Answer
$20\ ft$
Work Step by Step
Step 1. Using the figure given in the exercise, the ladder will need to be rotated around and pass the corner at $(8,6)$. Thus we need to find the minimum of line segment $\bar {AB}$. This is because the length of $\bar {AB}$ can be very large and only the minimum length will allow the ladder to pass.
Step 2. Let $k\lt0$ be the slope of line segment $\bar {AB}$. As it passes the point at $(8,6)$, the line equation can be written as $y-6=k(x-8)$ or $y=kx-8k+6$
Step 3. We can find the y-intercepts as $y=6-8k$, which gives $A(0, 6-8k)$. And the x-intercept is $x=8-\frac{6}{k}$, which gives $B(8-\frac{6}{k},0)$
Step 4. The square of the ladder length is given by $D=\bar{AB}^2=(8-\frac{6}{k})^2+(6-8k)^2$. Letting $D'=0$, we have $2(8-\frac{6}{k})(\frac{6}{k^2})+2(6-8k)(-8)=0$, which gives $16k^4-12k^3+12k-9=0$ or $(4k-3)(4k^3+3)=0$
Step 5. As $k\lt0$, we have the solution $k=-\sqrt[3] {\frac{3}{4}}=-\frac{\sqrt[3] 6}{2}\approx-0.9086$, which gives $D=(8+\frac{12}{\sqrt[3] 6})^2+(6+4\sqrt[3] 6)^2\approx389.325$ and $\bar{AB}\approx19.73\approx20\ ft$
Step 6. Test signs of $D'$ to have $..(-)..(-\frac{\sqrt[3] 6}{2})..(+)..(0)$ (e.g. use $x=-2, -\frac{1}{2}$), indicating a local minimum of the function.
