I calculate the length of the stick in terms of a, b and $\theta$. From the figure I observe that the length of the stick could be seen as the sum of two hypotenuses of two right-angled triangles. Its length is:

${\displaystyle l(\theta )=\frac{a}{\sin \theta }+\frac{b}{\cos \theta }.}$

Now, I have to calculate the minimum of this expression, in order to make the stick pass through the corner. For this, I calculate the derivative of $l(\theta )$ and equate it to 0. I must say from the beginning that derivatives are not so familiar to me. For a minimum value of the length:

${\displaystyle \frac{\mathrm{dl}}{d\theta }=\frac{-a\cos \theta }{{\sin }^2\theta }+\frac{b\sin \theta }{{\cos }^2\theta }=0.}$

So for a minimum value $a{\cos }^3\theta =b{\sin }^3\theta$ and

${\displaystyle \tan \theta ={\left(\frac{a}{b}\right)}^{1/3}.}$

Now, I have to calculate $\sin \theta$ and $\cos \theta$ as functions of $\tan \theta$. I know that:

${\displaystyle \cos x=\frac{1}{\sqrt{1+{\tan }^{2}x}}\hspace{0.5em}\mathrm{and}\sin x=\frac{\tan x}{\sqrt{1+{\tan }^{2}x}}}$

In the case of the problem, I have:

${\displaystyle \frac{1}{\cos \theta }=\sqrt{1+(\frac{a}{b}{)}^{2/3}}}$

and

${\displaystyle \frac{1}{\sin \theta }=(\frac{b}{a}{)}^{1/3}\sqrt{1+(\frac{a}{b}{)}^{2/3}}}$

So the minimum length is

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{\frac{a}{\sin \theta }+\frac{b}{\cos \theta }}& =(a^{2/3}{b}^{1/3}+b)\sqrt{\frac{a^{2/3}+b^{2/3}}{b^{2/3}}}\\ \multicolumn{1}{c}{}& =(a^{2/3}+b^{2/3}{)}^{3/2}\end{array}.}$

The result issymmetric in a and b.

If $a=65\mathrm{cm}$ and $b=75\mathrm{cm}$ then ${65}^{2/3}+{75}^{2/3}=16.16623563+17.78446652=33.95070215$ and $33.{951}^{3/2}=197.8213407$ so an object of about 197 cm could be manoeuvred around the bend but it is not possible to manoeuvre a 200 cm object around this bend.