1. ## Carpentry problem

Two carpenters are trying to carry a board down a corridor and around a corner in a building. The corner is at a 90degree angle and the corridor is 9 feet wide. What is the length of the longest board that can be carried parallel to the floor through the corridor, to the nearest tenth of a foot?

I'm pretty sure this involves the law of cosines and sines but I have no idea how to even begin doing this one.

2. ## Re: Carpentry problem

I used Pythagoras to get an answer of 12.7 ft

On the diagram, the blue bar is the maximum distance that you can fit around the corner. We can construct a triangle (green) which, from the information given in the question is isosceles with the two sides of 9m.

To find the length of the blue line, which is the hypotenuse, use Pythagoras.

EDIT: Use Earboth's answer, it is better than mine

3. ## Re: Carpentry problem

Originally Posted by explodingtoenails
Two carpenters are trying to carry a board down a corridor and around a corner in a building. The corner is at a 90degree angle and the corridor is 9 feet wide. What is the length of the longest board that can be carried parallel to the floor through the corridor, to the nearest tenth of a foot?

I'm pretty sure this involves the law of cosines and sines but I have no idea how to even begin doing this one.
Here is how I would do this question:

1. Use proportions:

$\displaystyle \frac9x = \frac y9~\implies~y=\frac{81}x$

2. Use Pythagorean theorem:

$\displaystyle l=\sqrt{81+x^2}+\sqrt{81+y^2}$

$\displaystyle l=\sqrt{81+x^2}+\sqrt{81+\left(\frac{81}x\right)^2 }$

$\displaystyle l(x)=\sqrt{81+x^2}+\frac9x \sqrt{x^2+81}=\sqrt{81+x^2}\left(1+\frac9x\right)$

Actually you have to use absolute values. But since all lengthes in an building are positive I use a simple x.

3. Differentiate l wrt x and solve the equation l'(x) = 0. (Remark: This is the very tricky part of the calculations. Be careful!)

You'll get x = 9.

4. Therefore $\displaystyle l(9) = l_{max} = 18 \cdot \sqrt{2} \approx 25.5\ ft$