# Minimum ladder over wall optimization

A fence 6 feet tall runs parallel to a tall building at a distance of 2 feet from the building. We want to find the the length of the shortest ladder that will reach from the ground over the fence to the wall of the building. Here are some hints for finding a solution: Use the angle that the ladder makes with the ground to define the position of the ladder and draw a picture of the ladder leaning against the wall of the building and just touching the top of the fence.

If the ladder makes an angle 0.46 radians with the ground, touches the top of the fence and just reaches the wall, calculate the distance along the ladder from the ground to the top of the fence.

a.The distance along the ladder from the top of the fence to the wall is

b. Using these hints write a function L(x) which gives the total length of a ladder which touches the ground at an angle , touches the top of the fence and just reaches the wall. L(x) = . c. Use this function to find the length of the shortest ladder which will clear the fence.The length of the shortest ladder is

I know that the length of the ladder that touches the fence is 6/ sin(0.46). The function would probably be a positive quadratic that would involve the length of the ladder as a variable. The ladder cannot be less than 0 feet (one portion is already about 13 ft. long). How do I figure out the distance from the fence to the wall? How do I then figure out the function to get the length of the ladder?

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Anything wrong with this question? The way I asked it? The content? Or will there always be boo-birds who vote questions down no matter what? – cuabanana Apr 21 '13 at 19:56

Hint: (not using the angle) Put the fence top at $(2,6)$ so the wall is the $y$ axis, and the base of the ladder is on the $x$ axis somewhere to the right of $2$. The line connecting $(x,0)$ to the point $(0,y)$ where the top of the ladder just hits the wall must go through $(2,6)$. You can minimize for simplicity the squared length $x^2+y^2$ of the ladder, and use similar triangles to get a relation $(x-2)/6=x/y$ between $x,y$. From here you can get a one variable function to minimize, and then don't forget to take squareroot for the actual ladder length.

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We need a labelled picture. The wall of the building is on the left. The bottom of the fence is at $B$, the top of the fence is at $T$, the foot of the ladder is at $F$, and the ladder hits the building at $L$. It is clear that to make the ladder short, it should touch the fence at $T$, even though this risks damaging the fence.

I will use the trigonometric approach, because it is the suggested one. But I would not necessarily start that way otherwise.

Let $x$ be the angle $BFT$ that the ladder makes with the ground. Then $\frac{6}{FT}=\sin x$, so $FT=\frac{6}{\sin x}$.

The rest of the ladder has length $TL$. Note that $\frac{2}{TL}\cos x$, so $TL=\frac{2}{\cos x}$.

Thus the total length $f(x)$ of the ladder is given by $$f(x)=\frac{6}{\sin x}+\frac{2}{\cos x}.$$ Calculate $f'(x)$. For the critical point, you should end up with an equation of the shape $\tan^3 x=a$.

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