# Applied max and min problem

• Nov 9th 2009, 09:11 PM
NDHS
Applied max and min problem
Hi, I'm having a bit of trouble getting this one problem started:
Quote:

Painters are painting the second floor exterior wall of a building that adjoins a busy sidewalk. A corridor 2 m wide and 3 m high is built to protect pedestrians. What is the length of the shortest ladder that will reach from the ground over the corridor to the wall of the building?
I have tried using the pythagorean theorem as well as similar triangles but it doesn't seem to be working out. Any help would be appreciated.
• Nov 9th 2009, 10:53 PM
redsoxfan325
Quote:

Originally Posted by NDHS
Hi, I'm having a bit of trouble getting this one problem started:
I have tried using the pythagorean theorem as well as similar triangles but it doesn't seem to be working out. Any help would be appreciated.

There are two ways to calculate the length of the ladder. If the height where it touches the building is $\displaystyle y$ and the distance from the bottom of the ladder to the building is $\displaystyle x$, then the length of the ladder can be calculated using two different formulas.

$\displaystyle \sqrt{x^2+y^2}=\ell=\sqrt{2^2+(y-3)^2}+\sqrt{(x-2)^2+3^2}$

There's a bit of algebra involved, but, you end up with $\displaystyle y=\frac{3x}{x-2}$

So you want to minimize $\displaystyle \sqrt{x^2+y^2}$ (minimizing $\displaystyle x^2+y^2$ will be easier and will give the same result) given that $\displaystyle y=\frac{3x}{x-2}$.

Doing this calculation doesn't seem to lead into anything nice though. ($\displaystyle x$ ends up being the irrational solution to a cubic polynomial.) I got $\displaystyle x\approx4.62$ and $\displaystyle y\approx5.29$ so $\displaystyle \ell=\sqrt{x^2+y^2}\approx7.02$