MATH SOLVE

6 months ago

Q:
# A funnel is made up of a partial cone and a cylinder as shown in the figure. The maximum amount of liquid that can be in the funnel at any given time is 16.59375π cubic centimeters. Given this information, what is the volume of the partial cone that makes up the top part of the funnel?

Accepted Solution

A:

the funnel which is composed of the partial cone and the cylindrical bottom has a combined volume of 16.59375π.

ok, now, if we just get the volume the the cylindrical bottom holds, and subtract it from 16.59375π, what's leftover is the volume the partial cone holds, so let's do so, keeping in mind the cylindrical part has a diameter of 1.5, and therefore a radius of half that, or 0.75, and a height of 1.5.

[tex]\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h\quad \begin{cases} r=radius\\ h=height\\ -----\\ r=0.75\\ h=1.5 \end{cases}\implies V=\pi 0.75^2\cdot 1.5 \\\\\\ V=0.84375\pi \\\\ -------------------------------\\\\ \stackrel{\textit{volume of the funnel in total}}{16.59375\pi }~-~\stackrel{\textit{volume of the cylindrical bottom}}{0.84375\pi }=\stackrel{\stackrel{partial~cone}{volume}}{15.75\pi }[/tex]

ok, now, if we just get the volume the the cylindrical bottom holds, and subtract it from 16.59375π, what's leftover is the volume the partial cone holds, so let's do so, keeping in mind the cylindrical part has a diameter of 1.5, and therefore a radius of half that, or 0.75, and a height of 1.5.

[tex]\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h\quad \begin{cases} r=radius\\ h=height\\ -----\\ r=0.75\\ h=1.5 \end{cases}\implies V=\pi 0.75^2\cdot 1.5 \\\\\\ V=0.84375\pi \\\\ -------------------------------\\\\ \stackrel{\textit{volume of the funnel in total}}{16.59375\pi }~-~\stackrel{\textit{volume of the cylindrical bottom}}{0.84375\pi }=\stackrel{\stackrel{partial~cone}{volume}}{15.75\pi }[/tex]