The side length of a rectangular box with a square base is increasing at the rate of 2 ft/sec, while the height is decreasing at
the rate of 2 ft/sec. At what rate is the volume of the box changing when the side length is 10 ft and the height is 12 ft? Remember to use the product rule when you find the expression for dV/dt
To solve the question we proceed as follows: length=width=x ; g=height=h thus V=x*x*h v=x^2h dv/dt=2xh dx/dt+x^2 dh/dt but dx/dt=2 ft/sec dh/dt=-2 ft/sec thus plugging the values in the equation we get: dv/dt=2×10×12+10²(-2) dv/dt=240-200 dv/dt=40 ft³/sec
