3 1 point At a sand gravel plant, sand is falling off a conveyor belt and onto a conical pile at the rate of 10 cubic feet per minute. The diameter of the base of the cone is three times the altitude. At what rate is the height of the pile changing when the pile is 15 feet high? \( 6 /(4057) \mathrm{ft} / \mathrm{min} \) \( 8 /(405) \mathrm{ft} / \mathrm{min} \) \( 8 /(405 \Pi) \mathrm{ft} / \mathrm{min} \) None of these.

To solve this problem, we need to relate the volume of the conical pile to its height and then find the rate of change of height with respect to time. 1. **Volume of a cone**: The volume \( V \) of a cone is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] where \( r \) is the radius of the base and \( h \) is the height. 2. **Relationship between radius and height**: We are given that the diameter of the base of the cone is three times the altitude (height). Therefore, the radius \( r \) can be expressed in terms of the height \( h \): \[ d = 3h \implies r = \frac{d}{2} = \frac{3h}{2} \implies r = \frac{3}{2}h \] 3. **Substituting \( r \) in the volume formula**: \[ V = \frac{1}{3} \pi \left(\frac{3}{2}h\right)^2 h = \frac{1}{3} \pi \left(\frac{9}{4}h^2\right) h = \frac{3}{4} \pi h^3 \] 4. **Finding the rate of change of volume**: We know that the volume is increasing at a rate of \( \frac{dV}{dt} = 10 \) cubic feet per minute. We can differentiate the volume with respect to time: \[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{3}{4} \pi h^3\right) = \frac{3}{4} \pi \cdot 3h^2 \frac{dh}{dt} = \frac{9}{4} \pi h^2 \frac{dh}{dt} \] 5. **Setting the rates equal**: \[ 10 = \frac{9}{4} \pi h^2 \frac{dh}{dt} \] 6. **Solving for \( \frac{dh}{dt} \)** when \( h = 15 \): \[ 10 = \frac{9}{4} \pi (15^2) \frac{dh}{dt} \] \[ 10 = \frac{9}{4} \pi (225) \frac{dh}{dt} \] \[ 10 = \frac{2025}{4} \pi \frac{dh}{dt} \] \[ \frac{dh}{dt} = \frac{10 \cdot 4}{2025 \pi} = \frac{40}{2025 \pi} \] 7. **Simplifying**: \[ \frac{dh}{dt} = \frac{8}{405 \pi} \text{ ft/min} \] Thus, the rate at which the height of the pile is changing when the pile is 15 feet high is: \[ \frac{8}{405 \pi} \text{ ft/min} \] The correct answer is: \[ \boxed{\frac{8}{405 \pi} \text{ ft/min}} \]