Optimization - Calculus -Beam?

Question

Please help. I do not understand the problem. Specifically, the scenario. I understand optimization but I just need to know how to set it up. Any help would be appreciated, thanks!

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

Re: Similar Triangles
Sorry, but I'm still confused. I thought you had to optimize it.

Ah, I see now. Shouldn't it be b^2 + y^2 = x^2 though?

Re: gp4rts

So, you're basically solving for one part of the beam length. After that, would you find the other part of the beam length, add it together to find the total length?

Scratch that last comment. Care to walk me through it to make sure that I got the right answer?

I differentiated, set the derivative equal to zero and found A = 47.1289 degrees. Then I substituted that value in the general equation and found x = 12.7. Could you please confirm this? Thank you.

Well, I worked it both ways and both of them gave me the correct answer (I hope...can anyone confirm x = 12.7?) I would like to thank you both!

Answer 1

Let b = distance of base of beam from wall, x = length of beam, y = height at which beam touches wall.
Applying similar triangles,
5/(b-4) = y/b
5b = y(b-4)
y = 5b/(b-4)
+Yes, that's right, sorry: Applying pythagorean theorem,
x^2 = b^2 + y^2
x = sqrt(b^2(1 + 25/(b-4)^2)), so you must find a value of b that minimises x.

Answer 2

The beam touches the ground, the top of the fence, and the wall. Say it touches the ground at a point x feet from the building, and call y the height it touches the building.

Then y/x = 5/(x - 4) by similar triangles, so

y = 5x/(x - 4).

Now for r, r^2 = x^2 + y^2. Forget the square root, since minimizing for r^2 also minimizes for r.

x^2 + y^2 = x^2 + (5x/(x - 4))^2 = r^2, now you have an equation for r in terms of x and you can find the minimum.

Answer 3

The beam will rest on the top of the fence: for any given location of the base of the beam, its shortest length will always be if it touches the top of the beam. Let the angle the beam makes with the ground be A; then the length of the beam is given by (4+d)/[cos(A)], where d is the distance from the fence to the beam/ground contact point. Also, d/5 = cot(A). This gives the equation for L (beam length) as a function of T as

L(A) =[ 4 + 5*cot(A)]/cos(A);

Differentiate this to find A that minimizes L. Then compute L from the formula.

Answer 4

It's so easy!! The answer is in the question!! Open your eyes as well as your mind!!!