Grain is falling from a chute onto the ground, forming a conical pile whose diameter is always three times its height. How high is the pile (to the nearest hundredth of a foot) when it contains 1060 ft3 of grain?

Answer 1

Let $r$ be the radius of the base of the pile (in feet), and let $h$ be the height of the pile (in feet). Then the radius of the top of the pile is $3h$ (in feet). [asy]
size(180);

real r = 0.16657;

draw(Circle((0,0),r));
draw((0,0)--(0.85*r,0));
draw((0,0)--(0.85*r,0.85*r*3));
draw((0,0)--(-0.85*r,0.85*r*3));
draw((-0.85*r,0)--(-0.85*r,0.85*r*3),dashed);

label("$r$", (0.43*r,-0.15*r), UnFill);
label("$3h$", (-0.99*r,0.85*r), UnFill);

dot((0,0));
[/asy]

The area of the base of the pile is $\pi r^2$ (in square feet), and the area of the top of the pile is $\pi (3h)^2 = 9\pi h^2$ (in square feet). Because the grain forms a cone, the volume of the pile is \[\frac{1}{3}\cdot\pi r^2 h = \frac{1}{3}\cdot 9\pi h^2 h = 3\pi h^3.\]We are given that $3\pi h^3 = 1060$, so $h^3 = \frac{1060}{3\pi}$. Since $\pi \approx 3.14,$ $h^3 \approx \frac{1060}{3\cdot 3.14} = \frac{1060}{9.42} \approx 112.67.$

Therefore, $h\approx\boxed{4.93}$ (to the nearest hundredth of a foot).