Answer:
Approximately 0.0012 ft/min when the pile is 16 ft high.
Explanation:
To find how fast the circumference of the base is increasing, we need to use related rates. Let's denote the height of the conical pile as 'h' and the radius of the base as 'r'. Given that the bottom radius is always half the altitude, we have r = 0.5h.
We are given that the wheat is poured through the chute at a rate of 22 ft³/min. This implies that the volume of the cone is increasing at a rate of 22 ft³/min.
The volume of a cone can be expressed as V = (1/3)πr²h. Substituting r = 0.5h, we have V = (1/3)π(0.5h)²h = (1/12)πh³.
Now, let's differentiate both sides of the equation with respect to time 't':
dV/dt = (1/12)π * 3h² * dh/dt.
Since dV/dt is given as 22 ft³/min, and we are looking for how fast the circumference of the base is increasing, we need to find dh/dt when h = 16 ft.
Plugging in the given values, we have:
22 = (1/12)π * 3(16)² * dh/dt.
Simplifying this equation, we can solve for dh/dt:
dh/dt = 22 / [(1/12)π * 3(16)²].
Calculating the value, we find:
dh/dt ≈ 0.0012 ft/min.
Therefore, the circumference of the base will be increasing at a rate of approximately 0.0012 ft/min when the pile is 16 ft high.