Should I use trig?

Let $L$ be the length of the plank and $x$ the distance from the base of the plank to the base of the fence. Since the triangles formed by: (i) the plank, ground and fence and (ii) the plank, ground and wall, are similar, we see that

$\dfrac{L}{x + 1} = \dfrac{\sqrt{x^{2} + 8^{2}}}{x} \Longrightarrow L = \left(1 + \dfrac{1}{x}\right)\sqrt{x^{2} + 64}.$

Then by the product rule we have that

$\dfrac{dL}{dx} = \left(-\dfrac{1}{x^{2}}\right)\sqrt{x^{2} + 64} + \left(1 + \dfrac{1}{x}\right)*\dfrac{x}{\sqrt{x^{2} + 64}} =$

$\dfrac{1}{\sqrt{x^{2} + 64}}\left(-\dfrac{(x^{2} + 64)}{x^{2}} + (1 + x)\right) = \dfrac{1}{\sqrt{x^{2} + 64}}*\left(-\dfrac{64}{x^{2}} + x\right).$

Now $\dfrac{dL}{dx} = 0$ when $\dfrac{64}{x^{2}} = x \Longrightarrow 64 = x^{3} \Longrightarrow x = 4.$

As we could have the plank stretch out to infinity, it is clear that the critical point at $x = 4$ will yield a minimum, and this minimum is

$L(4) = \left(1 + \dfrac{1}{4}\right)\sqrt{16 + 64} = \dfrac{5}{4}*\sqrt{5*16} = 5\sqrt{5} = \boxed{11.2}$ ft. to one decimal place.

maxima and minima would also help. where a^(2/3) + b^(2/3) = c^(2/3) a = 8 b = 1 c = 11.18