Originally Posted by

**galactus** There are various ways to tackle this. I will present a method using trig.

$\displaystyle L=L_{1}+L_{2}$

Then, $\displaystyle L=8csc(t)+sec(t)$ (look at the two triangles made by the ladder, wall, and ground).

$\displaystyle \frac{dL}{dt}=\frac{-8cos^{3}(t)+sin^{3}(t)}{sin^{2}(t)cos^{2}(t)}$ *****

$\displaystyle \frac{dL}{dt}=0$ if $\displaystyle sin^{3}(t)=8cos^{3}(t), \;\ tan^{3}(t)=8$ *****

$\displaystyle tan(t)=2$ which gives the absolute minimum for L because

$\displaystyle \lim_{t\to\ 0^{+}}L=\lim_{t\to \frac{{\pi}^{-}}{2}}L={\infty}$

If $\displaystyle tan(t)=2\Rightarrow csc(t)=\frac{\sqrt{5}}{2} \;\ and \;\ sec(t)=\sqrt{5}$

Hence, $\displaystyle L=8(\frac{\sqrt{5}}{2})+\sqrt{5}=5\sqrt{5} \;\ feet$