We've Sprung a Leak!

Let $V_b$ and $V_c$ be the volumes of the cylindrical basin and conical reservoir respectively. The important connection to make in solving this problem is that the volumetric increase of water in the basin is equivalent to the volumetric decrease of water in the cone, or $\frac{dV_b}{dt} = -\frac{dV_c}{dt}$ . An additional necessary observation is that the cone's structure is such that the height of any level of water is half that of the radius of the water at that point, or $\frac{h}{2} = r$ .

$V_c = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{1}{12}\pi h^3$

$\frac{d}{dt}\left(V_c\right) = \frac{d}{dt}\left(\frac{1}{12}\pi h^3\right)$

$\frac{dV_c}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt} = \frac{1}{4}\pi (16)^2 (-2) = -128\pi \longrightarrow \frac{dV_b}{dt} = 128\pi$

Now we shift focus to the cylindrical basin, whose volume is given by $V_b = \pi r^2 h = 36\pi h$ (we can substitute $r = 6$ because the radius of the cylinder doesn't change as the water level rises). Differentiating and plugging in $\frac{dV_b}{dt}$ :

$\frac{dV_b}{dt} = 36\pi \frac{dh}{dt} \longrightarrow 128\pi = 36\pi \frac{dh}{dt} \longrightarrow \frac{dh}{dt} = \frac{32}{9} \approx 3.5556 \text{ feet per minute}$

By similar triangles, we have

$\dfrac{x}{y}=\dfrac{10}{20}=\dfrac{1}{2}$

$x=\dfrac{y}{2}$

$V=\dfrac{\pi x^2y}{3}=\dfrac{\pi \left(\dfrac{y}{2}\right)^2y}{3}=\dfrac{\pi y^3}{12}$

$\dfrac{dV}{dt}=\dfrac{\pi}{12}(3)(y^2) \left(\dfrac{dy}{dt}\right)=\dfrac{\pi}{4}(y^2)\left(\dfrac{dy}{dt}\right)$

When $y=16$ , $\dfrac{dy}{dt}=2$ . Substitute:

$\dfrac{dV}{dt}=\dfrac{\pi}{4}(16)^2(2)=128 \pi$

The volume of the basin is

$V= \pi r^2 h=\pi (6)^2 h= 36\pi h$

$\dfrac{dV}{dt} = 36 \pi \dfrac{dh}{dt}$

Substitute:

$128 \pi =36 \pi \dfrac{dh}{dt}$

$\dfrac{dh}{dt}=\dfrac{128}{36} \approx \boxed{3.556}$