# A Simpler Solution to an Optimization Problem.

The Problem:

My Solution: Let me model this situation in the first quadrant of the $$x$$-$$y$$ plane. Let us take the line $$y=mx+c$$ as the ladder (moving in the first quadrant). This is how the ladder moving in the corridor looks:

The $$y$$-intercept of the line is $$c$$ and the $$x$$-intercept is $$\left(-\frac{c}{m}\right)$$. Since the ladder is moving in the first quadrant, we have $$m<0 \text{ and } c>0$$ Let $$L$$ be the length of the ladder. Then, $$L^2=c^2+\left(\frac{c}{m}\right)^2$$

or

$$c=-\frac{mL}{\sqrt{1+m^2}} \text{ (Negative sign because m<0)}$$

and therefore, the equation of the ladder becomes $$y=mx-\frac{mL}{\sqrt{1+m^2}}$$

As the ladder is transported through the corridor, the distance between the ladder and the corner becomes smaller up to a minimum and then start increasing again. The goal here is to choose $$L$$ such that the ladder just touches the corner as it clears the corridor. This $$L$$ would be the max length of the ladder.

Let the gap along the $$y$$-direction between the corner and the ladder be $$s$$. Let the co-ordinates of the corner be $$(a,b)$$. Then, $$s=b-ma+\frac{mL}{\sqrt{1+m^2}}$$

For a given ladder, this distance hits a minimum value for a certain value of $$m$$. Let's find that out $$\frac{ds}{dm}=-a+\frac{L}{\sqrt{1+m^2}}-\frac{Lm^2}{(1+m^2)^\frac{3}{2}}=0$$

Solving which we obtain the value of $$m$$ which makes $$s$$ hit its minimum value $$m=-\left\{\left(\frac{L}{a}\right)^\frac{2}{3}-1\right\}^\frac{1}{2} \text{ (Negative because m<0)}$$

Therefore, $$s_{\text{min}}=b+a\left\{\left(\frac{L}{a}\right)^\frac{2}{3}-1\right\}^\frac{1}{2}+L\left\{1-\left(\frac{a}{L}\right)^\frac{2}{3}\right\}$$

The value of $$s_{\text{min}}$$ in the case of this problem is zero. Therefore, solving the above equation, we get $$L=a\left\{\left(\frac{b}{a}\right)^\frac{2}{3}+1\right\}^\frac{3}{2}=\left(b^\frac{2}{3}+a^\frac{2}{3}\right)^\frac{3}{2}$$

Plugging, $$a=8$$ and $$b=6$$ as asked in the question we get $$L=19.7313 \approx 20 \text{ feet}$$

Is this correct? If yes, can we make an easier model to solve this problem? (Only in the context of calculus)

Your answer is correct. Note that the question says to round $$\color{red}{down}$$ to the nearest foot.

Consider

You have two right-angled triangles, and they share the same angle $$\theta$$ since the horizontal black and green lines are parallel. Then you can shown that the length of the ladder is given by

$$L(\theta)=\frac{6}{\sin(\theta)}+\frac{8}{\cos(\theta)}$$

for $$0<\theta<\frac{\pi}{2}$$. Then

$$L'(\theta)=0\implies \tan^3(\theta)=\frac{3}{4}\implies \theta \approx 0.73752$$ $$\implies L\approx 19.7313$$

• This model assumes the ladder to be in contact with the corner at all times. But the ladder touches the corner only momentarily. Sep 18 at 16:56
• This is the length of the longest ladder you can carry horizontally around the corner of the corridor. Sep 18 at 19:35

Your animation of the sliding ladder is alright. The answer given here cannot get any simpler. There is corner contact only at a single instant.

Easier modeling is by using symbols with trig. Start with making a sketch of the corner.

$$\text{For easy trig typing using shorthand}\;$$ $$s = \sin \phi, c = \cos \phi \text{, slant ladder length = }$$

$$L=\frac {b}{s}+ \frac{a}{c} \tag 1$$

Differentiate with respect to $$\phi$$

$$\frac{-bc}{s^2}+\frac{as}{c^2}=0 \tag2$$

Simplify to find $$\tan \phi$$

$$\frac{s}{c}=\tan \phi= {\left( \frac {b}{a}\right)}^ {\frac13} \tag 3$$

Construct right triangle with Pythagorean thm ( even if the dimension does not tally to linear dimension ) resolving $$(s,c)$$ to conveniently plug their values into (1) and then simplify

$$L={\left( a^{\frac23}+ b^{\frac23}\right)}^ {\frac32} \tag 4$$

Actually it is an Astroid envelope of the sliding ladder having equation

$$x^{\frac23}+ y^{\frac23}= L^{\frac23}$$

on to which you juxtaposed a sharp corner from right side.

The 3 lines ( two blue,one red) have the same length 19.7313 units. Only the red line touches the corner $$(8.6)$$ . The blue lines do not touch the corner, there is considerable gap/clearance.

Now plug in numerical values for symbols

$$a=6,\; b=8,\;L \approx 19.7313 \;ft, \; \phi= 42.2568^{\circ}; \tag 5$$

If corridor widths are changed you know what to do.

• This model assumes the ladder to be in contact with the corner at all times. But the ladder touches the corner only momentarily. Sep 18 at 16:57
• Not at all. The model does not assume contact with corner always but assumes contact only with the wall and floor along which the ladder slides. This is ensured in the formulation. Your misconception can be hopefully cleared by sketch just now added. To drive home this point I mentioned already about the Astroid that touches the corner at a single point. All other positions of the ladder keeps it clear from re-entrant corner without any contact or interference as shown. Sep 18 at 20:58

$$x = {b\over \sin \varphi}$$ $$\ell- x = {a\over \cos \varphi}$$ so $$\boxed{\ell = {b\over \sin \varphi}+ {a\over \cos \varphi}}$$

Now take the derivative of it with respect to $$\varphi$$

• This model assumes the ladder to be in contact with the corner at all times. But the ladder touches the corner only momentarily. Sep 18 at 16:57
• So?..............................
– Aqua
Sep 18 at 17:08