For my algebra class, this is the only question im stuck on this section:

Grain silos can be described as a hemisphere sitting atop a cylinder. The interior volume V of the silo can be modeled by V = 2/3(pi)r3 + (pi)r2h Where h=height of a cylinder with radius r. For a cylinder 6m tall, what radius would give the silo a volume that is numerically equal to 24(pi) times this radius?

That's the problem, word-for-word verbatim. 2/3 is a fraction and pi is the symbol for (pi). I'm lost How do I use the Latex thing so I can type it out exactly as it appears?

Any help would be greatly appreciated.

Originally Posted by Rtedmonston
For my algebra class, this is the only question im stuck on this section:

Grain silos can be described as a hemisphere sitting atop a cylinder. The interior volume V of the silo can be modeled by V = 2/3(pi)r3 + (pi)r2h Where h=height of a cylinder with radius r. For a cylinder 6m tall, what radius would give the silo a volume that is numerically equal to 24(pi) times this radius?

That's the problem, word-for-word verbatim. 2/3 is a fraction and pi is the symbol for (pi). I'm lost How do I use the Latex thing so I can type it out exactly as it appears?

Any help would be greatly appreciated.
You only have to "translate" the text into an equation:

$\frac23 \pi r^3+\pi \cdot r^2 \cdot h = 24 \pi \cdot r$

Replace h by 6:

$\frac23 \pi r^3+6 \pi \cdot r^2 = 24 \pi \cdot r$

Collect all terms on the LHS and factor out $\pi r$ :

$\pi r \left( \frac23 r^2+6 \cdot r - 24 \right)=0$

A product of 2 factors equals zero if at least one factor equals zero. By the 1st factor you'll get r = 0 which is a very, very slim cylinder, nearly invisible

With the 2nd factor you have to solve a quadratic equation in r:

$\frac23 r^2+6 \cdot r - 24 = 0$

You'll get 2 solutions for r. The negative value isn't very plausible here.