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)
 A: 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
$$L'(\theta)=0\implies \tan^3(\theta)=\frac{3}{4}\implies \theta \approx 0.73752$$
$$\implies L\approx 19.7313$$
A: 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.
A: 
$$x = {b\over \sin \varphi}$$ $$\ell- x = {a\over \cos \varphi}$$  so $$\boxed{\ell = {b\over \sin \varphi}+ {a\over \cos \varphi}}$$
Now take the derivative of it with respect to $\varphi$
