1. **State the problem:** Gravel is dumped forming a cone-shaped pile with volume increasing at 30 cubic feet per minute. The base diameter equals the height at all times. We need to find how fast the height is increasing when the height is 23 feet. 2. **Given:** - Rate of volume change: $\frac{dV}{dt} = 30$ ft³/min - Height $h$ and diameter $d$ satisfy $d = h$ so radius $r = \frac{d}{2} = \frac{h}{2}$ - Volume formula for cone: $$V = \frac{1}{3} \pi r^2 h$$ 3. **Express volume in terms of height only:** $$V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \pi \frac{h^2}{4} h = \frac{\pi}{12} h^3$$ 4. **Differentiate volume with respect to time $t$:** $$\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$$ 5. **Plug in known values and solve for $\frac{dh}{dt}$:** $$30 = \frac{\pi}{4} (23)^2 \frac{dh}{dt}$$ 6. **Isolate $\frac{dh}{dt}$:** $$\frac{dh}{dt} = \frac{30}{\frac{\pi}{4} \times 23^2} = \frac{30}{\frac{\pi}{4} \times 529} = \frac{30}{\frac{529\pi}{4}} = \frac{30 \times 4}{529 \pi} = \frac{120}{529 \pi}$$ 7. **Final answer:** $$\frac{dh}{dt} = \frac{120}{529 \pi} \approx 0.0723 \text{ feet per minute}$$ The height of the pile is increasing at approximately 0.0723 feet per minute when the height is 23 feet.