# 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?

• You've misunderstood. I think the ladder has to be a single line. Otherwise you could do it with length 9 - straight up the wall and then horizontal to the house. You don't have to reach the house at ground level. – Scott Burns Nov 26 '16 at 3:07
• I think you need to consider $L^2=1+x)^2+8^2$ and find maxima of that – Learnmore Nov 26 '16 at 3:09
• @learnmore. Not quite. You need $L^2 = {(1+x)^2(x^2+64)\over x^2}$ – Scott Burns Nov 26 '16 at 3:13
• $L^2=(1+x)^2+y^2$ and $y:(1+x)=8:x$ – MattG88 Nov 26 '16 at 3:15
• This probably isn't the OP's fault, but I can think of at least three different scenarios given the poor wording of the question. – The Count Nov 26 '16 at 3:16

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 $L=5\sqrt{5}$.

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

$$L = \sqrt{f(4)} = 5\sqrt{5}\;{\rm ft}$$

• I prefer the square root form. It is easier to solve, and its derivative, though seemingly difficult, can easily be obtained by using the product rule and the power rule. – W. Zhu Nov 26 '16 at 4:16