Figure 3.8.14(b) animates the passage of a ladder through the corridors. The accompanying slider controls the angle $\mathrm{\θ}$ (see Figure 3.8.14(a)) measured in degrees. As the slider is moved, the corresponding value of $L$, the length of the ladder, is displayed. As $\mathrm{\θ}\to 0°$ or $\mathrm{\θ}\to 90°$, the length of the ladder becomes arbitrarily large, but of course, such large ladders would not fit around the corner of the passageway. Hence, it is the length of the shortest segment that determines the length of the longest ladder that can be brought around the corner of the passageway.

Figure 3.8.14(c) is a graph of $L$ against the angle $\mathrm{\θ}$, measured in degrees. For the particular values of $a$ and $b$ used in Figure 3.8.14(b), there appears to be a minimum value for $L$ near $\mathrm{\θ}\doteq 44°$.

From Figure 3.8.14(a), $L\=$ $∥\mathrm{AB}∥$$\+$$∥\mathrm{BC}∥$, so that $L\left(\mathrm{\θ}\right)\=\frac{a}{\sin \left(\mathrm{\theta }\right)}\+\frac{b}{\cos \left(\mathrm{\theta }\right)}$.

Additional manipulations (performed by hand) put the length of the longest ladder into the form ${\left(a^{2\/3}\+b^{2\/3}\right)}^{3\/2}$.

An alternate solution can be constructed along the following lines. Let point $B$ in Figure 3.8.14(a) be the origin of a Cartesian coordinate system. Then the equation of the green line representing the ladder is $y\=m x$, where $m$, the slope of the line, is to be determined. The intersection of this line with $y\=b$ is the point $\left(b\/m\,b\right)$; the intersection with the vertical line $x\=-a$ is the point $\left(-a\,-a m\right)$, where $a$ is taken as positive.

The length of the segment $\mathrm{AC}$ is then $L\left(m\right)\=\sqrt{{\left(b\/m\+a\right)}^2\+{\left(b\+a m\right)}^2}$. The equation $L\prime \left(m\right)\=0$ has multiple solutions for $m\=\stackrel{\^}{m}$ ; the appropriate real, positive solution gives $L\left(\stackrel{\^}{m}\right)\={\left(a^{2\/3}\+b^{2\/3}\right)}^{3\/2}$.