approx 76.3454153°$, this value being reached for isosceles triangles ($b=c$).
Write the formula for inradius as $$r=\frac{Area}{s}$$
where $Area$ is the area of the triangle and $s$ is the semiperimeter. In general case, you know $b$ and $c$, and you want to write all these is terms of angle $A$.
$$Area=\frac12bc\sin A$$
$$s=\frac12\left(b+c+\sqrt{b^2+c^2-2bc\cos A}\right)$$
You can then plot $r(A)$. This is what I get from google:
The maximum value is indeed around $80^\circ$.
If you want the exact minimum, take the derivative, and set it to $0$.
