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Closing Room, related rates?

  1. Apr 26, 2013 #1
    This is an extra credit question in my class. We didn't get any picture or diagram, so my assumptions might be as good as yours.

    1. Problem
    Consider a room where the walls can close in on each other. The room has a height of 9 feet, width of 12 feet and a length of 20 feet. For simple terms, the side walls are 20 feet long and the front and back walls have a length of 12 feet. To keep this room from closing in on the top, a man places a pole diagonally on the back wall, wedging it into the corners. However, this only causes the right wall to close in on the left at a rate of 2 ft/min and for the height to increase at some constant rate. While all this is occurring, the pole maintains to be wedged into the back wall diagonally.

    How fast is the total area of the back wall changing when the base of the wall is at 10 feet?

    2. Known

    I know by the Pythagorean theorem that before the room starts closing, the diagnol of the back wall is 15 feet, but according to the problem, this length never changes so even when the back wall base is 10 feet, that diagnol is the same. I'm also given a rate of 2ft/min.

    3. Attempt

    I'm only interested in the area of the back wall. I have a little rectangle drawn with a diagnol of 15 feet, width of 10 and length of sqrt125 (again by Pythagorean thoerem).

    Since we want to know the rate of change of the area, I need to use

    A = xy. However taking the derivative of this function requires that I know the rate of change of the height.

    I can do this by looking at the triangle within the rectangle x^2 + y^2 = c^2
    Where y= sqrt125, x=10 and c=15. Since I'm letting y = sqrt125, I need Dy/Dt.

    Taking the derivative of both sides with respect to time I get

    2x DX/DT + 2y DY/DT = 0
    Since the diagnol length c is constant taking the derivative makes it 0. DX/DT = 2ft/min (since the length x is decreasing, this value is going to be negative in the following calculation) , x = 10 and y = sqrt125. Solving for DY/DT I get

    DY/DT = (-2)(x)(DX/DT)/(2y)

    = (-2)(10)(-2)/(2sqrt125) =
    1.78 ft/min.

    I now have the rate of change of the height. So I can use A = xy

    Taking the derivative of both sides and applying the product rule I have..

    DA/DT = (-2)(sqrt125) + (1.78)(10) = -4.56. Ft^2/min.

    I'm almost positive the answer cannot be negative. Any help would be appreciated.
     
  2. jcsd
  3. Apr 26, 2013 #2

    Dick

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    I think you did that rather well. Why do you think the rate can't be negative? I might have made a mistake but I agree with you.
     
  4. Apr 27, 2013 #3

    Dick

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    Try to think of an intuitive reason why the area rate must be negative. Imagine a long thin rectangle with a constant diagonal closing along the length direction. When does it hit a maximum area and when does it start contracting again?
     
  5. Apr 27, 2013 #4
    Hmmm I suppose the more I think about it, it does make sense. But this also raises the question in my mind, lets say if I wanted to find out when the area is a maxium(from the point the walls started to move). Would it be the time where the rectangle creates a near square? And how could I mathematically go about that?
     
  6. Apr 27, 2013 #5

    Dick

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    Use what you've already done to find critical points. What conditions on x and y do you need for dA/dt to be zero?
     
  7. Apr 27, 2013 #6
    Hmm okay,

    So I can take A = xy

    Write Y in terms of X, take the derivative, and find the critical point(s). And then use that X value to find the particular Y value to give me the dimensions for the max area.

    Makes sense, but how could I go about finding out what time this happens at?

    T = (12 - x)/(-2) + (Y in terms of X)/(1.78) ?

    Or instead, possibly I could use the critical point of X found above, and divide it by the rate of X to get the time that it happens at.
     
  8. Apr 27, 2013 #7

    Dick

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    Your x is given by 12-2T, right? Express y in terms of T and look for critical points of A in terms of T.
     
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