Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 20 / b

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 givesIts squared length is $F=(a+X{)}^2+(b+Y{)}^2$. Apply Lagrange multipliers to this function subject to $G=XY-ab=0$:Multiplying by $X$ and $Y$ respectively eliminates the multiplier, giving $X(a+X)=Y(b+Y)$. Substitute $Y=ab/X$ and multiply by $X^2$: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 $B(A^2+B^2)$, soAlong $XY=ab$ in the positive quadrant, $F\to \infty$ as $X\to 0^{+}$ 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.