# Math Help - Minimum Length

1. ## Minimum Length

The wall of a building is to be braced by a beam that must pass over a parallel fence 5 feet high and 4 feet from the building. Find the length of the shortest beam that can be used.

I got an answer of 10.91 feet for the beam, and was wondering if this is correct, number seems a bit odd (though looking into the back of my book that seems common on these problems..) sadly, this is an even problem, and i did alot of work wanna know if i got it right.

2. Originally Posted by Lammalord
The wall of a building is to be braced by a beam that must pass over a parallel fence 5 feet high and 4 feet from the building. Find the length of the shortest beam that can be used.

I got an answer of 10.91 feet for the beam, and was wondering if this is correct, number seems a bit odd (though looking into the back of my book that seems common on these problems..) sadly, this is an even problem, and i did alot of work wanna know if i got it right.
Hello,

let l be the length of the beam and x the distance from the fence to the foot of the beam (see attachment)
To calculate l² use Pythagorean theorem:

$l^2 = (x+4)^2+(5+\frac{20}{x})^2$ . Expand the brackets:

$l^2=x^2+8x+16+25+\frac{200}{x} + \frac{400}{x^2}$

l get an extreme value if (l²)' = 0

$(l^2)'=2x+8-\frac{200}{x^2} - \frac{800}{x^3}$ . Now solve (l²)' = 0

$2x+8-\frac{200}{x^2} - \frac{800}{x^3}=0~\implies~$ $2x^4+8x^3-200x-800=0~\implies~ 2x^3(x+4)-200(x+4)=0$

$(2x^3-200)(x+4)=0~\iff~x=-4~\vee~x=\sqrt[3]{100}\approx 4.6416$ . Plug in the positive value into the equation for l² and calculate afterwards the square root:

$l^2\approx 161.3321~\implies~l \approx 12.7$

3. oh wow, i finally ended up getting that answer later the next day when i relized a simple mistake, you cant take $x^2+(\frac{5x}{x-4})^2=c^2$ take the square root of it all then, and just remove the squares.. but the way you did it is alot better.. cause im pretty sure trying to take the dervivitive of $\sqrt{x^2+(\frac{5x}{x-4})^2}$ then having to set it to 0 and solve for X isnt easy/fun... a bad move to use the Pythagorean theorem as the primary equation and the simular triangle theorem as the secondary (to you pic) make the bottom be X for the large triangle, and X-4 as the small one so $\frac{x}{x-4} = \frac{y}{5}$ -solve for y and substitute when $x^2+y^2=c^2$ where y = how far the ladder is up the wall and c = the ladder and x = distance from the bottem of the wall