In a shopping mall there is a hall of cuboid shape with dimensions $800 imes 800 imes 720$ units, which needs to have the facility of an escalator added along the path shown by the dotted line in the figure (the escalator winds once around the four walls of the hall while ris

Question

In a shopping mall there is a hall of cuboid shape with dimensions $800 	imes 800 	imes 720$ units, which needs to have the facility of an escalator added along the path shown by the dotted line in the figure (the escalator winds once around the four walls of the hall while rising from the floor to the ceiling). Find

(i) the minimum total length of the escalator

(ii) the heights at which the escalator changes its direction

(iii) the slopes of the escalator at the turning points.

Accepted Answer

> [!TLDR]
>
> Unfold the four walls of the square-base hall into one flat rectangle; the escalator's path (once around while rising floor-to-ceiling) becomes the straight-line diagonal of that rectangle, so its length, its height at each wall-seam, and its slope all follow from straight-line geometry.

The hall is a cuboid $800	imes800	imes720$; the escalator winds once around the four walls of the square base while climbing from the floor ($y=0$) to the ceiling ($y=720$). "Unfolding" the four vertical walls (each $800$ units wide) into one flat strip turns this 3-D winding path into a straight line on a flat rectangle of width $4	imes800=3200$ and height $720$ — going once around corresponds to moving all the way across this unfolded width.

**Step 1. Part (i) — minimum total length.** On the unfolded rectangle, the escalator runs from one bottom corner $(0,0)$ to the diagonally opposite top corner $(3200,720)$ (rising the full height exactly once it has gone all the way around). The straight-line diagonal is the shortest such path, of length
$$L=\sqrt{3200^2+720^2}=\sqrt{10{,}240{,}000+518{,}400}=\sqrt{10{,}758{,}400}$$
Checking $3280^2=(3300-20)^2=3300^2-2(3300)(20)+20^2=10{,}890{,}000-132{,}000+400=10{,}758{,}400$, so
$$L=3280	ext{ units}$$ …

Data given	Equation
Slope $m$ , $y$ -intercept $b$	$y=mx+b$
Slope $m$ , point $(x_1,y_1)$	$y-y_1=m(x-x_1)$
Two points $(x_1,y_1),(x_2,y_2)$	$\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}$ , equivalently $\begin{vmatrix}x-x_1 & y-y_1\\ x_2-x_1 & y_2-y_1\end{vmatrix}=0$
$x$ -intercept $a$ , $y$ -intercept $b$ (both $\ne0$ )	$\dfrac{x}{a}+\dfrac{y}{b}=1$
Normal length $p$ , angle $\alpha$ of the normal with the $x$ -axis	$x\cos\alpha+y\sin\alpha=p$
Parametric, through $(x_1,y_1)$ at inclination $\theta$ , parameter $r$ = signed distance from $(x_1,y_1)$	$\dfrac{x-x_1}{\cos\theta}=\dfrac{y-y_1}{\sin\theta}=r$

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