The longest ladder around a corner

Idea. For each angle $\theta$, the ladder wedged between the two outer walls and touching the inner corner has length $L(\theta)$. A ladder passes the corner if it is shorter than $L(\theta)$ for *every* $\theta$, so the answer is $\min L$.

Derivative.
$$L'(\theta)=-\frac{a\cos\theta}{\sin^2\theta}+\frac{b\sin\theta}{\cos^2\theta}=0\iff\tan^3\theta=\frac ab.$$
$L'$ is negative for small $\theta$ and positive near $\frac\pi2$, and $L\to+\infty$ at both ends, so this critical point is the global minimum.

Value. Let $t=\tan\theta=(a/b)^{1/3}$. Then $\sin\theta=\frac t{\sqrt{1+t^2}}$ and $\cos\theta=\frac1{\sqrt{1+t^2}}$, so
$$L=\sqrt{1+t^2}\left(\frac at+b\right).$$
With $\frac at=a^{2/3}b^{1/3}$ and $\sqrt{1+t^2}=\dfrac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}$, this simplifies to
$$L_{\min}=\left(a^{2/3}+b^{2/3}\right)^{3/2}.$$

Check. For $a=b=1$: $L_{\min}=2^{3/2}=2\sqrt2\approx2.83$. For $a=8$, $b=1$: $\tan\theta=2$ and $L=\frac8{2/\sqrt5}+\sqrt5=5\sqrt5\approx11.18$, and $(4+1)^{3/2}=5\sqrt5$ as well.