Answer to Question #168985 in Calculus for Vishal

Question

Two hallways, of widths 'a' and 'b', meet at right angles. What is the greatest possible length of a ladder which can be carried horizontally around the corner ?

Answer 1

Solution:

Let's draw a figure according to the question and label it.

Let AB be the ladder with length L. C is the point of intersection of hallways.

So, $L=AC+BC$

Given, $BE=b, AD=a$

Let $CD=y, CE=x$

Then, both triangles ACD and CEB are right-angled triangles.

By pythagoras theorem, in triangle ACD,

$AD^2+CD^2=AC^2 \\\Rightarrow a^2+y^2=AC^2$

$\Rightarrow AC=\sqrt {a^2+y^2}$

By pythagoras theorem, in triangle CEB,

$CE^2+EB^2=BC^2 \\\Rightarrow x^2+b^2=BC^2$

$\Rightarrow BC=\sqrt {x^2+b^2}$

Next, $\angle CAD=\angle BCE$ [Corresponding angles]

We take their tangent as: $\dfrac{CD}{AB}=\dfrac{BE}{CE}$

$\Rightarrow \dfrac{y}{a}=\dfrac{b}{x}$

$\Rightarrow y=\dfrac{ab}{x}$ ...(i)

Now, $L=AC+BC=\sqrt {a^2+y^2}+\sqrt {x^2+b^2}$

$=\sqrt {a^2+(\dfrac{ab}{x})^2}+b\sqrt {(\dfrac{x}{b})^2+1}$ [Using (i)]

Put $\dfrac{x}{b}=t$

$L=a\sqrt {1+(\dfrac{1}{t})^2}+b\sqrt {(t)^2+1}$

$=a\sqrt {\dfrac{t^2+1}{t^2}}+b\sqrt {t^2+1}$

$=\dfrac at\sqrt {t^2+1}+b\sqrt {t^2+1}$

$=\sqrt {t^2+1}(\dfrac at+b)$ ...(ii)

Now differentiating w.r.t. $t$ ,

$L'=(\sqrt {t^2+1})'(\dfrac at+b)+\sqrt {t^2+1}(\dfrac at+b)'$

$=(\dfrac{1}{2\sqrt {t^2+1}}\times 2t)(\dfrac at+b)+\sqrt {t^2+1}(\dfrac {-a}{t^2}+0)$

$=(\dfrac{t}{\sqrt {t^2+1}})(\dfrac at+b)-(\dfrac {a\sqrt {t^2+1}}{t^2})$

Put $L'=0$

$\Rightarrow (\dfrac{t}{\sqrt {t^2+1}})(\dfrac at+b)-(\dfrac {a\sqrt {t^2+1}}{t^2})=0$

$\Rightarrow (\dfrac{t}{\sqrt {t^2+1}})(\dfrac at+b)=\dfrac {a\sqrt {t^2+1}}{t^2}$

$\Rightarrow {a+bt}=\dfrac {a(t^2+1)}{t^2}$

$\Rightarrow t^2( {a+bt})= a(t^2+1)$

$\Rightarrow at^2+bt^3= a(t^2+1) \\ \Rightarrow bt^3=a \\ \Rightarrow t^3=\dfrac ab \Rightarrow t=(\dfrac ab)^{1/3}$

Put this value of t in (ii)

$L=\sqrt {[(\dfrac ab)^{1/3}]^2+1}\times(\dfrac a{(\dfrac ab)^{1/3}}+b)$

$=\sqrt {[(\dfrac ab)^{2/3}]+1}\times(\dfrac a{(\dfrac {a^{1/3}}{b^{1/3}})}+b)$

$=\sqrt {\dfrac {a^{2/3}+b^{2/3}}{b^{2/3}}}\times( {a^{2/3}b^{1/3}}+b)$

$= {\dfrac {\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}}\times b^{1/3}( {a^{2/3}}+b^{2/3})$

$=( {a^{2/3}}+b^{2/3})^{3/2}$

This is required length of the ladder.