Finding the length of the shortest ladder 
A wall 8 feet high is 1 foot from a house. Find the length $L$ of the
  shortest ladder over the wall to the house. Draw a triangle with
  height $y$, base $1 + x$, and hypotenuse $L$.

The shortest ladder is placed straight up close to the wall. It has length $8+8=16$ for going up and down. Does this even need calculus? Or have I misunderstood the problem?
 A: 
The triangles $ABC$ and $AB'C'$ satisfy the relation
$$
\frac{y}{1 + x} = \frac{8}{x}
$$ 
So that 
$$
y = \frac{8}{x}(1 +x) \tag{1}
$$
The length of the ladder (red line) follows from
$$
L^2 = (1 + x)^2 + y^2 \stackrel{(1)}{=} (1 + x)^2 + \frac{64}{x^2}(1 + x)^2 = (1 + x)^2\frac{x^2 + 64}{x^2}
$$ 
Call $f(x) = L^2$, and note that minimizing $f(x)$ is equivalent to minimize $L$, therefore we want to find the minimum of 
$$
f(x) = (1 + x)^2\frac{x^2 + 64}{x^2}
$$
Which can be done by solving the problem
$$
\frac{df}{dx} = 2\frac{-64 - 64 x + x^3 + x^4}{x^3} = 0
$$
whose solution is $x = 4$. It is easy to see that for this value $f$ has a minimum since 
$$
\left.\frac{d^2f}{dx^2}\right|_{x = 4} = \frac{15}{2} > 0
$$
The length of the ladder is then
$$
L = \sqrt{f(4)} = 5\sqrt{5}\;{\rm ft}
$$
A: If the distance of the bottom of the ladder from the wall is $x$, then the distance from the house is $x+1$ and the height at which the ladder hits the house is $8\frac{1+x}{x}$. So the length of the ladder must be:
$$L=\sqrt{(1+x)^2 +8^2\frac{(1+x)^2}{x^2}}=\frac{1+x}{x}\sqrt{x^2+8^2}$$
Wolfram alpha says the derivative of this is:
$$\frac{x^3-64}{x^2\sqrt{x^2+64}}$$
Which means the minimum must be at $x=4$ and $L=5\sqrt{5}$.
