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.