Problem source

The diagram shows a coffee filter consisting of an inverted hollow
right circular cone of height $H$ cm and base radius $a$ cm. 

\noindent \begin{center}
\psset{xunit=1.0cm,yunit=0.8cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-1.67,-2.3)(2.85,3.85) \rput{0}(0,3){\psellipse(0,0)(1.23,0.72)} \rput{0.69}(0,0.01){\psellipse(0,0)(0.49,0.23)} \psline(-1.23,2.95)(0,-2) \psline(0,-2)(1.23,2.96) \psline{->}(0,3)(0.66,3.61) \psline{->}(0.66,3.61)(0,3) \rput[tl](0.35,3.27){$a$} \psline{<->}(1,0)(1,-2) 
 \rput[tl](1.05,-0.86){$x$} \psline{<->}(2,3)(2,-2) \rput[tl](2.09,0.97){$H$} \end{pspicture*}
\par\end{center}

When the water level is $x$ cm above the vertex, water leaves the
cone at a rate $Ax$ $\mathrm{cm}^{3}\mathrm{sec}^{-1},$ where $A$
is a positive constant. Suppose that the cone is initially filled
to a height $h$ cm with $0 < h < H.$ Show that it will take $\pi a^{2}h^{2}/(2AH^{2})$
seconds to empty. 

Suppose now that the cone is initially filled to a height $h$ cm,
but that water is poured in at a constant rate $B$ $\mathrm{cm}^{3}\mathrm{sec}^{-1}$
and continues to drain as before. Establish, by considering the sign
of $\mathrm{d}x/\mathrm{d}t$, or otherwise, what will happen subsequently
to the water level in the different cases that arise. (You are not
asked to find an explicit formula for $x$.)