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**HallsofIvy** Volume is changing at a constant rate: "water is flowing into the tank at the rate of 2 cubic feet per minute" so dV/dt= 2.

Write the volume as a function of height of the water: if V= f(h) the dV/dt= f'(h) dh/dt= 2 and you can solve for dh/dt.

To find the volume as a function of the height, volume of a pyramid of height h with base area A is $\displaystyle \frac{1}{3}Ah$. Here, of course, the base will be a square with side length depending on height. You can get the length of an edge of that square as a function of height by remembering that the pyramid of water will be **similar** to the entire pyramid. It has edge length 12 and height 10 so x/h= 12/10 or x= (6/5)h. The area of the base is [/tex]A= (36/25)h^2[/tex] so the volume is $\displaystyle V= \frac{12}{25}h^3$.