= Solution
Assume positive corridor widths $a,b$. Write the line through the two wall-contact endpoints in intercept form. Requiring it to pass through the inner corner gives
$$
\frac a{a+X}+\frac b{b+Y}=1
\quad\Longleftrightarrow\quad XY=ab,
\qquad X,Y>0.
$$
Its squared length is $F=(a+X)^2+(b+Y)^2$. Apply <Lagrange multipliers> to this <function> subject to $G=XY-ab=0$:
$$
2(a+X)=\lambda Y,\qquad2(b+Y)=\lambda X.
$$
Multiplying by $X$ and $Y$ respectively eliminates the multiplier, giving $X(a+X)=Y(b+Y)$. Substitute $Y=ab/X$ and multiply by $X^2$:
$$
X^3(a+X)=ab^2(a+X),\qquad
\boxed{(X^3-ab^2)(X+a)=0.}
$$
The branch $X=-a$ is geometrically inadmissible. Thus $X=a^{1/3}b^{2/3}$ and $Y=a^{2/3}b^{1/3}$. If $A=a^{1/3}$ and $B=b^{1/3}$, the two intercepts become $A(A^2+B^2)$ and $B(A^2+B^2)$, so
$$
F_{\min}=(A^2+B^2)^3,
\qquad\boxed{L=(a^{2/3}+b^{2/3})^{3/2}.}
$$
Along $XY=ab$ in the positive quadrant, $F\to\infty$ as $X\to0^+$ or $X\to\infty$, and this is its only <stationary point>. It is therefore the minimum of the longest sticks that fit at each orientation. A stick being turned must pass through every intermediate orientation, so this narrowest clearance is precisely the maximum carryable length, rather than a maximum of $F$. This is the <ladder around a right-angled corner> problem, with a zero-width stick and wall contact allowed as in the ideal geometry.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1-ladder-corner.png]
{title=The critical ladder position touching the two outer walls and the inner corridor corner}
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