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Minimization problem

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data


    An 8 foot tall wall is 27 feet from a building. What is the smallest sized beam that could be placed against the wall, sit on top of the 8 foot wall, and touch the ground on the other side of the wall?

    2. Relevant equations

    I want to minimize the beam length, so will differentiate and set to zero, then find the minimum value.

    3. The attempt at a solution

    I don't know how to derive the relevant equation and what beam length would be a function of. Can you help point me in the right direction?
     
  2. jcsd
  3. Oct 9, 2009 #2

    Mark44

    Staff: Mentor

    Differentiate what? Before doing any of this, draw a sketch of the wall and the building, and show the beam. Since you don't know the length of the beam, call its length, oh, maybe, L. Put in the known dimensions and see if you can figure out the unknown dimensions. You need to do all this before you get any kind of function that you can differentiate.
     
  4. Oct 9, 2009 #3
    I am at a loss for how to find the unknown dimensions.

    Thinking in terms of a picture, the "beam" crosses the Y-axis at (0,Y), runs through the point (27,8), and crosses the X-axis at (X,0). This forms a rectangle with points (0,0) and (27,8), and two like triangles: one sitting on top of the rectangle with coordinates (0,Y), (0,Y-8), and (27,8). The second triangle is adjacent to the rectangle with coordinates (27,8), (27,0), and (X-27, 0). If the interior angles of these triangles are a=90 degrees, b, and c, then b+c=90 degrees. Also, tan b=27/(Y-8)=(X-27)/8 and tan c=(Y-8)/27=8/(X-27).

    I probably should be able to find the unknown dimensions with this information but I'm at a loss.
     
  5. Oct 9, 2009 #4

    Mark44

    Staff: Mentor

    OK, that's a good start. It looks like you have the short wall running up the y-axis. If so, the beam touches the short wall at (0, 8), and touches the building wall at (27, y). Does that help?
     
  6. Oct 9, 2009 #5
    So the slope is (8-y)/27 right? I'm not sure what the next step is though, what am I missing?
     
  7. Oct 10, 2009 #6

    Mark44

    Staff: Mentor

    I don't think the slope of the beam enters into the problem. It's probably easier to put the low end of the beam at the origin, and the low wall x units to the right, and the wall of the building 27 units more to the right. You know the height of the low wall, and you can assume that the beam touches the building at a point y units up.

    From this information, and the fact that you're dealing with two similar right triangles, you should be able to get a relationship (an equation) between x and y.

    Also, since L (the length of the beam) is the hypotenuse of a right triangle, can you get an equation that represents this length? It will involve both x and y.

    Now, you should have L as a function of x and y, and with the first equation, you should be able to solve for y in terms of x. From this, you can write L as a function of x alone.
     
  8. Oct 10, 2009 #7
    I know the length of the large triangle, L=sqrt[(X+27)^2+(Y)^2]. If I denote the hypotenuse of the smaller triangle by l, then (l)^2=(X)^2+64, and (L)^2=(X+27)^2+(Y)^2. However, this seems like two equations and four unknowns. If this is solvable, then I don't know how to do it.
     
  9. Oct 10, 2009 #8

    Mark44

    Staff: Mentor

    OK, that's one of them. Another equation that can help you rewrite L(x, y) as a function of only one variable involves the two similar right triangles. For the smaller triangle, the altitude is 8, and the base is x. For the other, the altitude is y and the base is x + 27. Can you get an equation from this information, and solve it for y? Once you do that you can replace y in your L(x, y) function.
     
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