Question

A candle in the shape of a circular cone has a base of radius r and a height of h that is the same length as the radius. Which expresses the ratio of the volume of the candle to its surface area (including the base)?
For a cone, ( I will attach an image of the formulas)

Answer 1

\[V = \frac{ \pi }{ 3 } * r^2h \space and \space SA = (\pi)r^2 + (\pi)r \sqrt{r^2 + h^2}\]

Answer 2

So they told us that the height of the candle \(h\) is the same length as the radius \(r\).
Let's go ahead and express our formulas without the \(h\)'s, replacing them with \(r\)'s.

\[\large V=\frac{\pi}{3}r^2h \qquad \rightarrow \qquad V=\frac{\pi}{3}r^3\]
\[\large SA=\pi r^2+ \pi r \sqrt{r^2+h^2} \qquad \rightarrow \qquad SA=\pi r^2+\pi r \sqrt{r^2+r^2}\]\[\large SA=\pi r^2+\pi r \sqrt{2r^2} \qquad \rightarrow \qquad SA=\pi r^2+\sqrt2 \pi r^2\]\[\large SA=(1+\sqrt2)\pi r^2\]

Ok so that's how we express them in terms of \(r\).

Now we need to find the ratio \(\dfrac{V}{SA}\).

\[\large \frac{V}{SA}=\frac{\dfrac{\pi}{3}r^3}{(1+\sqrt2)\pi r^2}\]
We get some nice cancellations,
\[\large \frac{V}{SA}=\frac{\dfrac{\cancel{\pi}}{3}r^{\cancel{3}1}}{(1+\sqrt2)\cancel{\pi} \cancel{r^2}}\]

Which gives us,\[\large \frac{V}{SA}=\frac{r}{3(1+\sqrt2)}\]

Answer 3

Hopefully that's what you were looking for :O

Answer 4

Yes that helped a lot, but I'm a bit confused because after looking in the book to check what I got the right answer is apparently:
\[\frac{ r(1 - \sqrt{2}) }{ -3 }\]

Answer 5

Oh so they rationalized it.
It's bad to leave square roots in the bottom of a fraction.
So what we can do is, multiply the top and bottom by the conjugate of the term in the brackets.\[\large \frac{r}{3(1+\sqrt2)}\left(\frac{1-\sqrt2}{1-\sqrt2}\right) \qquad = \qquad \frac{r(1-\sqrt2)}{3(1-2)}\]

Answer 6

Ohh right, I forgot about that. Thanks for your help!

Answer 7

@zepdrix HOW U DO THAT

Answer 8

The correct answer was:
rh/3r + sqrt/h^2 +r^2
For anyone who was doing Algebra 2 B Unit 3: ational Functions the answers are
1. B
2. C
3. D
4. D
5. B