I checked this question but didn't fully understand it.
I know that the volume of a right circular cone is $V = \frac{1}{3}\pi x^2h$
I know that I must take the first derivative and set it equal to zero, which will find the maximum.
My problem is how to deal with the variable $h$, the height?
How can I rewrite that in terms of $x$ or $r$.
I have tried to implicitly differentiate this equation, but that wasn't helpful for me.
Best Answer
Ok. In the image above, there are two sketches. The one on the left is in 3d, but is kinda hard to refer to. The one on the right is in 2d and easier to see what's going on, so I'm going to use that (just imagine it's a cross-section of the 3d pic).
The secret of the whole problem is to relate $h$ and $b$. This is done using the equation for the right half of the circle: $$x = \sqrt{r^2-y^2}$$
$b$ is the $x$ value when $y$ is offset from the top of the sphere by $h$. Thus, the equation in terms of $b$, $h$, and $r$ is: $$b = \sqrt{r^2 - (r-h)^2}$$
Now to relate to the volume of a cone: $$V_{cone} = \frac\pi 3 \cdot b^2 h$$ $$V_{cone} = \frac\pi 3 \cdot (r^2 - (r - h)^2) h$$ Simplifying: $$V_{cone} = \frac \pi 3 \cdot (2h^2 r-h^3)$$ Differentiate: $$\frac{dV_{cone}}{dh} = \frac \pi 3 \cdot (4r - 3h)h$$ Solve: $$h = \left\{0, \frac{4r}{3}\right\}$$ We want the maximum, which obviously will occur at the second value of $h$. Finding the maximal volume is a simple algebra problem now.